Show $T$ is an invertible linear map iff $\operatorname{rank} T = \dim V$ Lemma: Let $W$ be a subspace of $V$. $\dim V= \dim W \iff V=W$.
Proof:
If $V=W$ then clearly $\dim V=\dim W$.
Assume $\dim V=\dim W$. Let $B= \{w_1,...…,w_n\}$ be a basis for $W$ then since $W \subset V$ it is a set of $n$ linearly independent vectors in $V$ hence $\operatorname{span} B=V=W$.
Now, WTS: Let $V, W$ be finite dimensional vector spaces of dimension $n$. $T: V \rightarrow W$ is an invertible linear map $\iff$ $\operatorname{rank} T=\dim V$.
My Proof: Assume $T: V \rightarrow W$ is an invertible linear map. Therefore by definition it is bijective, hence injective, and so $\ker T = \{ \textbf{0} \} \implies \dim \ker T = 0$. Therefore by the rank nullity theorem, $\dim V= \dim(\operatorname{im} T)$.
Assume $\operatorname{rank}(T) =\dim V \implies \dim \ker(T)=0$ so $\ker T = \{\textbf{0}\}$. Hence the linear transformation is injective. Since $\dim(\operatorname{im} T)=\dim V=\dim W$ and $\operatorname{im} T \subset W$ it follows that from the lemma above that $\operatorname{im}T=W$, and so the linear map is surjective. Since the map is both surjective and injective it is bijective, therefore invertible.
Is the proof good? May someone tell me how to improve this proof, please?


Is the proof good? May someone tell me how to improve this proof, please?

I read you proof and it looks completely correct.
The only stylistic thing I would suggest is that you make it clearer what you are showing. You started out with a lemma and proved it before you even stated the question! And you only stated the question as "WTS". So you should put at the top of your post: "Question: I am supposed to show X Y Z. First I prove the following lemma."
