## Two finite dimensional vector spaces over the same field are isomorphic iff they are of the same dimensions.

# Proof :

__Claim__: $\dim U=\dim V$

__Claim__: $U\cong V$

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Friday, February 3, 2023

Theorem :

Let $U$ and $V$ be two finite dimensional real vector spaces which are isomorphic.

i.e. $\exists$ a function $f:U\to V$ which is one-one, onto and linear transformation.

Consider, $\dim U=n$

Let $S=\{\alpha_1, \alpha_2, … , \alpha_n\}$ be a basis of U.

We prove that $S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$ is a basis of V.

For this first we prove that, $S’$ is linearly independent.

Consider,

$a_1f(\alpha_1)+a_2f(\alpha_2)+ … +a_nf(\alpha_n)=0$, where $a_i\in \mathbb{R}$

$\implies f(a_1\alpha_1)+f(a_2\alpha_2)+ … +f(a_n\alpha_n)=0$, ($\because$ f is linear)

$\implies f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)=0$, ($\because$ f is linear)

$\implies a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n=0$, ($\because$ f is linear and one-one)

Since, $S=\{\alpha_1, \alpha_2, … , \alpha_n\}$ is a basis of U,

$\implies a_1=a_2= … =a_n=0$

$\therefore S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$ is linearly independent. ……. (1)

Now, to prove that, $S’=\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$ spans V.

i.e. to prove that, every vector in $V$ can be expressed as a linear combination of $\{f(\alpha_1), f(\alpha_2), … , f(\alpha_n)\}$.

Let $v$ be any arbitrary element of $V$.

As $f:U\to V$ is onto, for $v\in V, \exists \ \alpha\in U$ such that $f(\alpha)=v$.

As $\alpha\in U$ and $S=\{\alpha_1, \alpha_2, … , \alpha_n\}$ is a basis of U, $\therefore \exists \ a_1, a_2, , … , a_n\in \mathbb{R}$ such that

$\alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n$

Now, $v=f(\alpha)$

$=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$

$=a_1f(\alpha_1)+a_2f(\alpha_2)+ … +a_nf(\alpha_n)=0$, ($\because$ f is linear)

$\therefore$ every vector in $V$ can be expressed as a linear combination of elements in $S’$.

$\therefore L(S’)=V$ …….. (2)

From (1) and (2),

The set $S’$ forms basis of V.

$\therefore \dim V=n$

Hence, $\dim U=n=\dim V$

Conversely,

Suppose that, $\dim U=n=\dim V$

Let $B=\{\alpha_1, \alpha_2, … , \alpha_n\}$ be a basis of U and $B’=\{\beta_1, \beta_2, … , \beta_n\}$ be a basis of V.

As $\alpha \in U$ and $\{\alpha_1, \alpha_2, … , \alpha_n\}$ is a basis of U,

$\alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n$ , $\forall \ a_i\in \mathbb{R}$

For this $\alpha$, define $f:U\to V$ as $f(\alpha)=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n$, $\forall \ \alpha\in U$

$f(\alpha)=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$

$=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n$

Let $\alpha, \beta$ be any two vectors in $U$ and $\{\alpha_1, \alpha_2, … , \alpha_n\}$ is a basis of U.

$\therefore \alpha$ and $\beta$ can be expressed as linear combination of $\{\alpha_1, \alpha_2, … , \alpha_n\}$.

Let $\alpha=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n$ and $\beta=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n$, where $a_i, b_i \in \mathbb{R}$

To prove that, f is well defined.

Let $\alpha=\beta$

$\implies a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n$

$\implies (a_1-b_1)\alpha_1+(a_2-b_2)\alpha_2+ … +(a_n-b_n)\alpha_n=0$

$\implies a_1-b_1=0, a_2-b_2=0, … , a_n-b_n=0$,

($\because \{\alpha_1, \alpha_2, … , \alpha_n\}$ is linearly independent)

$\implies a_1=b_1, a_2=b_2, … , a_n=b_n$

$\therefore f(\alpha)=f(a_1\alpha_1+ … +a_n\alpha_n)$

$=f(b_1\alpha_1+ … +b_n\alpha_n)$

$=f(\beta)$

$\therefore$ f is well defined.

Now, to prove that f is one-one.

Consider, $f(\alpha)=f(\beta)$

$\implies f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)=f(b_1\alpha_1+b_2\alpha_2+ … b_n\alpha_n)$

$\implies a_1\beta_1+a_2\beta_2+ … +a_n\beta_n=b_1\beta_1+b_2\beta_2+ … +b_n\beta_n$

$\implies (a_1-b_1)\beta_1+(a_2-b_2)\beta_2+ … +(a_n-b_n)\beta_n=0$

$\implies a_1-b_1=0, a_2-b_2=0, … , a_n-b_n=0$

$\implies a_1=b_1, a_2=b_2, … , a_n=b_n$

$\implies \alpha=\beta$

$\therefore$ f is one-one.

Now, to prove that f is onto.

Consider, $v\in V, \implies v=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n$, for some $a_i\in \mathbb{R}$

Let $u=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n\in U$

$\therefore f(u)=f(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$

$=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n$

$=v$

$\therefore$ f is onto.

Now, to prove that, f is linear transformation.

Let $x, y$ be any arbitrary elements of $U$.

To prove that, $f(x+y)=f(x)+f(y)$

As $\{\alpha_1, \alpha_2, … , \alpha_n\}$ is a basis of $U$,

$x=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n$ and $y=b_1\alpha_1+b_2\alpha_2+ … +b_n\alpha_n$, where $a_i, b_i \in \mathbb{R}$

$\therefore x+y=(a_1+b_1)\alpha_1+(a_2+b_2)\alpha_2+ … +(a_n+b_n)\alpha_n$

$\implies f(x+y)$

$=f[(a_1+b_1)\alpha_1+(a_2+b_2)\alpha_2+ … +(a_n+b_n)\alpha_n]$

$=(a_1+b_1)\beta_1+(a_2+b_2)\beta_2+ … +(a_n+b_n)\beta_n$

$=a_1\beta_1+b_1\beta_1+a_2\beta_2+b_2\beta_2+ … +a_n\beta_n+b_n\beta_n$

$=a_1\beta_1+a_2\beta_2+ … +a_n\beta_n+b_1\beta_1+b_2\beta_2+ … +b_n\beta_n$

$=f(x)+f(y)$

Now, to prove that, $f(cx)=cf(x)$

Consider, $c\in \mathbb{R}$ and $x=a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n\in U$

$\therefore f(cx)=f[c(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)]$

$=f(ca_1\alpha_1+ca_2\alpha_2+ … +ca_n\alpha_n)$

$=ca_1\beta_1+ca_2\beta_2+ … +ca_n\beta_n$

$=c(a_1\beta_1+a_2\beta_2+ … +a_n\beta_n)$

$=cf(a_1\alpha_1+a_2\alpha_2+ … +a_n\alpha_n)$

$=cf(x)$

Hence, $U\cong V$

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