# 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?