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

Two finite dimensional vector spaces over the same field are isomorphic iff they are of the same dimensions
Theorem : 

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

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, $U\cong V$ 

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