Algebra

Table of Contents

Notation

  • $G$ denotes a group
  • CANI stands for:
    • Commutative
    • Associative
    • Neutral element (e.g. 0 for addition)
    • Inverse

Terminology

almost all
is an abbreviation meaning "all but finitely many"

Definitions

Homomorphisms

A homomorphism $\varphi: G \to H$ is a structure-preserving map between the groups $G$ and $H$, i.e.

\begin{equation*} \varphi(ab) = \varphi(a) \varphi(b), \quad \forall a, b \in G \end{equation*}

A endomorphism is a homomorphism from the group $G$ to itself, i.e. $\varphi: G \to G$

An automorphism is an invertible homomorphism, denoted $\text{Aut}(V)$.

In the case of a vector space $V$, we have

\begin{equation*} \text{GL}(V) = \text{Aut}(V) \end{equation*}

Linear isomorphism

Two vector spaces $V_1$ and $V_2$ are said to be isomorphic if and only if there exists a linear bijection $\phi: V_1 \to V_2$.

Field

An (algebraic) field $(K, +, \cdot)$ is a set $K$ and the maps

  • $+: K \times K \to K$
  • $\cdot: K \times K \to K$

that satisfy

\begin{equation*} \underbrace{C^+ \ A^+ \ N^+ \ I^+}_{\text{over } K} \qquad \underbrace{C^\cdot \ A^\cdot \ N^\cdot \ I^\cdot}_{\text{over } K \backslash \{ 0 \}} \end{equation*}

Vector space

A vector space $V$ over a field $F$ is a pair consisting of an abelian group $V = (V, \dot{+})$ and a mapping

\begin{equation*} \dot{+}: F \times V \to V \qquad (\lambda, \mathbf{v}) \mapsto \lambda \mathbf{v} \end{equation*}

such that for all $\lambda, \mu \in F$ and $\mathbf{v}, \mathbf{w} \in V$ we have A D D U:

  • Associativity: $\lambda (\mu \mathbf{v}) = (\lambda \mu) \mathbf{v}$
  • Distributivity over field-addition: $(\lambda + \mu)(\mathbf{v} = (\lambda \mathbf{v}) \dot{+} (\mu \mathbf{v})$
  • Distributivity over field-multiplication: $\lambda (\mathbf{v} \dot{+} \mathbf{w}) = (\lambda \mathbf{v}) \dot{+} (\lambda \mathbf{w})$
  • Uint: $1_F \mathbf{v} = \mathbf{v}$

Ring

A ring $(R, +, \cdot)$ over a set $R$ with the maps:

  • $+: K \times K \to K$
  • $\cdot: K \times K \to K$

which satisfy the same as a field, but without C and I for $\cdot$, i.e. multiplication.

A division ring is a ring where the multiplicative operation $\cdot$ allows an inverse, i.e. it's

\begin{equation*} \underbrace{C^+ \ A^+ \ N^+ \ I^+}_{\text{over } K} \qquad \underbrace{\cancel{C^\cdot} \ \cancel{A^\cdot} \ N^\cdot \ I^\cdot}_{\text{over } K \backslash \{ 0 \}} \end{equation*}

Equivalence relations

A equivalence relation on some set is defined as some relation betweeen $a$ and $b$, denoted $a \sim b$, such that the relation is:

  • reflexive: $a \sim a$
  • symmetric: $a \sim b \iff b \sim a$
  • transistive $a \sim b, \ b \sim c \implies a \sim c$

Why do we care about these?

  • Partitions the set it's defined on into unique and disjoint subsets

Unitary transformations

A unitary transformation is a transformation which preserves the inner product.

More precisely, a unitary transformation is an isomorphism between two Hilbert spaces.

Groups

Notation

  • $[G : H]$ or $\left| G : H \right|$ denotes the number of left cosets of a subgroupd $H$ of $G$, and is called the index

Definitions

The symmetric group $S_n$ of a finite set of $n$ symbols is the group whose elements are all the permutation operations that can be performed on the $n$ distinct symbols, and whose group operation is the composition of such permutation operations, which are defined as bijective functions from the set of symbols to itself.

An action of a group is a formal way of interpreting the manner in which the elements of the group correspond to transformations of some space in a way that preserves the structure of the space.

If $G$ is a group and $X$ is a set, then a (left) group action $\varphi$ of $G$ on $X$ is a function

\begin{equation*} \begin{split} \varphi: G \times X & \to X \\ (g, x) & \mapsto \varphi(g, x) \end{split} \end{equation*}

that satisfies the following two axioms (where we denote $\varphi(g, x)$ as $g \cdot x$):

  • identity: $e \cdot x = x$ for all $x \in X$ ($e$ denotes the identity element of $G$)
  • compatibility: $(g \circ h) \cdot x = g \cdot (h \cdot x)$ for all $g, h \in G$ and all $x \in X$

where $g \circ h$ denotes the result of first applying $h$ to $x$ and then applying $g$ to the result.

From these two axioms, it follows that for every $g \in G$, the function $\varphi$ which maps $x \in X$ to $g \cdot x \in X$ is a bijective map from $X$ to $X$. Therefore, one may alternatively define a group action of $G$ on $X$ as a group homomorphism from $G$ into the symmetric group $S(X)$ of all bijections from $X$ to $X$.

The action of $G$ on $X$ is called transistive if $X$ is non-empty and if:

\begin{equation*} \forall x, y \in X, \ \exists g \in G \quad \text{such that} \quad g \cdot x = y \end{equation*}

Faithful (or effective ) if

\begin{equation*} \forall g, h \in G, g \ne h, \ \exists x \in X \quad \text{such that} \quad g \cdot x \ne h \cdot x \end{equation*}

That is, in a faithful group action, different elements of $G$ induce different permutations of $X$.

In algebraic terms, a group $G$ acts faithfully on $X$ if and only if the corresponding homomorphism to the symmetric group, $G \to \text{Sym}(X)$, has a trivial kernel.

If $G$ does not act faithfully on $X$, one can easily modify the group to obtain a faithful action. If we define:

\begin{equation*} N = \{ g \in G \mid g \cdot x = x, \forall x \in X \} \end{equation*}

then $N$ is the normal subgroup of $G$; indeed, it is the kernel of the homomorphism $G \to \text{Sym}(X)$. The factor group $G / N$ acts faithfully on $X$ by setting $(gN) \cdot x = g \cdot x$.

We say a group action is free (or semiregular or fixed point free ) if, given $g, h \in G$,

\begin{equation*} g \cdot x = h \cdot x \quad \implies \quad g = h \end{equation*}

We say a group action is regular if and only if it's both transitive and free; that is equivalent to saying that for every two $x, y \in X$ there exists precisely one $g \in G$ s.t. $g \cdot x = y$.

Consider a group $G$ acting on a set $X$. The orbit of an element $x$ in $X$ is the set of elements in $X$ to which $X$ can be moved by the elements of $G$. The orbit of $x$ is denoted by $G \cdot x$:

\begin{equation*} G \cdot x = \{ g \cdot x \mid g \in G \} \end{equation*}

An abelian group, or commutative group, is simply a group where the group operation is commutative!

A monoid is an algebraic structure with a single associative binary operation and an identity element, i.e. it's a semi-group with a binary operation.

Cosets

Let $G$ be a group and $H$ be a subgroup of $G$. Let $g \in G$. The set

\begin{equation*} g H = \{ gh : h \in H \} \end{equation*}

of products of $g$ with elements of $H$, with $g$ on the left is called a left coset of $H$ in $G$.

The number of left cosets of a subgroup $H$ of $G$ is the index of $H$ in $G$ and is denoted by $|G : H|$ or $[G : H]$. That is,

\begin{equation*} [G : H] = \left| G / H \right| \end{equation*}

Center

Given a group $G$, the center of $G$, denoted $Z(G)$ is defined as the set of elements which commute with every element of the group, i.e.

\begin{equation*} Z(G) := \left\{ g \in G: gh = hg, \quad \forall h \in G \right\} \end{equation*}

We say that a subgroup $H$ of $G$ is central if it lies inside $Z(G)$, i.e. $H \le Z(G)$.

Abelianisation

Given a group $G$, define a abelianisation of $G$ to be the quotient group

\begin{equation*} G^{ab} = G / \comm{G}{G} \end{equation*}

with $\comm{G}{G} \triangleleft G$ is the normal subgroup generated by the commutators $\comm{a}{b}$ for $a, b \in G$, i.e.

\begin{equation*} \comm{G}{G} := \left\langle a b a^{-1} b^{-1} \mid a, b \in G \right\rangle \end{equation*}

Theorems

Let $G$ be a group of order $n$ and let $p$ be a prime divison of $n$.

Then $G$ has an element of order $p$.

Fermat's Little Theorem

Let $p$ be a prime number.

Then for $a \in \mathbb{Z}$, the number $a^p - a$ is an integer multiple of $p$. That is

\begin{equation*} a^p \equiv a \mod p \end{equation*}

Isomorphism theorems

Let

Then $N := \text{ker}(\varphi)$ is a normal subgroup of $G$, and $\text{im}(\varphi) \le H$. Furthermore, there is an isomorphim

\begin{equation*} \bar{\varphi}: G / \text{ker}(\varphi) \overset{\cong}{\longrightarrow} \text{im}(\varphi) \end{equation*}
\begin{equation*} \begin{split} \overline{\varphi}: \quad & G / \text{ker}(\varphi) \to \text{im}(\varphi) \\ & gN \mapsto \overline{\varphi}(g N) = \varphi(g) \end{split} \end{equation*}

In particular, if $\varphi$ is surjective, then

\begin{equation*} G / \text{ker}(\varphi) \cong H \end{equation*}

Le $G$ be a group, $N \triangleleft G$ and $H \le G$.

  1. $HN = \left\{ hn : h \in H, n \in N \right\} \le G$
  2. $H \cap N \triangleleft H$
  3. $N \triangleleft HN$
  4. $H / H \cap N \cong HN / N$

  1. First $ee \in HN$ so clearly $H N \ne \emptyset$. Let $h_i, h_2 \in H$ and $n_1, n_2 \in N$. We then want to show that $(h_1, n_1) (h_2 n_2) \in HN$. Observe that

    \begin{equation*} N h_2 = h_2 N \end{equation*}

    since $N \triangleleftG$. So $\exists n' \in N$ so that $n_1 h_2 = h_2 n'$,

    \begin{equation*} h_1 n_1 h_2 n_2 = \underbrace{h_1 h_2}_{H} \underbrace{n' n_2}_{\in N} \in HN \end{equation*}

    And now we check that the $\big( h_1 n_1 \big)^{-1} \in HN$

    \begin{equation*} \big( h_1 n_1 \big)^{-1} = n_1^{-1} h_1^{-1} \in H h_1^{-1} = h_1^{-1} N \subseteq HN$ \end{equation*}

  2. Let $a \in H \cap N$, $h \in H$. Wan to show that $h a h^{-1} \in H \cap N$.

    \begin{equation*} \begin{split} a &\in H \implies h a h^{-1} \in H \\ a &\in N \triangleleft G \implies h a h^{-1} \in N \end{split} \end{equation*}
  3. Let $n \in N$, $g \in HN$. Then $g n g^{-1} \in N$ since $N \triangleleft G$, so $g n g^{-1} \in N$, for all $g \in G$.

