Chapter 1 extra note 7

isomorphism; isomorphic
subspace
dimension of a vector space
replacement theorem

Selected lecture notes:

Isomorphic

Definition:
An invertible linear transformation is called an isomorphism.

Definition:
Two vector spaces

$V$ and

$W$ are called isomorphic (

$V\cong W$) if

$\mathrm{\exists }$ an isomorphism

$T:V\to W$.

Theorem:
Let

$T:V\to W$ be an isomorphism and

$\{v_1,\cdots ,v_n\}$ is a basis of

$V$, then

$\{T(v_1),\cdots ,T(v_n)\}$ is a basis of

$W$.

• Proof:

  claim 1:

  $\{T(v_1),\cdots ,T(v_n)\}$ is linearly independent

  pf:
  

  $\because$

  $T$ is an isomorphism,

  $\mathrm{\exists }{T}^{-1}:W\to V$.
  Let us assume
  

  $$c_{1}T(v_1)+\cdots +c_{n}T(v_n)=0.$$
  By applying

  $T^{-1}$ to both side of the equation, the right hand side gives

  $T^{-1}(0)=0$, while the left hand side gives
  

  $$T^{-1}(c_{1}T(v_1)+\cdots +c_{n}T(v_n))=c_1{v}_1+\cdots c_n{v}_n,$$
  where we have used the fact that

  $T^{-1}$ is linear.

  Since

  $\{v_1,\cdots ,v_n\}$ is a basis, it is linearly independent, so

  $c_1=\cdots =c_n=0$.
  Therefore,

  $\{T(v_1),\cdots ,T(v_n)\}$ is linearly independent.

  claim 2:

  $\{T(v_1),\cdots ,T(v_n)\}$ is generating

  pf:
  Given

  $w\in W$,

  $T^{-1}(w)\in V$.
  Since

  $\{v_1,\cdots ,v_n\}$ is a basis, there exists

  $c_1,\cdots ,c_n$ such that
  

  $$T^{-1}(w)=c_1{v}_1+\cdots c_n{v}_n.$$
  By applying

  $T$ to both side we have
  

  $$w=c_{1}T(v_1)+\cdots c_{n}T(v_n).$$
  So that

  $w$ can be written as a linear combination of

  $\{T(v_1),\cdots ,T(v_n)\}$.

Corollary:
Let

$T:V\to W$ be linear and

$\{v_1,\cdots ,v_n\}$ be a basis of

$V$, if

$\{T(v_1),\cdots ,T(v_n)\}$ is not a basis of

$W$, then

$T$ is not invertible.

Theorem:
Let

$V,W$ be vector spaces with bases

$\beta =\{v_1,\cdots ,v_n\}$ and

$\gamma =\{w_1,\cdots ,w_n\}$, respectively. We define

$T:V\to W$ by

$$T(c_1{v}_1+\cdots c_n{v}_n)=c_1{w}_1+\cdots c_n{w}_n.$$
Then

$T$ is an isomorphism.

• Proof:

  $T$ is clearly linear.

  Define

  $S:W\to V$ by
  

  $$S(c_1{w}_1+\cdots c_n{w}_n)=c_1{v}_1+\cdots c_n{v}_n$$
  Since

  $\gamma$ is a basis,

  $S$ is well-defined.
  Also we have

  $T\circ S=I_w$ and

  $S\circ T=I_v$.
  Therefore,

  $T$ is both right and left invertible, and is thus an isomorphism.

Remark:

$$[T{]}_{\beta }^{\gamma }=I_n:\text{the }n\times n\text{ identity matrix}.$$

---

Vector subspace

Definition:
Let

$V_0\subseteq V$ where

$V$ is a vector space.

$V_0$ is a vector subspace is it is a vector space under the same field, vector addition and multiplication.

Theorem:
Let

$V$ be a vector space and

$V_0\subseteq V$, then

$V_0$ is a vector subspace if and only if

$\alpha \mathbf{u}+\beta \mathbf{v}\in V_0$ for all

$\alpha ,\beta \in F$ and

$\mathbf{u}\mathbf{,}\mathbf{v}\in V_0$.

Example 1

Let

$T:V\to W$ be a linear transformation,

$\text{Ker}(T)\subseteq V$ is a subspace.

• Proof:

  Given

  $\mathbf{u}\mathbf{,}\mathbf{v}\in \text{Ker}(T)$,

  $\alpha ,\beta \in F$, we have

  $T(\mathbf{u})=\mathbf{0}$ and

  $T(\mathbf{v})=\mathbf{0}$.
  Since

  $T$ is linear and

  $V$ is a vector space,
  

  $$T(\alpha \mathbf{u}+\beta \mathbf{v})=\alpha T(\mathbf{u})+\beta T(\mathbf{v})=\alpha \cdot \mathbf{0}+\beta \cdot \mathbf{0}=\mathbf{0}.$$
  So

  $\alpha \mathbf{u}+\beta \mathbf{v}\in \text{Ker}(T)$.

