Let $V$ and $W$ be finite dimensional vector spaces over the field $F$. Prove that $V$ is isomorphic to $W$ iff $dimV=dimW$. Problem Let $V$ and $W$ be  finite dimensional  vector spaces over the field $F$. Prove that $V$ is isomorphic to $W$ iff $\operatorname{dim}V=\operatorname{dim}W$.
\operatorname{dim}
Attempt
$\Rightarrow$ Define a linear transformation $T$ from $V$ to $W$. Suppose $V$ is isomorphic to $W$ but $\operatorname{dim}V\neq \operatorname{dim}W$.Let $\operatorname{dim}V=m$ and $\operatorname{dim}W=n$ provided $m\neq n$. If $m<n$ then $T$ is not onto and if $m>n$ then $T$ is not one-one. Contradiction ,Thus $\operatorname{dim}V=\operatorname{dim}W$.
$\Leftarrow$ Suppose $\operatorname{dim}V=\operatorname{dim}W$. Let $(a_1,...,a_n)$ and $(b_1,...,b_n)$ be basis of $V$ and $W$ respectively. Define a linear transformation $T:V\rightarrow W$ such that $T(a_i)=b_i$ ,where $1\leq i\leq n$.

*

*$T$ is injective iff $T$ sends linearly independent set to linear independent. Let $a_1,...,a_n$ be vectors in $V$ and $a\in V$, then $$a=c_1a_1+...+c_n a_n$$
therefore
$$c_1T(a_1)+...+c_n T(a_n)=0$$
$$T(c_1a_1+...+c_n a_n)=0=T(0)$$
Thus ,$c_1a_1+...+c_na_n=0$ and $c_1=...=c_n=0$. Thus the image set of $T$ linearly independent.


*$T$ is onto. Since the nullity of $T$ is $0$.


*$T$ is linear transformation: $$T(ca_i+a_j)=cb_i+b_j=cT(a_i)+T(a_j)$$.
Q.E.D.
Is the proof correct?
 A: See this is more precise.
$(\Rightarrow)$ Let $m=\operatorname{dim}(V)$ and $n=\operatorname{dim}(W)$. Let $T$ be a isomorphism from $V$ onto $W$. Since $T$ is  one one, $\operatorname{Nullity}(T)=0$. Next, since $T$ is onto, so $\operatorname{Im}(T)=W$, which implies $\operatorname{rank}(T)=n$. Therefore from Rank Nullity theorem we get, $m=n$.
$(\Leftarrow)$ Let $\{v_1,\dots,v_n\}$ and $\{w_1,\dots,w_n\}$ be bases of $V$ and $W$ respectively. Let $T:V\rightarrow W$ be the linear transformation induced by the map $v_i\mapsto w_i$ for all $i=1,2,\dots,n$. Then $T$ is an isomorphism from $V$ to $W$.
A: Proof seems OK to me. Only the lack of precise writing.
Further
For one-one, we need to prove that  $Ker T= \{0\}$ (as $T(a_i) = T(a_j)$ provided $i\neq j$) is same as $a_i-a_j \in Ker T$). Assume the contrary. Suppose $a\neq 0 \in Ker T$. Let $a = c_1 a_1 + \dots + c_n a_n$ not all $a_i$s zero.
$T(a) = T(c_1 a_1 + \dots + c_n a_n) = c_1 T(a_1) + \dots + c_n T(a_n) = c_1 b_1 + \dots + c_n b_n = 0$ which implies  $c_i = 0$ for all $1 \leq i \leq n$. Since $b_1,...,b_n$ are basis for $W$.
For more details .You can see  here