  4. Want to show that $H / H \cap N \cong H N / N$. First Isomorphism Theorem tells us that

    \begin{equation*} \text{im}(\varphi) \cong A / \text{ker}(\varphi) \end{equation*}

    Therefore, letting

    \begin{equation*} \varphi : H \subseteq HN \overset{\text{can}_N}{\rightarrow} HN / N \end{equation*}

    where we simply factor out the $n$ from every element in $HN$.

    \begin{equation*} \varphi(h) = h N \in \text{can}_N(H) \le \text{can}_N(HN) = HN / N \le G / N \end{equation*}

    So $\varphi$ maps into $HN / N$, but we need it to be surjective, i.e.

    \begin{equation*} \text{im}(\varphi) = HN / N \end{equation*}

    An element of $HN / N$ is a coset $hn N$ for $h \in H, n \in N$, which is clearly $h N = \varphi(h)$. And finally,

    \begin{equation*} \begin{split} \text{ker}(\varphi) &= \left\{ h \in H : h N = N \right\} \\ &= \left\{ h \in H : h \in \text{ker}(\text{can}_n) = N \right\} \\ &= H \cap N \end{split} \end{equation*}

    Hence, by the First Isomorphism Theorem,

    \begin{equation*} HN / N = \text{im}(\varphi) \cong H / \text{ker}(\varphi) = H / H \cap N \end{equation*}

Notice $H N / N = \text{can}_N(H)$. In fact, $H N = \text{can}_N^{-1} \big( \text{can}_N(H) \big)$.

As an example, $\theta: \mathbb{Z} \to \mathbb{Z} / 12 \mathbb{Z}$. Then

\begin{equation*} \theta(15 \mathbb{Z}) = 15 \mathbb{Z} / \text{ker}(\theta |_{15 \mathbb{Z}}) = 15 \mathbb{Z} / \big( 15 \mathbb{Z} \cap 12 \mathbb{Z} \big) = 15 \mathbb{Z} / 60 \mathbb{Z} \end{equation*}

by the First Isomorphism Theorem and since $60 = \text{lcm} \big( 12, 15 \big)$. And, we can also write

\begin{equation*} \theta (15 \mathbb{Z}) = \frac{\theta^{-1} \big( \theta(15 \mathbb{Z} \big)}{12 \mathbb{Z}} = \frac{12 \mathbb{Z} + 15 \mathbb{Z}}{12 \mathbb{Z}} = 3 \mathbb{Z} / 12 \mathbb{Z} \end{equation*}

since $3 = \text{gcd}(12, 15) = \text{hcf}(12, 15)$.

Want to show that both are isom. to $\mathbb{Z} / 4 \mathbb{Z}$. We do this by constructing map:

\begin{equation*} \begin{split} \pi: \quad & \mathbb{Z} \to 3 \mathbb{Z} / 12 \mathbb{Z} \\ & n \mapsto 3n + 12 \mathbb{Z} \end{split} \end{equation*}

and just take the coset. And,

\begin{equation*} \text{ker}(\pi) = \left\{ n : 12 \mid 3n \right\} = 4 \mathbb{Z} \end{equation*}

Hence by 1st Isom. Thm.

\begin{equation*} \mathbb{Z} / 4 \mathbb{Z} \cong 3 \mathbb{Z} / 12 \mathbb{Z} \end{equation*}

Kernels and Normal subgroups

Arising from an action $\bullet$ of a group $G$ on a set $X$, is the homomorphism

\begin{equation*} \begin{split} f: G & \to S_X \\ f: g & \mapsto f_g \end{split} \end{equation*}

the permutation of $X$ corresponding to $g$.

The kernel of this homomorphism is also called the kernel of the action $\bullet$ and is denoted $\text{Ker}(\bullet)$.

Remember,

\begin{equation*} f(g) = f_g \end{equation*}

is the permutation $x \mapsto g \bullet x$. Therefore, $f(g) = \text{id}_X$ if and only if $g \bullet x = x$ for all $x \in X$, and so

\begin{equation*} \text{Ker}(\bullet) = \{ g \in G : g \bullet x = x, \forall x \in X \} \end{equation*}

consists of those elements which stabilize every element of $X$.

Let $H$ be a subgroup of a group $G$, and let $a \in G$.

The conjugate of $H$ by $a$, written $aHa^{-1}$, is the set

\begin{equation*} aHa^{-1} = \{ aha^{-1} : h \in H \} \end{equation*}

of all conjugate of elements of $H$ by $a$.

This is the image $f(H)$ of $H$ under the conjugation homomorphism $f: G \to G$ where $f: g \mapsto aga^{-1}$.

Hence, $aHa^{-1}$ is a subgroup of $G$.

Let $H$ be a subgroup of $G$.

If $aHa^{-1} \subseteq H$ for all $a \in G$, then

\begin{equation*} aHa^{-1} = H, \quad \forall a \in G \end{equation*}

A subgroup $N$ of a group, $G$, is called a normal subgroup if it is invariant under conjugation; that is, the conjugation of an element of $N$ by an element of $G$ is still in $N$:

\begin{equation*} N \lhd G \quad \iff \quad \forall h \in N, \forall g \in G: ghg^{-1} \in N \end{equation*}

$A \le G$. Then $A \triangleleft G$ if and only if $g A g^{-1} \le A$ for all $g \in G$.

Let $G$ and $H$ be groups.

The kernel of any homomorphism $f: G \to H$ is normal in $G$.

Hence, the kernel of any group action of $G$ is normal in $G$.

Let $G$ be a group acting on the set $X = \{ aH : a \in G \}$ of left cosets of a subgroup $H$ of $G$.

  1. The stabilizer of each left coset $aH$ is the conjugate $a H a^{-1}$
  2. If $H = a H a^{-1}$ for all $a \in G$ then $H = \text{Ker}(\bullet)$

Factor / Quotient groups

Let $G$ be a group acting on a set $X$ and $K$ be the kernel of the action.

The set of cosets of $K$ in $G$ is a group with binary operation

\begin{equation*} (aK) (bK) = (ab) K \end{equation*}

which defines the factor group or quotient group of $G$ by $K$ and is denoted

\begin{equation*} G \ / \ K \end{equation*}

But I find the following way of defining a quotient group more "understandable":

Let $\varphi: G \to G / H$ be a group homomorphism such that

\begin{equation*} \begin{split} \forall g_1, g_2 \in G, & \quad g_1 \ne g_2, \\ \varphi(g_1) = \varphi(g_2) & \iff g_1 \sim g_2 \end{split} \end{equation*}

That is, $\varphi$ maps all distinct $g_1, g_2 \in G$ which are equivalent under $\sim$ to the same element in $G / H$, but still preserves the group structure by being a homomorphism of groups.

There is a function $f: G \to G / N$ with

\begin{equation*} f(a) = aN, \quad \forall a \in G \end{equation*}

For $a, b \in G$ we have that

\begin{equation*} f(a) f(b) = (aN) (bN) = (ab) N = f(ab) \end{equation*}

thus $f$ is a homomorphism.

Then, clearly

\begin{equation*} \text{Ker}(f) = N \end{equation*}

$f$ is called the natural homomorphism from $G$ to $G / N$.

The order of a factor group $G / N$ is the number of distinct cosets of $N$ in $G$.

If $G$ is finite, then

\begin{equation*} \bigg| \frac{G}{N} \bigg| = \frac{|G|}{|N|} \end{equation*}

Group presentations

Notation

  • $\left\langle x \mid x^n = e \right\rangle$ or $\left\langle x \mid x^n \right\rangle$ refers to the group generated by $x$ such that $x^n = e$
  • $\left\langle x \right\rangle$ denote is the free group as generated by $x$

  • In general: $\left\langle \text{group element} \mid \text{defining the unit} \right\rangle$, e.g.

    \begin{equation*} \left\langle x, y \mid xy x^{-1} y^{-1} \right\rangle \cong \mathbb{Z} \times \mathbb{Z} \end{equation*}

    where the "unit-condition" simply specifies that the group is commutative

Free groups

$\left\langle x_1, \dots, x_m \right\rangle$ is the free group generated by $x_1, \dots, x_m$.

Elements are symbols in $x_1, \dots, x_m, x_1^{-1}, \dots, x_m^{-1}$, subject to

  • group axioms
  • " and all logical consequences :) "

Let $r_1, \dots, r_n \in \left\langle x_1, \dots, x_m \right\rangle$.

The group with presentation

\begin{equation*} \langle \ \overbrace{x_1, \dots, x_m}^{\text{generators}} \mid \underbrace{r_1, \dots, r_n}_{\text{relations}} \ \rangle \end{equation*}

is the group generated by $x_1, \dots, x_m$ subject to

  • group axioms
  • $r_1 = r_2 = \dots = r_n = e$
  • " and all logical consequences :) "

There's no algorithm for deciding whether $\left\langle x_1, \dots, x_m \mid r_1, \dots, r_n \right\rangle$ is the trivial group.

Let $F = \left\langle A_1, \dots, A_n \right\rangle$ and let $a_1, \dots, a_n \in G$, where $G$ is a group.

Then there is a unique homomorphism:

\begin{equation*} \begin{split} \pi: \quad & F \to G \\ & A_i \mapsto a_i \end{split} \end{equation*}

And the image of $\pi$ is the subgroup of $G$ generated by $\{ a_1, \dots, a_n \}$.

Central Extensions of Groups

Let

  • $A$ be an abelian group
  • $G$ be an arbitrary group

An extension of $G$ by the group $A$ is given an exact sequence of group homomorphisms

\begin{equation*} 1 \longrightarrow A \overset{\iota}{\longrightarrow} E \overset{\pi}{\longrightarrow} G \longrightarrow 1 \end{equation*}

The extension is called central if $A$ is abelian and its image $\im(\iota) \subseteq Z(E)$, where $Z(E)$ denotes the center of $E$, i.e.

\begin{equation*} a \in A, b \in E \quad \implies \quad \iota(a) b = b \iota(a) \end{equation*}

For a group $G$ acting on another group $H$ by a homomorphism $\tau: G \to \Aut(H)$, the semi-direct product group $G \rtimes H$ is the set $H \times G$ with the multiplication given by the formula

\begin{equation*} (x, g) \cdot (x', g') := \big( x \tau(g)(x'), g g' \big) \quad \text{for} \quad (x, g), (x', g') \in H \times G \end{equation*}

This is a special case of a group extension with $\pi(g, a) = x$ and $\iota(x) = (a, x)$:

\begin{equation*} 1 \longrightarrow H \overset{\iota}{\longrightarrow} G \rtimes H \overset{\pi}{\longrightarrow} G \longrightarrow 1 \end{equation*}

Exact sequence

An exact sequence of groups is given by

\begin{equation*} G_0 \overset{f_0}{\longrightarrow} G_1 \overset{f_1}{\longrightarrow} \cdots \overset{f_n}{\longrightarrow} G_n \end{equation*}

of groups and group homomorphisms, where exact refers to the fact that

\begin{equation*} \ker(f_{n + 1}) = \im(f_n) \end{equation*}

Linear Algebra

Notation

  • $\text{Mat}(n \times m; F)$ denotes the set of all matrices on the field $F$

  • $M_\mathcal{B}^\mathcal{A} = {}_\mathcal{B}[f]_\mathcal{A}$ denotes the representing matrix of the mapping $f: V \to W$ wrt. bases $\mathcal{A}$ and $\mathcal{B}$, where $\mathcal{A}$ is ordered basis for $V$ and $\mathcal{B}$ ordered basis for $W$:

    \begin{equation*} {}_\mathcal{B}[f]_\mathcal{A} = {}_\mathcal{B}[f]_{\mathcal{B}} \circ {}_{\mathcal{B}}\text{id}_{\mathcal{A}} \circ {}_\mathcal{A}[f]_\mathcal{A} \end{equation*}
  • ${}_\mathcal{B}[f] = {}_{\mathcal{B}}\text{id} [f]$ and $[f]_{\mathcal{B}} = [f] \text{id}_{\mathcal{B}}$ where ${}_{\mathcal{B}}\text{id}_{\mathcal{A}}$ denotes the "identity-mapping" from elements represented in the basis $\mathcal{A}$ to the representation in $\mathcal{B}$.
  • $\mathcal{S}(n)$ denotes the n-dimensional standard basis
  • $_R \langle T \rangle = \{ r_1 t_1 + \dots r_m t_m \mid t_1, \dots, t_m \in T, r_1, \dots, r_m \in R \}$
  • $F^\times = \{ x \in F \mid x \ne 0 \}$, i.e. the set of non-zero elements of $F$

Vector Spaces

Notation

  • $X$ is a set
  • $F$ is a field
  • $\text{Maps}(X, F) := \{ f: X \to F \}$
  • $F \langle X \rangle = \{ f: X \to F \mid f(x) = 0_F, \forall x \in X \}$

Basis

A subset of a vector space is called a generating set of the vector space if its span is all of the vector space.

