# Random math: 08

Source of most content: Gunhee Cho

# Classification of VS

Def. Given two vector spaces $V,W/F$,
(1) we call $\Phi:V\rightarrow W$ a linear transform (L.T.) if $\Phi$ is a function from $V$ to $W$ and preserves a linearity, i.e., $\forall a,b \in F,\forall v,w, \in V, \ \Phi (av+bw) = a\Phi(v)+b \Phi(w)$.
(2) If $\Phi^{-1}: W \rightarrow V$ is also a linear transform, i.e., $\Phi$ is bijective, then we call $\Phi$ vector space isomorphism.

## Proposition

$\text{Given two finite dimensional VSs } V,W/F, \ dimV=dimW \text{ iff } \exists \text{ VS isomorphism } \Phi:V\rightarrow W, \ V\stackrel{\Phi}{\cong}W\nonumber$

### Proof

1. $(\Leftarrow)$ Suppose there exists a VS isomorphism $\Phi:V \rightarrow W$. Take any basis $\beta=\{v_1, \cdots, v_n \}$ of $V$. Define $\gamma :=\{\Phi(v_1), \cdots, \Phi(v_n) \} \subseteq W$.
• Let $a_1 \Phi(v_1) + \cdots + a_n \Phi(v_n) = 0 = \Phi(a_1 v_1 + \cdots + a_n v_n)$. Since $\Phi$ is injective, $\Phi(0)=\Phi(0\cdot 0) = 0\Phi(0)=0$ $\leadsto a_1 v_1 + \cdots a_n v_n =0 \Leftrightarrow a_1=a_2=\cdots=a_n=0$. So, $\gamma$ is linearly independent.
• Take any $w \in W$, then $\exists v \in V$ s.t. $\Phi(v)=w$ since $\Phi$ is surjective. Since $\beta$ is a basis, by Prop, $\exists! a_1, \cdots, a_n \in F$ s.t. $a_1 v_1 + \cdots + a_n v_n =v \leadsto w = \Phi(a_1v_1 + \cdots+ a_n v_n) \leadsto w = a_1 \Phi(v_1) + \cdots a_n \Phi(v_n)$, which implies that $w \in span(\{\Phi(v_1), \cdots, \Phi(v_n) \})$.
$\implies$ Therefore, $\gamma$ is a basis of $W$ and has same dimension by definition.
2. $(\Rightarrow)$ Suppose $dimV=dimW=n<\infty$. Take any basis $\beta=\{v_1, \cdots, v_n \}$ of $V$ and $\gamma=\{w_1, \cdots, w_n\}$ of $W$. Define $\varphi:V\rightarrow W \ (v:=a_1 v_1 + \cdots + a_n v_n \mapsto a_1 w_1 + \cdots + a_n w_n)$. Then, $\varphi$ is well-defined since if $v_1 = v_2 \rightarrow \varphi(v_1)=\varphi(v_2)$.
• Let $\varphi(v) = a_1 w_1 + \cdots + a_n w_n = \varphi(v’) = b_1 w_1 + \cdots + b_n w_n$. Then, by Prop, $\forall i = 1, \cdots n, a_i = b_i$. Therefore, $v = v’$ which implies that $\varphi$ is injective.
• Let $v=a_1 v_1 + \cdots+ a_n v_n \in$ <$\beta$>. By definition, $\varphi(a_1 v_1+\cdots+ a_n v_n)=a_1 w_1 + \cdots + a_n w_n$, which spans $W$ by definition of $\gamma$. Therefore, $\varphi$ is surjective.
• Let $\forall u, u’ \in V$, then there exists a unique expression by a basis $\beta$, i.e., $u = a_1 v_1 + \cdots + a_n v_n,\ u’= b_1 v_1 + \cdots + b_n v_n$, where $\forall i=1, \cdots, n, \ a_i, b_i \in F$. Then, $\forall x,y \in F,\ \varphi(xu+yu’)=\varphi((xa_1+yb_1)v_1 + \cdots + (xa_n+yb_n)v_n)= (xa_1+yb_1)w_1 + \cdots + (xa_n+yb_n)w_n$$= x\varphi(u)+y\varphi(u’). Therefore, \Phi:V\rightarrow W is a linear transform. \implies There exists a VS isomorphism \varphi. $dim(V/W) = dimV -dimW \text{ where } W \text{ is a subspace of } V \text{and } V/W \text{ is quotient space.} \nonumber$ ### Proof Let \alpha= \{ w_1, \cdots, w_m\} be a basis of W, i.e., dim W = m and \beta = \{ w_1, \cdots, w_m, v_1, \cdots, v_s \} be a basis of V, i.e., dim V = m+s. Then, it is sufficient to show that \gamma := \{v_1 + W, \cdots, v_s +W\} is a basis of V/W. 1. (Linearity) Let c_1(v _1 + W) + \cdots + c _s (v _s + W) = 0 _{V/W}. Then, ( c _1v _1 + \cdots+ c _s v _s) + W = 0 _{V/W} \leadsto (c_1v_1 + \cdots + c_sv_s) + W + W = W \leadsto (c_1v_1 + \cdots + c_sv_s) + W = W \Leftrightarrow w’=(c_1v_1 + \cdots+ c_sv_s) \in W. Since \alpha is a basis of W, w= (c_1v_1 + \cdots + c_sv_s) = a_1 w_1 + \cdots + a_m w_m \leadsto c_1v_1 + \cdots + c_sv_s - a_1 w_1 - \cdots - a_m w_m =0. Since \beta is a basis, this implies that c_1 = \cdots = c_s = a_1 = \cdots = a_m =0. So, \gamma is linearly independent. 2. (Span) Let x\in V/W, i.e., x= v+W, where v \in V. And v can be expressed by \beta, i.e., x = a_1 w_1 + \cdots + a_m w_m + c_1 v_1 + \cdots + c_s v_s + W. Here, a_1 w_1 + \cdots + a_m w_m \in W$$ \leadsto x = c_1 v_1 + \cdots + c_s v_s + W = c_1 v_1 + W + \cdots + c_s v_s + W = c_1 (v_1 +W) + \cdots + c_s (v_s +W)$. Therefore, $span(\gamma) = V/W.$
$\implies dim V/W = s$.

