Linear algebra - Prove that for any isomorphism there is an "identity basis(?)"

Question

I'd appreciate if you could help me with this question. What I thought about so far is showing that for any basis $B$, we know that $[T]_B$ is invertible so there are $k$ elementary matrices such that $E^k*...*E^1*[T]_B=I$.

I'm not sure if it's a good direction. Here is the question:

Let $V$ be a vector space over $F$. Prove that for any isomorphism $T : V \rightarrow F^n$, there exists a basis $B$ for $V$ such that $T=Is_B$, meaning $T(v)=[v]_B$

Note: if you see a solution I'd be happy if you could write a hint in addition to your solution, so i can try it myself first.

Also, I'm not a native english speaker so if anything is not clear please tell me and I'll fix it.

Answer 1

Hint

Let $B=\{ v_1,..., v_n \}$ be the basis you seek. Since $T(v)=[v]_B$ for all $v$ yoou must have for $v=v_i \in B$: $$T(v_i)=[v_i]_{B}$$

Now use the fact that $$[v_i]_B=(0,0,..., 0,1 ,0,..0)=e_i$$