A vector space that has a finite generating set is said to be finitely generated.

$L \subseteq V$ is called linearly independent if for all pairwise different vectors $\mathbf{v}_1, \dots, \mathbf{v}_r \in L$ and arbitrary scalars $\alpha_1, \dots, \alpha_r \in F$,

\begin{equation*} \alpha_1 \mathbf{v}_1 + \dots + \alpha_n \mathbf{v}_n = 0 \implies \alpha_i = 0, \forall i \end{equation*}

A basis of a vector space $V$ is a linearly independent generating set in $V$.

The following are equivalent for a subset of $E \subseteq V$ for a bector space $V$:

  1. $E$ is a basis, i.e. linearly independent generating set
  2. $E$ is minimal among all generating sets, i.e. $E \setminus \left\{ \mathbf{v} \right\}$ does not generate $V$ for any $\mathbf{v} \in E$
  3. $E$ is maximal among all linearly independent subsets, i.e. $E \cup \left\{ \mathbf{v} \right\}$ is not lineraly independent for any $\mathbf{v} \in V$.

"Minimal" and "maximal" refers to the inclusion and exclusion.

Let $V$ be a vector space containing vector subspaces $U, W \subseteq V$. Then

\begin{equation*} \dim \big( U \cup W \big) + \dim \big( U \cap W \big) = \dim U + \dim W \end{equation*}

Linear mappings

Let $V, W$ be vector spaces over a field $F$. A mapping $f: V \to W$ is called linear or more precisely $F$ linear if

\begin{equation*} \begin{split} f(\mathbf{v}_1 + \mathbf{v}_2) &= f(\mathbf{v}_1) + f(\mathbf{v}_2) \\ f(\lambda \mathbf{v}_1) &= \lambda f(\mathbf{v}) \end{split} \end{equation*}

This is also a homomorphism of vector spaces.

Let $f : V \to W$ be a linear mapping between vector spaces. Then,

\begin{equation*} \dim V = \dim \Big( \text{Ker}(f) \Big) + \dim \Big( \text{Im}(f) \Big) \end{equation*}

where usually we use the terminology:

  • rank of $f$ is $\dim \big( \text{Im}(f) \big)$
  • nullity of $f$ is $\dim \big( \text{Ker}(f) \big)$

Linear Mappings and Matrices

Let $F$ be a field and let $m, n \in \mathbb{N}$.

There exists a bijection $\phi: F^m \to F^n$ and set of matrices with $n$ rows and $m$ columns:

\begin{equation*} \begin{split} M: \text{Hom}_F(F^M, F^n) & \overset{\sim}{\to} \text{Mat}(n \times m; F) \\ f &\mapsto [f] \end{split} \end{equation*}

Which attaches to each linear mapping $f$, its representing matrix $M(f) := [f]$, defined

\begin{equation*} [f] := \begin{pmatrix} f(\mathbf{e}_1) & f(\mathbf{e}_2) & \dots & f(\mathbf{e}_m) \end{pmatrix} \end{equation*}

i.e. the matrix-representation of $f$ is a defined by how $f$ maps the basis of the target space.

Observe that the matrix product between two matrices $A = [f]$ and $B = [g]$,

\begin{equation*} f \circ g = \big( AB \big)_{ik} = \sum_{j=1}^{m} A_{ij} B_{jk} \end{equation*}

An elementary matrix is any square matrix which differs from the identity matrix in at most one entry.

Any matrix whose only non-zero entries lie on the diagonal, and which has the first 1's along the diagonal and then 0's elsewhere, is said to be in Smith Normal Form

\begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{equation*}

$\forall A \in \text{Mat}(n \times m; F)$ there exists invertible matrices $P$ and $Q$ s.t. $PAQ$ is a matrix in Smith Normal Form.

A linear mapping $\phi: V \to W$ is injective if and only if

\begin{equation*} \text{ker}(\phi) = \left\{ 0 \right\} \end{equation*}

Let $\mathbf{v}, \mathbf{w} \in V$, then

\begin{equation*} \phi(\mathbf{v}) = \phi(\mathbf{w}) \iff \phi(\mathbf{v} - \mathbf{w}) = \mathbf{0} \iff \mathbf{v} - \mathbf{w} \in \text{ker}(\phi) \end{equation*}

Hence, if

\begin{equation*} \phi \text{ injective} \iff \text{ker}(\phi) = \left\{ \mathbf{0} \right\} \end{equation*}

as claimed.

Let $A, B \in \text{Mat}(n; R)$ be square matrices over some commutative ring $R$ are conjugate if

\begin{equation*} B = P^{-1} A P \end{equation*}

for an invertible P ∈ (n; R).

Further, conjugacy is an equivalence relation on $\text{Mat}(n; R)$.

Trace of linear map

The trace of a matrix $A$ is defined

\begin{equation*} \tr(A) = \sum_{i=1}^{n} \tensor{A}{^{i}_{i}} \end{equation*}

The trace of a finite product of matrices $A_1, \dots, A_k$ is independent of the order of the product (given that the products are valid). In other words, trace is invariant under cyclic permutations.

To see that trace is a invariant under cyclic permutations, we observe

\begin{equation*} \begin{split} \tr(AB) &= \tr \big( \tensor{(AB)}{^{i}_{j}} \big) \\ &= \tr \big( \tensor{A}{^{i}_{a}} \tensor{B}{^{a}_{j}} \big) \\ &= \tensor{A}{^{i}_{a}} \tensor{B}{^{a}_{i}} \\ &= \tensor{B}{^{a}_{i}} \tensor{A}{^{i}_{a}} \\ &= \tr \big( BA \big) \end{split} \end{equation*}

This case of two matrices can easily be generalized to case of products of multiple matrices.

Rings and modules

A ring is a set with two operatiors $(R, + , - )$ that satisfy:

  1. $(R, + )$ is an abelian group
  2. $(R, \cdot)$ is a monoid

  3. The distributive laws hold, meaning that $\forall a,b,c \in R$:

    \begin{equation*} \begin{split} a \cdot (b + c) &= (a \cdot b) + (a \cdot c) \\ (a + b) \cdot c &= (a \cdot c) + (b \cdot c) \end{split} \end{equation*}

Important: in some places, e.g. earlier in your notes, they use a slightly less restrictive definition of a ring, and in that case we'd call this definition a unitary ring.

Polynomials

A field $F$ is algebraically closed if each non-constant polynomial $P \in F[X] \backslash F$ with coefficients in $F$ has a root in $F$.

E.g. $\mathbb{C}[X]$ is algebraically closed, while $\mathbb{R}[X]$ is not.

If a field $F$ is algebraically closed, then every non-zero polynomial $P \in F[X] \backslash \left\{ 0 \right\}$ decomposes into linear factors

\begin{equation*} P = c \big( X - \lambda_1 \big) \cdots \big( X - \lambda_n \big) \end{equation*}

with $n \ge 0$, $c \in F^{\times}$ and $\lambda_1, \dots, \lambda_n \in F$.

This decomposition is unique up to reordering of the factors.

Let $R$ be an integral domain and let $P, Q \in R[x]$ with $Q$ monic.

Then there exists unique $A, B \in R[X]$ such that

\begin{equation*} P = A Q + B \end{equation*}

and $\deg(B) < \deg(Q)$ or $B = 0$.

Ideals and Subrings

Let $R$ and $S$ be rings. A linear map $f: R \to S$ is a ring homomorphism if the following hold for all $x, y \in R$:

\begin{equation*} \begin{split} f(x + y) &= f(x) + f(y) \\ f(xy) &= f(x) f(y) \end{split} \end{equation*}

A subset $I$ of a ring $R$ is an ideal, written $I \trianglelefteq R$, if the following hold:

  1. $I \ne \emptyset$
  2. $I$ is closed under subtraction
  3. $ri, ir \in I$ for all $i \in I$ and $r \in R$, i.e. $I$ closed under multiplication by elements of $R$
    • I.e. we stay in $I$ even when multiplied by elements from outside of $I$

Ideals are sort of like normal subgroups for rings!

Let $R$ be a commutative ring and let $T \subset R$.

Then the ideal of $R$ generated by $T$ is the set

\begin{equation*} _R \langle T \rangle = \{ r_1 t_1 + \dots r_m t_m \mid t_1, \dots, t_m \in T, r_1, \dots, r_m \in R \} \end{equation*}

together with the zero element in the case $T = \emptyset$.

If $T = \{ t_1, \dots, t_n \}$, a finite set, we will often write

\begin{equation*} _R \langle t_1, \dots, t_n \rangle \end{equation*}

Let $R$ be a commutative ring. An ideal $I$ of $R$ is called a principal ideal if

\begin{equation*} I = \langle t \rangle \end{equation*}

for some $t \in R$.

Let $R'$ be a subset of a ring $R$. Then $R'$ is a subring if and only if

  1. $R'$ has a multiplicative identity
  2. $R'$ is closed under subtraction: $a, b \in R' \implies a - b \in R'$
  3. $R'$ is closed under multiplication

It's important to note that $R'$ and $R$ does not necessarily have the same identity element, even though $R'$ is a subring of $R$!

Let $R$ be a ring. An element $a \in R$ is a called a unit if it's invertible in $R$ or in other words has a multiplicative inverse in $R$, i.e.

\begin{equation*} \forall a \in R, \ \exists a^{-1} \in R : \quad a a^{-1} = 1 = a^{-1} a \end{equation*}

We will use the notation $R^\times$ for the group of units of a ring $R$.

In a ring $R$ a non-zero element $a$ is called a zero-divisor or divisor of zero if

\begin{equation*} \exists b \ne 0 : \quad ab = 0 \quad \text{or} \quad ba = 0 \end{equation*}

An integral domain is a non-zero commutative ring that has no zero-divisors, i.e.