This result and 1st isomorphism theorem are very useful to prove rank-nullity theorem; $dim V/ \ker (T) = dim \text{Im} T$, i.e., $dim V = rk(T) + N(T)$, where $T:V\rightarrow V$.

## Remark

For the classification of all VS, consider $F:=\{V \mid \text{vector space } V/\mathbb{F}\}$, which is a collection of all VS. Give the relation $\sim$ on $F$ by $V \sim W$ if $\exists \text{VS isomorphism } \Phi: V \rightarrow W$. Then $\sim$ is an equivalence relation, which implies that there exists a partition of $F$ by one-to-one correspondence.

NB1; Partition of $F$ is based on the dimension of VS since the relation exists between VS which have same dimension.

### Proof

Show that $\sim$ on $F$ is the equivalent relation.

1. (Reflexivity) $dimA=dimA \Leftrightarrow A \sim A$.
2. (Symmetricity) If $A \sim B \Leftrightarrow dimA=dimB$, then $dimB =dimA \Leftrightarrow B \sim A$.
3. (Transitivity) If $A \sim B \land B \sim C \Leftrightarrow dimA=dimB \land dimB=dimC$, then $A \sim C \Leftrightarrow dimA=dimC$.

In particular, consider $\mathbb{F}^n := \mathbb{F} \times \cdots \times \mathbb{F}$, which is a vector space over $\mathbb{F}$ with operation $+$ as $(a_1, \cdots, a_n)+(b_1, \cdots, b_n)= (a_1+b_1, \cdots, a_n+b_n)$ and scalar multiplication as $\forall k \in \mathbb{F},$ $k(a_1, \cdots, a_n)=(ka_1, \cdots, ka_n)$. Then, there exists standard basis $\epsilon=\{ e_1, \cdots, e_n\}$, where $e_j$ is 0’s except j-th element=1. Then, $dim\mathbb{F}^n=n$. By proposition and above remark, for any n-dimensional VS $V/\mathbb{F}$, $V \cong \mathbb{F}^n$.
e.g.) $V=Mat_{n\times m}(\mathbb{R})/\mathbb{R} \cong \mathbb{R}^{nm}$.

# Matrix representation

Def. Let $V, W$ be vector spaces over $F$ with $dimV=n,\ dimW=m$. Let $T:V \rightarrow W$ be a linear transform and fix a basis $\beta=\{v_1, \cdots, v_n \}$ of $V$ and $\gamma = \{w_1, \cdots, w_n \}$ of $W$. Then, $T(v_k)=a_{1k}w_1 + \cdots + a_{mk}w_m, \forall k \in [1, n]$.
$\begin{bmatrix} T(v_1)& \cdots & T(v_n) \end{bmatrix} = \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \cdots & & \cdots \\ a_{m1} & \cdots & a_{mn} \end{bmatrix} := [ T ]^\gamma_\beta$ is called matrix representation of $\gamma$ w.r.t. $\beta$.

e.g.) When $\Phi: V \rightarrow \mathbb{F}^n$, then $\begin{bmatrix} \Phi(v_1)& \cdots & \Phi(v_n) \end{bmatrix}=Id_{n\times n}$

NB2; matrix representation depends on the choice of basis of $V$ and $W$.
NB3; Let $V,W$ be VS $V/F$ with $dimV=n, dimW=m$. Then, $V \stackrel{\Phi_n}{\cong} F^n$ and $W \stackrel{\Phi_m}{\cong}F^m$. Let $T:V\rightarrow W$, then $f=(\Phi_m \circ T \circ \Phi^{-1} _n )$ is a linear transformation from $F^n$ to $F^m$.

