# Proof :

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.
Claim : $\dim U=\dim V$
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.
Claim : $U\cong V$
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, $U\cong V$