  1. If $ab = 0$ then $a = 0$ or $b = 0$
  2. $a \ne 0$ and $b \ne 0$ then $ab \ne 0$

Factor Rings

Let $I \trianglelefteq R$ be an ideal in a ring $R$. The set

\begin{equation*} x + I := \{ x + i : i \in I \} \subseteq R \end{equation*}

is a coset of $I$ in $R$ or the coset of $x$ wrt. $I$ in $R$.

Let $R$ be a ring and $I$ and ideal of $r$.

The mapping

\begin{equation*} \begin{split} \text{can}: \quad & R \to R / I \\ & r \mapsto \text{can}(r) \\ \forall r \in R, \quad & \text{can}(r) = r + 1 \end{split} \end{equation*}

Has the following properties:

  1. $\text{can}$ is surjective

  2. If $f: R \to S$ is a ring homomorphism with $f(I) = \{ 0_S \}$ so that

    \begin{equation*} I \subseteq \text{Ker}(f) \end{equation*}

    then there is a unique ring homomorphism:

    \begin{equation*} \overline{f} : R / I \to S \quad \text{such that} \quad f = \overline{f} \circ \text{can} \end{equation*}

Where the second point states that $f$ factorizes uniquely through the canonical mapping to the factor whenever the ideal $I$ is sent to zero.

Let $R$ and $S$ be rings.

Then every ring homomorphism $f: R \to S$ induces a ring isomorphism

\begin{equation*} \overline{f} = R / \text{Ker}(f) \overset{\sim}{\to} \text{Im}(f) \end{equation*}

Modules

We say $(M, \oplus, \odot)$ is a R-module, $R$ being a ring, if

\begin{equation*} \begin{split} \oplus:& M \times M \to M \\ \odot:& R \times M \to M \end{split} \end{equation*}

satisfying

\begin{equation*} C^{\oplus} \ A^{\oplus, \otimes} \ N^{\oplus} \ I^{\oplus} \quad A \ D \ D \ U \end{equation*}

Thus, we can view it as a "vector space" over a ring, but because it behaves wildly different from a vector space over a field, we give this space a special name: module.

Important: $M$ denotes a module here, NOT manifold as usual.

A unitary module is in the case where we also have $I^{\otimes}$, i.e. the ring is a unitary ring and contains a multiplicative identity-element.

Let $R$ be a ring and let $M$ be an R-module. A subset $M'$ of $M$ is a submodule if and only if

  1. $0_M \in M'$
  2. $a, b \in M' \implies a - b \in M'$
  3. $r \in r, a \in M' \implies ra \in M'$

Let $R$ be a ring, let $M$ and $N$ be R-modules and let $f: M \to N$ be an R-homomorphism.

Then $f$ is injective if and only if

\begin{equation*} \text{ker}(f) = \left\{ 0_M \right\} \end{equation*}

Let $T \subseteq M$. Then ${}_{R}\langle T \rangle$ is the smallest submoduel of $M$ that contains $T$.

The intersection of any collection of submodules of $M$ is a submodule of $M$.

Let $M_1$ and $M_2$ be submodules of $M$. Then

\begin{equation*} M_1 + M_2 = \left\{ a + b: a \in M_1, b \in M_2 \right\} \end{equation*}

is a submodule of $M$.

Let $R$ be a ring, $M$ and R-module and $N$ submodule of $M$.

For ever $a \in M$ the coset of $a$ wrt. $N$ in $M$ is

\begin{equation*} a + N = \left\{ a + b : b \in N \right\} \end{equation*}

It is a coset of $N$ in the abelian group $M$ and so is an equivalence class for the equivalence relation

\begin{equation*} a \sim b \iff a -b \in N \end{equation*}

The factor of $M$ by $N$ or quotient of $M$ by $N$ is the set

\begin{equation*} \big( M / \sim_N \big) \end{equation*}

of all cosests of $N$ in $M$.

Equipped with addition and s-multiplication

\begin{equation*} \begin{split} (a + N) + (b + N) & = (a + b) + N \\ r (a + N) &= ra + N \end{split} \end{equation*}

for all $a, b \in M$ and $r \in R$.

The R-module $M / N$ is the factor module of $M$ by submodule $N$.

Let $R$ be a ring, let $L$ and $M$ be R-modules, and $N$ a submodule of $M$.

  1. The mapping $\text{can}: M \to M / N$ defined by

    \begin{equation*} \text{can}(a) = a + N, \quad \forall a \in M \end{equation*}

    is surjective.

  2. If $f: M \to L$ is an R-homomorphism with

    \begin{equation*} f(N) = \left\{ 0_L \right\} \end{equation*}

    so that $N \subseteq \text{ker}(f)$, then there is a unique homomorphism $\bar{f}: M / N \to L$ such that $f = \bar{f} \circ \text{can}$.

Let $R$ be a ring and let $M$ and $N$ be R-module. Then every R-homomorphism $f: M \to N$ induces an R-isomorphism

\begin{equation*} \overline{f}: M / \text{ker}(f) \overset{\sim}{\to} \text{im}(f) \end{equation*}

Let $N, K$ be submodules of a R-module $M$.

Then $K$ is a submodule of $N + K = \left\{ b + c : b \in N, c \in K \right\}$ and $N \cap K$ is a submodule of $N$.

Also,

\begin{equation*} \frac{N + K}{K} \cong \frac{N}{N \cap K} \end{equation*}

Let $N, K$ be submodules of an R-module $M$, where $K \subseteq N$.

Then $N / K$ is a submodule of $M / K$.

Also,

\begin{equation*} \frac{M / K}{N / K} \cong M / N \end{equation*}

Determinants and Eigenvalue Reduction

Definitions

An inversion of a permutation $\sigma \in \mathfrak{S}_n$ is a pair $(i, j)$ such that $1 \le i < j \le n$ and $\sigma(i) > \sigma(j)$.

The number of inversions of the permutation $\sigma$ is called the length of $\sigma$ and writtein $\ell(\sigma)$.

The sign of a permutation $\sigma$ is defined to be the parity of the number of inversions of $\sigma$:

\begin{equation*} \sgn(\sigma) = \big( -1 \big)^{\ell(\sigma)} \end{equation*}

where $\ell$ is the length of the permutation.

For $n \in \mathbb{N}$, the set of even permutations in $\mathfrak{S}$ forms a subgroup of $\mathfrak{S}$ because it is the kernel of the group homomorphism $\text{sgn}: \mathfrak{S}_n \to \left\{ -1, 1 \right\}$.

This group is the alternating group and is denoted $A_n$.

(X̃ / ker(p))

Let $R$ be a ring.

The determinant is a mapping $\det : \text{Mat}(n ; R) \to R$ given by

\begin{equation*} A \mapsto \det(A) = \sum_{\sigma \in \mathfrak{S}_n} \text{sgn}(\sigma) a_{1 \sigma(1)} \cdots a_{n \sigma(n)} \end{equation*}

Let $R$ be a commutative ring, then

\begin{equation*} \det \big( A^T \big) = \sum_{\sigma \in \mathfrak{S}_n} \text{sgn}(\sigma) a_{\sigma(1) 1} \cdots a_{\sigma(n) n} = \det(A) \end{equation*}

Let $A \in \text{Mat}(n, R)$ for some ring $R$, and let $M_{ij}$ be the $(n - 1) \times (n - 1)$ defined by removing the i-th row and j-th column of $A$.

Then the cofactor matrix $C$ of $A$ is defined (component-wise)

\begin{equation*} C_{ij} = (-1)^{i + j} \det(M_ij) \end{equation*}

The adjugate matrix of $A$ is defined

\begin{equation*} \adj(A) = C^T \end{equation*}

where $C$ is the cofactor matrix.

The reason for this definition is so that we have the following identity

\begin{equation*} \adj(A) A = \det(A) I_n = A \adj(A) \end{equation*}

To see this, recall the "standard" approach to computing the determinant of a matrix $A$, where we do so by choosing some row $i$ and some column $j$, cut out the rest of the matrix, and write the determinant as a expansion in these coefficients. The cofactor matrix has exactly the determinant of that $(n - 1) \times (n - 1)$ matrix as a entries, hence the above expression is simply going to give us the same expression as the "expansion-method" for computing the determinant.

Let $A = \big( a_{ij} \big)$ be an $(n \times n)$ matrix with entries from a commutative ring $R$.

For a fixed $i$ the i-th row expansion of the determinant is

\begin{equation*} \det(A) = \sum_{j=1}^{n} a_{ij} C_{ij} \end{equation*}

and for a fixed $j$ the j-th column expansion of the determinant is

\begin{equation*} \det( A) = \sum_{i=1}^{n} a_{ij} C_{ij} \end{equation*}

where $C_{ij}$ is the $(i, j)$ cofactor of $A$

\begin{equation*} C_{ij} = (-1)^{i + j} \det \big( A\langle i, j \rangle \big) \end{equation*}

where $A \langle i, j \rangle$ is the matrix obtained from deleting the i-th row and j-th column.

Cayley-Hamilton Theorem

Let $A \in \text{Mat}(n, R)$ be a square matrix with entries in a commutative ring $R$.

Then evaluating its characteristic polynomial $\chi_A(x) \in R[x]$ at the matrix $A$ gives zero.

Let

\begin{equation*} B = \adj \big( t I_n - A \big) \end{equation*}

where $\adj$ denotes the adjugate matrix of $t I_n - A$ and $t \in R$.

Observe then that

\begin{equation*} (t I_n - A) B = \det(t I_n - A) = \chi_A(t) \end{equation*}

by the propoerty of the adjugate matrix.

Since $B$ is a matrix whose entries are a polynomial, we can write $B$ as

\begin{equation*} B = \sum_{i=0}^{n - 1} t^i B_i \end{equation*}

for some matrices $\left\{ B_i \in \text{Mat}(n, R) \right\}$.

Therefore,

\begin{equation*} \begin{split} \chi_A(t) &= (t I_n - A) B \\ &= (t I_n - A) \sum_{i=0}^{n - 1} t^i B_i \\ &= \sum_{i=0}^{n - 1} t^{i + 1} B_i - t^i A B_i \\ &= t^n B_{n - 1} + \bigg( \sum_{i=1}^{n - 1} t^{i} (B_{i - 1} - A B_i) \bigg) - A B_0 \end{split} \end{equation*}

Now, letting

\begin{equation*} \chi_A(t) I_n = t^n I_n + c_{n - 1} t^{n - 1} I_n + \dots + c_1 t I_n + c_0 I_n \end{equation*}

we obtain the following relations

\begin{equation*} \begin{split} I_n &= B_{n - 1} \\ c_i I_n &= B_{i - 1} - A B_i \quad \forall i = 1, \dots, n - 1 \\ c_0 I_n &= - A B_0 \end{split} \end{equation*}

Multiplying the above relations by $A^i$ for each $i$ we get the following

\begin{equation*} \begin{split} A^n &= A^n B_n \\ c_i A^i &= A^i B_{i - 1} - A^{i + 1} B_i \quad \forall i = 1, \dots, n - 1 \\ c_0 I_n &= - A B_0 \end{split} \end{equation*}

Substituting back into the above series, we have

\begin{equation*} A^n B_{n - 1} + \bigg( \sum_{i=1}^{n - 1} \big( A^i B_{i - 1} - A^{i + 1} B_{i} \big) \bigg) - A^n B_0 = A^n + c_{n - 1} A^{n - 1} + \dots + c_1 A + c_0 I_n = \chi_X(A) \end{equation*}

we observe that the LHS vanishes, since we can group terms which cancel! Hence

\begin{equation*} \chi_A(A) = 0 \end{equation*}

Eigenvalues and Eigenvectors

Theorems

Each endomorphism of non-zero finite dimensional vector space over an algebraically closed field has an eigenvalue.