NB4; Let $\epsilon_n, \epsilon_m$ be standard basis of $F^n, F^m$. Then, $[f ]_ {\epsilon_n}^{\epsilon_m}= [ \Phi_m \circ T \circ \Phi^{-1} _n] ^{\epsilon_m} _{\epsilon_n}=[ \Phi _{m}]^\epsilon _{\gamma} [ T]^\gamma _{\beta} [\Phi^{-1} _{n} ]^{\beta} _{\epsilon _{n}}=[T ]^{\gamma} _{\beta}$.

NB5; Let $T_1: V \rightarrow W, T_2: W \rightarrow Z$ be linear transformations, i.e., $T_2 \circ T_1 : V \rightarrow Z$ is L.T.. And let $\beta$ be a basis of $V$, $\gamma$ of $W$ and $\delta$ of $Z$. Then, $[T_2 \circ T_1 ]^{\delta} _{\beta} = [ T_2]^{\delta} _{\gamma} [ T_1]^{\gamma} _{\beta}$. This can be proved by tedious calculation.

## Proposition

$\text{Given vector spaces } V,W/F, \text{ with } dimV=n, dimW=m. \text{ Let } \beta \text{ be a basis of } V \text{ and } \gamma \text{ of } W. \nonumber$ $\text{Define } \Phi: \{T: \text{L.T. } V\rightarrow W \} \rightarrow \{M \in Mat_{m \times n}(F) \} \ (T \mapsto [ T]^{\gamma} _{\beta}). \nonumber$ $\text{Then, } \Phi \text{ is bijective, i.e., there is an one-to-one correspondence between L.T. and matrices.} \nonumber$

### Proof

1. (Injective) Let $\beta = \{v_1, \cdots, v_n \}, \gamma = \{w_1, \cdots, w_n \}$ and $T_1, T_2$ be L.T.. Then, $\forall i \in [1, n],$ $T_1 (v_i) = a_{1i}w_1 + \cdots + a_{mi}w_m, T_2 (v_i)=b_{1i}w_1 + \cdots + b_{mi}w_m.$ If $T_1 \ne T_2$, then $\exists (i,j)$ s.t. $a_{ij} \ne b_{ij}$ since if $a_{ij}=b_{ij} \forall (i,j)$, this implies that $\forall v \in V, \ T_1(v) = T_2(v)$ which is a contradiction. Therefore, $[ T_1 ]^{\gamma} _{\beta} \ne [ T_2 ]^{\gamma} _{\beta}$, i.e., $\Phi(T_1) \ne \Phi(T_2)$.
2. (Surjective) Take any $(a_{ij}) \in Mat_{m \times n}(F)$. Define $T(v_1), \cdots, T(v_n)$ by $T(v_k)= a_{1k}w_1 + \cdots + a_{mk}w_m$ and extend $T$ to $V$ by linearity, i.e., $\forall v \in V, \ T(v) = T(c_1 v_1+ \cdots + c_n v_n)= c_1 T(v_1)+ \cdots + c_n T(v_n)$. Then, $T:V\rightarrow W$ is well-defined as linear transformation. By construction of $T$, $[T ]^{\gamma} _{\beta} = (a _{ij})$, which implies that $\Phi$ is surjective.

$\text{Given a vector space } V/F, \ W \subseteq V, \text{ then, } span(W) = \cap_{W \subseteq H \subseteq V} H, \text{where } H \text{ is a subspace of } V \nonumber$

### Proof

1. $(\subseteq)$ Take any $v \in span(W)$ and take any subspace $H$ in $V$ containing $W$, i.e., $W\subseteq H \subseteq V$. Write $v=a_1 v_1 + \cdots + a_n v_n,$ where $\forall i \in [1, n], \ v_i \in W, a_i \in F$. Since $H \subseteq H$ and $H$ is a vector space over $F$, $(v_i\in H)$, thus $a_1 v_1 + \cdots + a_n v_n = v \in H$. (since VS is closed in addition and scalar multiplciation)
2. $(\supseteq)$ $W \subseteq span(W) \subseteq V$

NB6; If $H$ is a vector space in $V$ containing $W$, then $span(W) \subseteq H$. Thus, $span(W)$ is smallest subspace of $V$ containing $W$.
NB7; VS $V/F \leadsto \exists$ basis $\beta$ of $V$ where $\beta$ is maximal L.I. set. Moreover, $span(\beta)=V$. This means that $V$ is the smallest vector space containing $\beta$.