Example 2

Let

$T:V\to W$ be a linear transformation,

$\text{Ran}(T)\subseteq W$ is a subspace.

---

Dimension of a vector space

Theorem (replacement theorem):
Let

$V$ be a vector space,

$G=\{v_1,\cdots ,v_n\},n\ge 1$ be a generating set of

$V$. Let

$L=\{w_1,\cdots ,w_m\}\subset V$ be linearly independent, then

$m\le n$ and

$\mathrm{\exists }H\subset G$ that has exactly

$n-m$ vectors such that

$H\cup L$ is a generating set.

• Proof:

  We prove by induction

  • $k=1$:

  $L=\{w_1\}$ is linearly independent. Then

  $k=1\le n$.
  Since

  $G$ is generating, we have,
  

  $$w_1=c_1{v}_1+\cdots c_n{v}_n.$$
  claim:

  $c_1,\cdots ,c_n$ not all zero

  pf:
  If

  $c_1=c_2=\cdots =c_n=0$, then

  $w_1=0$ and

  $\{w_1\}$ is not linearly independent.
  So

  $c_1,\cdots ,c_n$ can not be all zero.

  WLOG assume

  $c_1\ne 0$, then
  

  $$v_1=\frac{1}{c_1}{w}_1-\frac{c_2}{c_1}{v}_2-\cdots -\frac{c_n}{c_1}{v}_n.$$
  That is,

  $v_1\in \text{span}\{w_1,v_2,\cdots ,v_n\}$.
  It is then clear that

  $\text{span}\{v_1,\cdots ,v_n\}\subseteq \text{span}\{w_1,v_2,\cdots ,v_n\}$, so

  $\{w_1,v_2,\cdots ,v_n\}$ is a generating set.

  • Assume
    $k=m$ is true. Let's check
    $k=m+1$:

  Let

  $L=\{w_1,\cdots ,w_{m+1}\}\subset V$ be linearly independent, then

  $\tilde{L}=\{w_1,\cdots ,w_m\}\subset V$ is also linearly independent.
  We then have

  $m\le n$ and

  $\mathrm{\exists }\tilde{H}\subset G$ that has exactly

  $n-m$ vectors such that

  $\tilde{H}\cup \tilde{L}$ is a generating set. WLOG, we may assume

  $\tilde{H}=\{v_{m+1},\cdots ,v_n\}$.
  Since

  $\tilde{H}\cup \tilde{L}$ is a generating set, we have
  

  $$w_{m+1}=c_1{w}_1+\cdots c_m{w}_m+c_{m+1}{v}_{m+1}+\cdots c_n{v}_n.$$
  claim:

  $c_{m+1},\cdots ,c_n$ not all zero

  pf:
  If

  $c_{m+1}=c_{m+2}=\cdots =c_n=0$, then
  

  $$c_1{w}_1+\cdots c_m{w}_m-w_{m+1}=0.$$
  There exists non-trivial linear combination to

  $0$ and

  $\{w_1,\cdots ,w_{m+1}\}$ is not linearly independent.
  So

  $c_{m+1},\cdots ,c_n$ can not be all zero.

  A direct consequence of this claim is that

  $\tilde{H}$ can not be an empty set, that is,

  $n-m>0$, so

  $m+1\le n$.

  Since

  $c_{m+1},\cdots ,c_n$ can not be all zero, WLOG assume

  $c_{m+1}\ne 0$, we have
  

  $$v_{m+1}\in \text{span}\{w_1,\cdots ,w_{m+1},v_{m+2},\cdots ,v_n\}.$$
  Let us define

  $H=\{v_{m+2},\cdots ,v_n\}$, it is then clear that

  $\text{span}\{\tilde{H}\cup \tilde{L}\}\subseteq \text{span}\{H\cup L\}$, so

  $H\cup L$ is a generating set.

Corollary:
If

$\{v_1,\cdots ,v_n\}$ and

$\{w_1,\cdots ,w_m\}$ are both bases of a vector space

$V$, then

$n=m$.

• Proof:

  $\{v_1,\cdots ,v_n\}$ is a generating set and

  $\{w_1,\cdots ,w_m\}$ is linearly independent, so

  $m\le n$.
  Similarly,

  $\{w_1,\cdots ,w_m\}$ is a generating set and

  $\{v_1,\cdots ,v_n\}$ is linearly independent, so

  $n\le m$.
  Therefore,

  $n=m$.

Definition:
Let

$V$ be a vector space and

$\{v_1,\cdots ,v_n\}\subset V$ be a basis, we define

$\text{dim}(V)=n$.