Inner Product Spaces

Definitions

Inner product

An inner product is a (anti-)bilinear map $\left\langle \cdot, \cdot \right\rangle: V \times V \to \mathbb{K}$ which is

  1. Symmetric
  2. Non-degenerate
  3. Positive-definite
Forms

Let $B: V \times W \to \mathbb{R}$. We say $B$ is a bilinear form if

\begin{equation*} \begin{split} B \Big( \big( \lambda_1 v_1 + \lambda_2 v_2 \big), w \Big) &= \lambda_1 B(v_1, w) + \lambda_2 B(v_2, w) \\ B \Big( v, \big(\lambda_1 w_1 + \lambda_2 w_2\big) \Big) &= \lambda_1 B(v, w_1) + \lambda_2 B(v, w_2) \end{split} \end{equation*}

for $\lambda_1, \lambda_2 \in \mathbb{R}$, $v, v_1, v_2 \in V$ and $w, w_1, w_2 \in W$.

A symmetric bilinear form on $T_p D$ is a bilinear map $B : T_p D \times T_p D \to \mathbb{R}$ such that $B(v, w) = B(w, v), \quad \forall v, w \in T_p D$.

Given two 1-forms $\alpha, \beta$ at $p \in D$ define a symmtric bilinear form $\alpha \beta$ on $T_pD$ by

\begin{equation*} (\alpha \beta) (v, w) = \frac{1}{2} \Big( \alpha(v) \beta(w) + \alpha(w) \beta(v) \Big) \end{equation*}

where $v, w \in T_p D$.

Note that $\alpha \beta = \beta \alpha$ and we denote $\alpha \alpha = \alpha^2$.

REDEFINE WITHOUT THE ADDED TANGENT-SPACE STRUCTURE, BUT ONLY USING VECTOR SPACE STRUCTURE.

A symmetric tensor on $D$ is a map which assigns to each $p \in D$ a symmetric bilinear form on $T_p D$; it can be written as

\begin{equation*} B = B_{ij} dx^i dx^j \end{equation*}

where $B_{ij}$ are smooth functions on $D$.

Remember, we're using Einstein notation.

Skew-linear and sesquilinear form

We say the mapping $f: V \to W$ between complex vector spaces is skew-linear if

\begin{equation*} \begin{split} f(\mathbf{v}_1 + \mathbf{v}_2) &= f(\mathbf{v}_1) + f(\mathbf{v}_2) \\ f(\lambda \mathbf{v}_1 ) &= \overline{\lambda} f(\mathbf{v}_1) \end{split} \end{equation*}

Let $V$ be a vector space over $\mathbb{C}$ equipped with the inner product

\begin{equation*} \big( -, - \big): V \times V \to \mathbb{C} \end{equation*}

Since this mapping is skew-linear in the second argument, i.e.

\begin{equation*} \big( \mathbf{z}, \lambda \mathbf{x} + \mu \mathbf{y} \big) = \overline{\big( \lambda \mathbf{x} + \mu \mathbf{y}, \mathbf{z} \big)} = \overline{\lambda \cdot \big( \mathbf{x}, \mathbf{z} \big) + \mu \cdot \big( \mathbf{y}, \mathbf{z} \big)} = \overline{\lambda} \big( \mathbf{z}, \mathbf{x} \big) + \overline{\mu} \big( \mathbf{z}, \mathbf{y} \big) \end{equation*}

we say this is a sesquilinear form.

Hermitian

Let $V$ be a vector space over $\mathbb{C}$. If the sequilinear form $\omega: V \times V \to \mathbb{C}$ is symmetric, i.e.

\begin{equation*} \omega\big( \mathbf{x}, \mathbf{y} \big) = \omega \big( \mathbf{y}, \mathbf{x} \big) \end{equation*}

we say $\omega$ is a Hermitian.

Theorems

Let $\mathbf{v}, \mathbf{w}$ be a vectors in an inner product space. Then

\begin{equation*} |(\mathbf{v}, \mathbf{w})| \le ||\mathbf{v}|| \ ||\mathbf{w}|| \end{equation*}

with equailty if and only if $\mathbf{v}$ and $\mathbf{w}$ are linearly dependent.

Adjoints and Self-adjoints

Let $V$ be an inner product space. then two endomorphism $\varphi, \psi: V \to V$ are called adjoint to if the following holds:

\begin{equation*} \big( \varphi (v), w \big) = \big( v, \psi (w) \big) \quad \forall v, w, \in V \end{equation*}

Let $T: V \to V$ and $T^*$ be the adjoint of $T$. We say $T$ is self-adjoint if and only if

\begin{equation*} T = T^* \end{equation*}

Let $V$ be a finite-dimensional inner product space and let $T: V \to V$ be a self-adjoint linear mapping.

Then $V$ has an orthonormal basis consisting of eigenvectors of $T$.

We'll prove this using induction on $\dim V$. Let $V_k$ denote the an inner product space with $\dim V_k = k$ with the self-adjoint linear mapping $T$.

Base case: $k = 1$. Let $\mathbf{v}_1$ denote an eigenvector of $T$, i.e.

\begin{equation*} T \mathbf{v}_1 = \lambda_1 \mathbf{v}_1, \lambda_1 \in F \end{equation*}

where $F$ is the underlying field of the vector space $V$. Then clearly $v_1$ spans $V_1$.

General case: Assume the hypothesis holds for $1, 2, \dots, k$, then

\begin{equation*} V_{k + 1} = V_k^{\perp} \oplus V_k \end{equation*}

where

\begin{equation*} \dim V_k^{\perp} = 1 \quad \text{and} \quad \dim V_k = k \end{equation*}

Further, let $\mathbf{v}_i \in V_k$ and $\mathbf{u} \in V_k^{\perp}$ with $\mathbf{v}_i$ be an eigenvector of $T$. Then

\begin{equation*} \big( \mathbf{v}_i, T \mathbf{u} \big) = \big( T \mathbf{v}_i, \mathbf{u} \big) = \lambda_i \big( \mathbf{v}_i, \mathbf{u} \big) = 0 \end{equation*}

since $T$ is self-adjoint. This implies that

\begin{equation*} T \mathbf{u} \in V_k^{\perp}, \quad \forall \mathbf{u} \in V_k^{\perp} \implies T \big( V_k^{\perp} \big) \subseteq V_k^{\perp} \end{equation*}

Therefore, restricting $T$ to $V_k^{\perp}$ we have the mapping

\begin{equation*} T \big|_{V_k^{\perp}}: V_k^{\perp} \to V_k^{\perp} \end{equation*}

Where $T \big|_{V_k^{\perp}}$ is also self-adjoint, since $T$ is self-adjoint on the entirety of $V_{k + 1}$. Thus, by assumption, we know that there exists a basis of $V_k^{\perp}$ consisting of eigenvectors of $T \big|_{V_k^{\perp}}$, which are therefore orthogonal to the eigenvectors of $T$ in $V_k$. Hence, existence of basis of eigenvectors for $V_k$ implies existence of basis of eigenvectors of $V_{k + 1}$.

Thus, by the induction hypothesis, if $T$ is a self-adjoint operator on the inner product space $V$ with $\dim V = n$, $n \in \mathbb{N}$, then there exists a basis of eigenvectors of $T$ for $V$, as claimed.

Jordan Normal Form

Notation

  • $\phi: V \to V$ is an endomorphism of the finite dimensional F-vector space $V$.

  • Characteristic equation of $\phi$:

    \begin{equation*} \chi_{\phi}(x) = \big( \lambda_1 - x \big)^{a_1} \big( \lambda_2 - x \big)^{a_2} \cdots \big( \lambda_s - x \big)^{a_s} \in F[x] \end{equation*}
  • $V_i = \text{ker}\Big( (\phi - \lambda_i \text{id}_V)^{a_i} \Big) \subseteq V$

  • Polynomials

    \begin{equation*} P_i(x) = \prod_{j = 1, j \ne i}^s \big( x - \lambda_j \big)^{a_j} \end{equation*}

  • Generalized eigenspace of $\phi$ wrt. eigenvalue $\lambda_i$

    \begin{equation*} E^{\text{gen}}(\lambda_i, \phi) = \left\{ \mathbf{v} \in V \mid \big( \phi - \lambda_i \text{id}_V \big)^{a_i} (\mathbf{v}) = \mathbf{0} \right\} \end{equation*}
  • $a_i$ denotes the dimension of $E^{\text{gen}}(\lambda_i, \phi)$

  • Basis

    \begin{equation*} \mathcal{B}_i = \left\{ \mathbf{v}_{i t} \in V \mid 1 \le t \le a_i \right\} \end{equation*}

  • Restriction of $\phi$

    \begin{equation*} \phi_i = \phi \big|_{E^{\text{gen}}(\lambda_i, \phi)} : E^{\text{gen}}(\lambda_i, \phi) \to E^{\text{gen}}(\lambda_i, \phi) \end{equation*}

  • $\psi_m: W_m / W_{m - 1} \to W_{m - 1} / W_{m - 2}$ defined

    \begin{equation*} \psi_m(\mathbf{w} + W_{m - 1}) = \psi(\mathbf{w}) + W_{m - 2}, \quad \mathbf{w} \in W_m \end{equation*}

    is well-defined and injective.

  • $d_m = \dim W_m / W_{m - 1}$

Definitions

An endomorphism $f: V \to V$ of an F-vector space is called nilpotent if and only if $\exists d \in \mathbb{N}$ such that

\begin{equation*} f^d = 0 \end{equation*}
\begin{equation*} \begin{split} \exp : \quad & \text{Mat}(n; \mathbb{C}) \to \text{Mat}(n; \mathbb{C}) \\ & A \mapsto \sum_{k=0}^{\infty} \frac{1}{k!} A^k \end{split} \end{equation*}

Motivation

  • $V$ be a finite dimensional vector space
  • $f: V \to V$ an endomorphism
    • a choice of ordered basis $\mathcal{B}$ for $V$ determines a matrix $~_{\mathcal{B}} [f]_{\mathcal{B}}$ representing $f$ wrt. basis $\mathcal{B}$
  • Another choice of basis leads to a different representation; would like to find the simplest possible matrix that is conjugate to a given matrix

Theorem

Given an integer $r \ge 1$, we define the $(r \times r)$ matrix $J(r)$

\begin{equation*} J(r) = \begin{pmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ 0 & 0 & 0 & \dots & 0 \end{pmatrix} \end{equation*}

or equivalently,

\begin{equation*} \big( J(r) \big)_{ij} = \begin{cases} 1 & \text{if } j = i + 1 \\ 0 & \text{otherwise} \end{cases} \end{equation*}

which we call the nilpotent Jordan block of size $r$.

Given an integer $r \ge 1$ and a scalar $\lambda \in F$ define an $(r \times r)$ matrix $J(r, \lambda)$ as

\begin{equation*} J(r, \lambda) = J(r) + \lambda I_r = \begin{pmatrix} \lambda & 1 & 0 & \dots & 0 \\ 0 & \lambda & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ 0 & 0 & 0 & \dots & \lambda \end{pmatrix} \end{equation*}

which we is called the Jordan block of size $r$ and eigenvalue $\lambda$.

Let $F$ be an algebraically closed field, and $V$ be a finite dimensional vector space and $\phi: V \to V$ be an endomorphism of $V$ with characteristic polynomial

\begin{equation*} \chi_{\phi}(x) = \big( \lambda_i - x \big)^{a_1} \big( \lambda_2 - x \big)^{a_2} \cdots \big( \lambda_s - x \big)^{a_s} \in F[x] \end{equation*}

where $a_i \ge 1$ and $\sum_{i=1}^{s} a_i = n$, for distinct $\lambda_1, \lambda_2, \dots, \lambda_s \in F$.

Then there exists an ordered basis $\mathcal{B}$ of $V$ s.t.

\begin{equation*} ~_{\mathcal{B}}[\phi]_{\mathcal{B}} = \text{diag} \Big( J(r_{11}, \lambda_1), \dots, J(r_{1 m_1}, \lambda_1), J(r_{21}, \lambda_2), \dots, J(r_{s m_s}, \lambda_s) \Big) \end{equation*}

with $r_{11}, \dots, r_{1 m_1}, r_{21}, \dots, r_{s m_s} \ge 1$ such that

\begin{equation*} a_i = r_{i1} + r_{i2} + \dots + r_{i m_i} \end{equation*}

with $1 \le i \le s$.

That is, $\phi$ in the basis $\mathcal{B}$ is block diagonal with Jordan blocks on the diagonal!

Proof of Jordan Normal Form
  • Outline

    We will prove the Jordan Normal form in three main steps:

    1. Decompose the vector space $V$ into a direct sum

      \begin{equation*} V = \bigoplus_{i=1}^s V_i \end{equation*}

      according to the factorization of the characteristic polynomial as a product of linear factors:

      \begin{equation*} \chi_{\phi}(x) = \big( \lambda_1 - x \big)^{a_1} \big( \lambda_2 - x \big)^{a_2} \cdots \big( \lambda_s - x \big)^{a_s} \in F[x] \end{equation*}

      for distinct scalars $\lambda_1, \lambda_2, \dots, \lambda_s \in F$, where for each $i$:

      • $V_i = \text{ker}\Big( (\phi - \lambda_i \text{id}_V)^{a_i} \Big) \subseteq V$
      • $\phi(V_i) \subseteq V_i$
    2. Focus attention on each of the $V_i$ to obtain the nilpotent Jordan blocks.
    3. Combine Step 2 and 3
  • Step 1: Decompose

    Rewriting $\chi_{\phi}(x)$ as

    \begin{equation*} \chi_{\phi}(x) = (-1)^n \prod_{j = 1}^s \big( x - \lambda_j \big)^{a_j} \in F[x] \end{equation*}

    where $\lambda_j$ are the eigenvalues of $\phi$.

    For $1 \le i \le s$ define

    \begin{equation*} P_i(x) = \prod_{j = 1, j \ne i}^s \big( x - \lambda_j \big)^{a_j} \end{equation*}

    There exists polynomials $Q_i(x) \in F[x]$ such that

    \begin{equation*} \sum_{i=1}^{s} P_i(x) Q_i(x) = 1 \end{equation*}

    For each $1 \le i \le s$, let

    \begin{equation*} \mathcal{B}_i = \left\{ \mathbf{v}_{it} \in V \mid 1 \le t \le a_i \right\} \end{equation*}

    be a basis of $E^{\text{gen}}(\lambda_i, \phi)$, where $a_i$ is the algebraic multiplicity of $\phi$ with eigenvalue $\lambda_i$.

    1. Each $E^{\text{gen}}(\lambda_i, \phi)$ is stable under $\phi$, i.e. $\phi \Big( E^{\text{gen}}(\lambda_i, \phi) \Big) \subseteq E^{\text{gen}}(\lambda_i, \phi)$

    2. For each $\mathbf{v} \in V$, $\exists \mathbf{v}_i \in E^{\text{gen}}(\lambda_i, \phi)$ such that

      \begin{equation*} \mathbf{v} = \sum_{i=1}^{s} \mathbf{v}_i \end{equation*}

      In other words, there is a direct sum decomposition

      \begin{equation*} V = \bigoplus_{i = 1}^s E^{\text{gen}}(\lambda_i, \phi) \end{equation*}

    3. Then

      \begin{equation*} \mathcal{B} = \mathcal{B}_1 \cup \mathcal{B}_2 \cup \dots \cup \mathcal{B}_s = \left\{ \mathbf{v}_{it} \mid 1 \le i \le s, 1 \le t \le a_i \right\} \end{equation*}

      is a basis of $V$, so in particular $n = \sum_{i=1}^{s} a_i = \dim V$. The matrix of the endomorphism $\phi$ wrt. to basis $\mathcal{B}$ is given by the block diagonal matrix

      \begin{equation*} ~_{\mathcal{B}}[\phi]_{\mathcal{B}} = \left( \begin{array}{c|c|c|c} B_1 & 0 & 0 & 0 \\ \hline 0 & B_2 & 0 & 0 \\ \hline 0 & 0 & \ddots & 0 \\ \hline 0 & 0 & 0 & B_s \end{array} \right) \in \text{Mat}(n; F) \end{equation*}

      with $B_i = ~_{\mathcal{B}}[\phi_i]_{\mathcal{B}} \in \text{Mat}(a_i; F)$.

    1. Let $\mathbf{v} \in E^{\text{gen}}(\lambda_i, \phi)$ is that

      \begin{equation*} \big( \phi - \lambda_i \text{id}_V \big)^{a_i}(\mathbf{v}) = \mathbf{0} \end{equation*}

      Then

      \begin{equation*} \Big( \phi \circ (\phi - \lambda_i \text{id}_V)^{a_i} \Big)(\mathbf{v}) = \Big( (\phi - \lambda_i \text{id}_V)^{a_i} \circ \phi \Big)(\mathbf{v}) = \phi(\mathbf{0}) = \mathbf{0} \end{equation*}

      Hence $\phi(\mathbf{v}) \in E^{\text{gen}}(\lambda_i, \phi)$, i.e. $E^{\text{gen}}(\lambda_i, \phi)$ is stable under $\phi$.

    2. By Lemma 6.3.1 we have $1 = \sum_{i=1}^{s} P_i(x) Q_i(x)$ and so evaluating this at $\phi$, we get

      \begin{equation*} \text{id}_V = \sum_{i=1}^{s} P_i(\phi) \circ Q_i(\phi) \end{equation*}

      Therefore, $\forall \mathbf{v} \in V$, we have

      \begin{equation*} \mathbf{v} = \sum_{i=1}^{s} \Big( P_i(\phi) \circ Q_i(\phi) \Big)(\mathbf{v}) \end{equation*}

      Observe that

      \begin{equation*} \Big( \big( \phi - \lambda_i \text{id}_V \big)^{a_i} \circ P_i(\phi) \circ Q_i(\phi) \Big)(\mathbf{v}) = \Big( \chi_{\phi}(\phi) \circ Q_i(\phi) \Big)(\mathbf{v}) = 0 (\mathbf{v}) = \mathbf{0} \end{equation*}

      where we've used Cailey-Hamilton Theorem for the second equality. Let

      \begin{equation*} \mathbf{v}_i := \Big( P_i(\phi) \circ Q_i(\phi) \Big)(\mathbf{v}) \in E^{\text{gen}}(\lambda_i, \phi) \end{equation*}

      then

      \begin{equation*} \mathbf{v} = \sum_{i=1}^{s} \mathbf{v}_i \end{equation*}

      hence all $\mathbf{v} \in V$ can be written as a sum of $v_i \in E^{\text{gen}}(\lambda_i, \phi)$, or equivalently,

      \begin{equation*} V = \bigoplus_{i = 1}^s E^{\text{gen}}(\lambda_i, \phi) \end{equation*}

      as claimed.

    3. Since $\mathcal{B}_i = \left\{ \mathbf{v}_{it} : 1 \le t \le a_i \right\}$ is a basis of $E^{\text{gen}}(\lambda_i, \phi)$ for each $i$, and since

      \begin{equation*} \text{span} \left( \mathcal{B}_i \right) \cap \text{span} \left( \mathcal{B}_j \right) = \emptyset, \quad i \ne j \end{equation*}

      we have

      \begin{equation*} \mathcal{B} = \bigcup_{i = 1}^s \mathcal{B}_i \end{equation*}

      form a basis of $V$. Consider the ordered basis $\mathbf{v}_{11}, \mathbf{v}_{12}, \dots, \mathbf{v}_{1a_1}, \mathbf{v}_{21}, \dots, \mathbf{v}_{s1}, \dots, \mathbf{v}_{s a_s}$, then the $\phi(\mathbf{v}_{it})$ can be expressed as a linear combination of the vectors $\mathbf{v}_{it}$ with $1 \le t \le a_i$. Therefore the matrix is block diagonal with i-th block having size $(a_i \times a_i)$.

    From this, one can prove that any matrix $A$ can be written as a Jordan decomposition:

    Let $A \in \text{Mat}(n; F)$, then there exists a diagonalisable (NOT diagonal) matrix $D$ and nilpotent matrix $N$ such that

    \begin{equation*} A = N + D \end{equation*}

    In fact, this decomposition is unique and is called the Jordan decomposition.

  • Step 2: Nilpotent endomorphisms

    Let $W$ be a finite dimensional vector space and $\psi \in \text{End}(W)$ such that

    \begin{equation*} \psi^m = 0 \end{equation*}

    for some $m \in \mathbb{N}$, i.e. $\psi$ is nilpotent. Further, let $m$ be minimal, i.e. $\psi^m = 0$ but $\psi^{m - 1} \ne 0$.

    For $0 \le i \le m$ define

    \begin{equation*} W_i = \text{ker}(\psi^i) \end{equation*}

    If $\mathbf{w} \in W_i$ then

    \begin{equation*} \psi^{i + 1}(\mathbf{w}) = (\psi \circ \psi^i) (\mathbf{w}) = \psi(\mathbf{0}) = \mathbf{0} \end{equation*}

    i.e. $\mathbf{w} \in W_i \implies \mathbf{w} \in W_{i + 1}$, hence

    \begin{equation*} W_i \subseteq W_{i + 1} \end{equation*}

    Moreover, since $\psi^0 = \text{id}_W$ and $\psi^{m} = 0$, we have $W_0 = W$ and $W_m = W$. Therefore we get the chain of subspaces

    \begin{equation*} 0 = W_0 \subseteq W_1 \subseteq W_2 \subseteq \dots \subseteq W_{m - 1} \subseteq W_m = W \end{equation*}

    We can now develop an algorithm for constructing a basis!

    • Constructing a basis
      1. Choose arbitrary basis $\mathcal{B}_m$ for $W_m / W_{m - 1}$
      2. Choose basis of $\mathcal{B}_{m - 1}$ of $W_{m - 1} / W_{ m - 2}$ by mapping $\mathcal{B}_m$ using $\psi_m$ and choosing vectors linearly independent of $\psi_m(\mathcal{B}_m)$.
      3. Repeat!

      Or more accurately:

      1. Choose arbitrary basis for $W_m / W_{m - 1}$:

        \begin{equation*} \left\{ \mathbf{v}_m^{(1)} + W_{m - 1}, \mathbf{v}_m^{(2)} + W_{m - 1}, \dots, \mathbf{v}_m^{(d_m)} + W_{m - 1} \right\} \end{equation*}

      2. Since $\phi_m: W_m / W_{m - 1} \to W_{m-1} / W_{m - 2}$ is injective, by the fact that the image of a set of linear independent vectors is a linearly independent set if the map is injective, then

        \begin{equation*} \left\{ \psi\Big(v_m^{(1)}\Big) + W_{m - 2}, \psi\Big(v_m^{(2)}\Big) + W_{m - 2}, \dots, \psi\Big(\mathbf{v}_m^{(d_m)}\Big) + W_{m - 2} \right\} \end{equation*}

        is linearly independent.

      3. Choose vectors

        \begin{equation*} \left\{ \mathbf{v}_{m - 1}^{(i)}: d_m + 1 \le i \le d_{m - 1} \right\} \end{equation*}

        such that

        \begin{equation*} \left\{ \mathbf{v}_{m - 1}^{(i)} + W_{m - 2} : 1 \le i \le d_{m - 1} \right\} \end{equation*}

        is a basis of $W_{ m - 1} / W_{m - 2}$.

      4. Repeat!

      Now, the interesting part is this:

      Let $W$ be a finite dimensional vector space and $\psi \in \text{End}(W)$ such that

      \begin{equation*} \psi^m = 0 \end{equation*}

      for some $m \in \mathbb{N}$, i.e. $\psi$ is nilpotent.

      Let $\mathcal{B}$ be the ordered basis of $W$ constructed as above

      \begin{equation*} \left\{ \mathbf{v}_j^{(k)}: 1 \le j \le m, 1 \le k \le d_j \right\} \end{equation*}

      Then

      \begin{equation*} ~_{\mathcal{B}}[\psi]_{\mathcal{B}} = \diag \Big( \underbrace{J(m), \dots, J(m)}_{d_m \text{ times}}, \underbrace{J(m - 1), \dots, J(m - 1)}_{d_{m - 1} - d_m \text{ times}}, \dots, \underbrace{J(1), \dots, J(1)}_{d_1 - d_2 \text{ times}} \Big) \end{equation*}

      where $J(r)$ denotes the nilpotent Jordan block of size $r$.

      It follows from the explicit construction of the basis $\mathcal{B}$ that

      \begin{equation*} \psi\Big(\mathbf{v}_i^{(j)}\Big) = \begin{cases} \mathbf{v}_{i - 1}^{(j)} & \text{if } i > 1 \\ 0 & \text{otherwise} \end{cases} \end{equation*}

      Since $~_{\mathcal{B}}[\psi]_{\mathcal{B}}$ is defined by how it maps the basis vectors in $\mathcal{B}$, and in the basis $\mathcal{B}$ $\mathbf{v}_{i}^{(j)}$ becomes a vector of all zeros except the entry corresponding to the j-th basis-vector chosen for $W_i$, where it is a 1.

      Hence $~_{\mathcal{B}}[\psi]_{\mathcal{B}}$ is a nilpotent Jordan block as claimed.

      Concluding step 2; for all nilpotent endomorphisms there exists a basis such that the representing matrix can be written as a block diagonal matrix with nilpotent Jordan blocks along the diagonal.

  • Step 3: Bringing it together

    Again considering the endomorphisms $(\phi - \lambda_i \text{id}_V)$ restricted to $E^{\text{gen}}(\lambda_i, \phi)$, we can apply Proposition 6.3.9 to see that this endomorphism can be written as a block diagon matrix of the form stated for a suitable choice of basis.

    The endomorphism $\lambda_i \text{id}_V$ restricted to $E^{\text{gen}}(\lambda_i, \phi)$ is of course $\lambda_i \text{id}_{E^{\text{gen}}(\lambda_i, \phi)}$, thus the matrix wrt. the chosen basis is just $\lambda_i I_{a_i}$. Therefore the matrix for $\phi = \lambda_i \text{id}_V + (\phi - \lambda_i \text{id}_V)$ (when restricted to $E^{\text{gen}}(\lambda_i, \phi)$) is just $\lambda_i I_{a_i}$ plus $J(a_i)$, i.e. $J(\lambda_i, a_i)$.

    Thus, each $B_i \in \text{Mat}(a_i; F)$ appearing in this theorem is exactly of the form we stated in Jordan Normal form.

Algorithm

  1. Calculate the eigenvalues, $\lambda_1, \dots, \lambda_s$ of $A$

    \begin{equation*} \chi_A(x) = (\lambda_1 - x)^{a_1} (\lambda_2 - x)^{a_2} \cdots (\lambda_s - x)^{a_s} \end{equation*}

  2. For each $\lambda = \lambda_i$ and $a = a_i$, let

    \begin{equation*} B = A - \lambda \ \text{id} \end{equation*}

    1. Compute $B, B^2, \dots, B^a$ and

      \begin{equation*} \text{null}(B) \subseteq \text{null}(B^1) \subseteq \dots \subseteq \text{null}(B^a) \end{equation*}

    2. Let

      \begin{equation*} d_i = \dim \Big( \text{null}(B^i) \Big), \quad d_0 = 0 \end{equation*}

    3. Set

      \begin{equation*} e_i := d_i - d_{i - 1} \quad \text{and} \quad \mathcal{B} := \emptyset \end{equation*}

      i.e. the difference in dimension between each of the nullspaces.

      1. Let $j$ be the largest integer such that $e_j > 0$.
      2. If $e_j$ does not exist, stop. Otherwise, goto step 3.
      3. Let $\mathbf{v} \in \text{null}(B^j) \setminus \text{null}(B^{j - 1})$ and $\mathbf{v} \notin \mathcal{B}$.
      4. Let
      \begin{equation*} \mathcal{B} := \mathcal{B} \cup \Big( B^{j - 1} \mathbf{v}, B^{j - 2} \mathbf{v}, \dots, B \mathbf{v}, \mathbf{v} \Big) \end{equation*}
      1. Change $e_i$ to $e_{i - 1}$ for $1 \le i \le j$.
  3. Let the full basis be the union of all the $\mathcal{B}$, i.e. $\cup \mathcal{B}$.

Tensor spaces

The tensor product $V \otimes W$ between two vector spaces $V, W$ is a vector space with the properties:

  1. if $v \in V$ and $w \in W$, there is a "product" $v \otimes w \in V \otimes W$

  2. This product $\otimes$ is bilinear:

    \begin{equation*} \begin{split} \big( \lambda_1 v_1 + \lambda_2 v_2 \big) \otimes w &= \lambda_1 \big( v_1 \otimes w \big) + \lambda_2 \big( v_2 \otimes w \big) \\ v \otimes \big( \lambda_1 w_1 + \lambda_2 w_2 \big) &= \lambda_1 \big( v \otimes w_1 \big) + \lambda_2 \big( v \otimes w_2 \big) \end{split} \end{equation*}

    for $\lambda_1, \lambda_2 \in \mathbb{R}$, $v, v_1, v_2 \in V$ and $w, w_1, w_2 \in W$.

Let $U$ be a vector space.

If $\varphi: V \times W \to U$ is a bilinear map, then there exists a unique linear map $\beta: V \otimes W \to U$ such that

\begin{equation*} \varphi(v, w) = \beta(v \otimes w), \quad \forall v \in V, w \in W \end{equation*}

I find the following instructive to consider.

In the case of a Cartesian product, the vector spaces are still "independent" in the sense that any element can be expressed in the basis

\begin{equation*} \mathcal{B}_{V \times W} = \left\{ (v_i, 0) \mid v_i \in \mathcal{B}_V \right\} \cup \left\{ (0, w_j) \mid w_j \in \mathcal{B}_V \right\} \end{equation*}

which means

\begin{equation*} \dim \big( V \times W \big) = \dim(V) + \dim(W) \end{equation*}

Now, in the case of a tensor product, we in some sense "intertwine" the spaces, making it so that we cannot express elements as "one part from $V$ and one part from $W$". And so we need the basis

\begin{equation*} \mathcal{B}_{V \otimes W} = \left\{ v_i \otimes w_j \mid v_i \in \mathcal{B}_V, w_j \in \mathcal{B}_W \right\} \end{equation*}

Therefore

\begin{equation*} \dim(V \otimes W) = \dim(V) \dim(W) \end{equation*}

By universal property of tensor products we can instead define the tensor product between two vector spaces $V, W$ as the dual vector space of bilinear forms on $V \times W$.

If $v \in V$, $w \in W$, then $v \otimes w \in V \otimes W$ is defined as the map

\begin{equation*} \big( v \otimes w \big)(B) = B(v, w), \quad \forall B: V \times W \overset{\sim}{\longrightarrow} \mathbb{R} \end{equation*}

for every bilinear form $B$.

That is,

\begin{equation*} B \in \Hom(V \times W, \mathbb{R}) \end{equation*}

and

\begin{equation*} V \otimes W = \Big( \Hom(V \times W, \mathbb{R}) \Big)^* \end{equation*}

by

\begin{equation*} \big( v \otimes w \big)(B) = B(v, w) \end{equation*}

Observe that this satisfies the universal property of tensor product since if we are given some $\varphi: V \times W \to U$, for every $\alpha \in U^*$, i.e. $\alpha: U \to \mathbb{R}$, we have $\alpha \circ \varphi: V \times W \overset{\sim}{\longrightarrow} \mathbb{R}$, i.e. $\alpha \circ \varphi$ is a bilinear form on $V \times W$. Furthermore, this dual of bilinear forms on $V \times W$ is then $V \otimes W \overset{\sim}{\longrightarrow} \big( U^* \big)^* = U$ (since we are working with finite-dimensional vector spaces).

\begin{equation*} \Hom_{}(V \otimes W, U) \cong Hom_{}\Big(V, \Hom_{}(W, U) \Big) \end{equation*}

From universal property of tensor product, for any $\varphi: V \times W \to U$, there exists a unique $\beta: V \otimes W \to U$ such that

\begin{equation*} \beta(v \otimes w) = \varphi(v, w) \in U \end{equation*}

Letting the map

\begin{equation*} \Psi: \Hom(V \otimes W, U) \to \Hom\Big(V, \Hom(W, U) \Big) \end{equation*}

be defined by

\begin{equation*} \Psi(\beta)(v) = \varphi(v, \cdot) \end{equation*}

where $\varphi(v, \cdot): W \to U$ is defined by $w \mapsto \varphi(v, w)$. By the uniqueness of $\beta$, and linearity of the maps under consideration, this defines an isomorphism between the spaces.

Suppose

  • $\left\{ v_1, \dots, v_m \right\}$ is a basis for $V$
  • $\left\{ w_1, \dots, w_n \right\}$ is a basis for $W$

Then a bilinear form $B: V \times W \to \mathbb{R}$ is fully defined by $B^{ij} := B(v_i, w_j)$ (upper-indices since these are the coefficients of the co-vectors). Therefore,

\begin{equation*} \dim \Big( (V \times W)^* \Big) = m \cdot n \end{equation*}

Since we are in finite dimensional vector spaces, the dual space $V \otimes W$ then has dimension

\begin{equation*} \dim \big( V \otimes W \big) = \dim \Big( (V \times W)^* \Big) = m \cdot n \end{equation*}

as well.

Furthermore, $v_i \otimes w_j$ form a basis for $V \otimes W$. Therefore we can write the elements in $V \otimes W$ as

\begin{equation*} \sum_{i,j}^{} B^{ij} v_i \otimes w_j \end{equation*}
\begin{equation*} \Hom(V, W) \cong W \otimes V^* \end{equation*}

First observe that if $\varphi \in \Hom(V, W)$, then $\varphi$ is fully defined by how it maps the basis elements $v_i$, and

\begin{equation*} \varphi(v_i) \in W \quad \implies \quad \exists \tensor{A}{^{j}_{i}}: \quad \varphi(v_i) = \tensor{A}{^{j}_{i}} w_j \end{equation*}

Then observe that if $w \otimes \alpha \in W \otimes V^*$, then

\begin{equation*} \big( w \otimes \alpha \big) = \tensor{B}{^{j}_{i}} \big( w_j \otimes \alpha^i \big) \end{equation*}

Hence, we have a natural homomorphism which simply takes $\varphi$, with coefficients $\tensor{A}{^{j}_{i}}$ to the corresponding element $\big( w \otimes \alpha \big)$ with the same coefficients! That is, the isomorphism is given by

\begin{equation*} \varphi^i(v_i) = \tensor{A}{^{j}_{i}} w_j \quad \mapsto \quad \big( w \otimes \alpha \big) = \tensor{A}{^{j}_{i}} \big( w_j \otimes \alpha^i \big) \end{equation*}

Tensor algebra

Now we'll consider letting $W = V$. We define tensor powers as

\begin{equation*} V^{\otimes k} = \underbrace{V \otimes \cdots \otimes V}_{k \text{ times}} \end{equation*}

with $V^{\otimes 0} = \mathbb{R}$, $V^{\otimes 1} = V$, etc. We can think of $V^{\otimes k}$ as the dual vector space of k-multilinear forms on $V$.

Combining all tensor powers using the direct sum we have define the tensor algebra

\begin{equation*} T(V) = \bigoplus_{k = 0}^{\infty} V^{\otimes k} \end{equation*}

whose elements are finite sums

\begin{equation*} \lambda 1 + v_0 + \sum_{i, j}^{} v_i \otimes v_j + \cdots + \sum_{i_1, \dots, i_p}^{} v_{i_1} \otimes \cdots \otimes v_{i_p} \end{equation*}

of tensor products of vectors $v_i \in V$.

The "multiplication" in $T(V)$ is defined by extending linearly the basic product

\begin{equation*} \big( v_1 \otimes \cdots \otimes v_p \big) \big( w_1 \otimes \cdots \otimes w_q \big) = v_1 \otimes \cdots \otimes v_p \otimes w_1 \otimes \cdots \otimes w_q \end{equation*}

The resulting algebra $T(V)$ is associative, but not commutative.

Two viewpoints

There are mainly due two useful ways of looking at tensors:

  1. Using Lemma lemma:homomorphisms-isomorphic-to-1-1-tensor-product, we may view $(r, s)$ tensors as linear maps $V^{\otimes s} \to V^{\otimes r}$. In other words,

    \begin{equation*} T_s^r(V) \cong \Hom(V^{\otimes s}, V^{\otimes r}) \end{equation*}
    • Furthermore, we can view $T_s^r(V)$ as the dual of $T_r^s(V) = V^{\otimes s} \otimes (V^* )^{\otimes r}$, i.e. consider a $(r, s)$ tensor as a "multilinear machine" which takes in $s$ vectors and $r$ co-vectors, and spits out a real number!
  2. By explicitly considering bases $\left\{ e_a \right\}$ of $V$ and $\left\{ \theta^a \right\}$ of $V^*$, the $(r, s)$ tensors are fully defined by how they map each combination of the basis elements. Therefore we can view tensors as multi-dimensional arrays!

Tensors (mainly as multidimensional arrays)

Let $V$ be a vector space $(\mathbb{R}, +, \cdot)$ where $R$ is some field, $+$ is addition in the vector space and $\cdot$ is scalar-multiplication.

Then a tensor $T$ is simply a linear map from some q-th Cartesian product of the dual space $V^*$ and some p-th Cartesian product of the vector space $V$ to the reals $\mathbb{R}$. In short:

\begin{equation*} T \in T_q^p V = \{ \phi: (V^*)^q \times V^p \overset{\sim}{\to} \mathbb{R} \} \end{equation*}

where $T_q^pV$ denotes the (p, q) tensor-space on the vector space $V$, i.e. linear maps from Cartesian products of the vector space and it's dual space to a real number.

Tensors are geometric objects which describe linear relations between geometric vectors, scalars, and other tensors.

A tensor of type $(p, q)$ is an assignment of a multidimensional array

\begin{equation*} T_{j_1, \dots, j_q}^{i_1, \dots, i_p} [ \mathbf{f} ] \end{equation*}

to each basis $\mathbf{f} = (\mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n)$ of an n-dimensional vector space such that, if we apply the change of basis

\begin{equation*} \mathbf{f} \mapsto \mathbf{f} \cdot R = (\mathbf{e}_i R_1^i, \dots, \mathbf{e}_i R_n^i ) \end{equation*}

Then the multi-dimensional array obeys the transformation law

\begin{equation*} T_{j'_1 \dots, j'_p}^{i'_1, \dots, i'_q} [\mathbf{f} \cdot R] = \big( R^{-1} \big)_{i_1}^{i'_1} \dots \big( R^{-1} \big)_{i_p}^{i'_p} \ T_{j_1, \dots, j_q}^{i_1, \dots, i_p} [ \mathbf{f} ] \ R_{j'_1}^{j_1} \dots R_{j'_q}^{j_q} \end{equation*}

We say the order of a tensor is $n$ if we require an n-dimensional array to describe the relation the tensor defines between the vector spaces.

Tensors are classified according to the number of contra-variant and co-variant indices, using the notation $(p, q)$, where

  • $p$ is # of contra-variant indices
  • $q$ is # of co-variant indices

Examples:

The tensor product takes two tensors, $S$ and $T$, and produces a new tensor, $S \otimes T$, whose order is the sum of the orders of the original tensors.

When described as multilinear maps, the tensor product simply multiplies the two tensors, i.e.

\begin{equation*} ( S \otimes T ) (v_1, \dots, v_n, v_{n + 1}, \dots, v_{n + m}) = S(v_1, \dots, v_n) T(v_{n + 1}, \dots, v_{n + m}) \end{equation*}

which again produces a map that is linear in its arguments.

On the components, this corresponds to multply the components of the two input tensors pairwise, i.e.

\begin{equation*} (S \otimes T )_{j_1, \dots, j_k, j_{k+1}, \dots j_{k + m}}^{i_1, \dots, i_l, i_{l+1}, \dots i_{l + n}} = S_{j_1, \dots, j_k}^{i_1, \dots, i_l} T_{j_{k+1}, \dots, j_{k + m}}^{i_{l + 1}, \dots, i_{l + n}} \end{equation*}

where

  • $S$ is of type $(l, k)$
  • $T$ is of type $(n, m)$

Then the tensor product $S \otimes T$ is of type $(l + n, k + m)$.

Let $\left\{ e_n \right\}$ be a basis of the vector space $V$ with $\dim V = n$, and $\left\{ f^b \right\}$ be the basis of $V^*$.

Then $\left\{ e_{a_1} \otimes \cdots \otimes e_{a_r} \otimes f^{b_1} \otimes \cdots f^{b_s} \right\}$ is a basis for the vector space of $\big( r, s \big) \text{-tensor}$ over $V$, and so

\begin{equation*} \dim T^r_s V = n^{r + s} \end{equation*}

Given $(r, s) \text{-tensor}$ $T$, we define $(r - 1, t - 1) \text{-tensor}$ $CT$

\begin{equation*} CT = T(\cdot, \dots, f^a, \dots, \cdot, \dots, e_a, \dots, \cdot) \end{equation*}

Components:

\begin{equation*} \tensor{\big( CT \big)}{^{a_1, \dots, a_{r - 1}}_{b_1, \dots, b_{s - 1}}} = \tensor{T}{^{a_1, \dots, a, \dots, a_{r - 1}}_{b_1, \dots, a, \dots, b_{s - 1}}} \end{equation*}

A contraction is basis independent:

\begin{equation*} \begin{split} \widetilde{CT} &= T \big( \dots, \tilde{f}^a, \dots, \tilde{e}_a, \dots \big) \\ &= T \Big( \dots, \tensor{A}{^a_b} f, \dots, \tensor{\big( A^{-1} \big)}{^c_a} e_c, \dots \Big) \\ &= \tensor{A}{^a_b} \tensor{\big( A^{-1} \big)}{^c_a} T( \dots, f^b, \dots, e_c, \dots) \\ &= T \big( \dots, f^b, \dots, e_a, \dots \big) = CT \end{split} \end{equation*}
\begin{equation*} \begin{split} \tensor{T}{_{(a, b)}} &= \frac{1}{2} \big( \tensor{T}{_{ab}} + T_{ba} \big) \implies T_{(a, b)} = T_{(b, a)} \\ \tensor{T}{_{[a, b]}} &= \frac{1}{2} \big( \tensor{T}{_{ab}} - \tensor{T}{_{ba}} \big) \implies T_{[a, b]} = - T_{[b, a]} \end{split} \end{equation*}

Generalized to $(r, s) \text{-tensor}$ by summing over combinations of the indices, alternating sign if wanting anti-symmetric. See the general definition of a wedge product for more for example.

Examples
  • Linear maps - matrices

    A linear map is represented as a matrix, and we say this is a tensor of order 2, since it requires 2-dimensional array to describe the relation.

Clifford algebra

First we need the following definition:

Given a commutative ring $R$ and $R$ modules M$ and $N$, an $R$ quadratic function $q: M \to N$ s.t.

  • (cube relation): For any $x, y, z \in M$ we have

    \begin{equation*} q(x + y + z) - q(x + y) - q(x + z) - q(y + z) + q(x) + q(y) + q(z) = 0 \end{equation*}

  • (homegenous of degree 2): For any $x \in M$ and any $r \in R$, we have

    \begin{equation*} q(rx) = r^2 q(x) \end{equation*}

A quadratic R-module is an $R$ module $M$ equipped with a quadratic form: an R-quadratic function on $M$ with values in $R$.

The Clifford algebra $\text{Cl}(M, q)$ of a quadratic R-module $(M, q)$ can be defined as the quotient of the tensor algebra $T_R(M)$ by the ideal generated by the relations $x \bigotimes x - q(x)$ for all $x \in M$; that is

\begin{equation*} \rm{Cl}(M, q) := T_R(M) / I(M, q) \end{equation*}

where $I_R(M, q)$ is the ideal generated by $x \bigotimes x - q(x)$.

In the case we're working with a vector space $V$ (instead of a module), we have

\begin{equation*} \rm{Cl}(V, q) := T(V) / I(V, q) \end{equation*}

Since the tensor algebra $T(V)$ is naturally $\mathbb{Z}$ graded, the Clifford algebra $\rm{Cl}(V, q)$ is naturally $\mathbb{Z} / 2 \mathbb{Z} \cong \mathbb{Z}_2$ graded.

Examples

Exterior / Grassman algebra

  • Consider a Clifford algebra $\rm{Cl}(V, q)$ generated by the quadratic form

    \begin{equation*} \begin{split} q: \quad & V \times V \to \mathbb{R} \\ & (v_1, v_2) \mapsto q(v_1, v_2) = 0 \end{split} \end{equation*}

    i.e. identically zero.

  • This apparently gives you the Exterior algebra over $V$
    • A bit confused as to why