Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\mathcal{B}$ be an ordered basis for $V$. Prove that $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is an eigenvalue of $[T]_{\mathcal{B}}$.
MY ATTEMPT
We say that $v\in V$ is an eigenvector of $T$ iff $Tv = \lambda v$ for some $\lambda\in\textbf{F}$ where $v\neq 0$.
This is equivalent to say that $Tv - \lambda v = (T - \lambda I_{V})v = 0$ and $v\neq 0$.
The last statement means that the linear operator $T - \lambda I_{V}$ is not invertible, which happens iff $[T - \lambda I_{V}]_{\mathcal{B}}$ is not invertible too.
Finally, we conclude the proposed claim, that is to say, the eigenvalues of $T$ are the roots of the polynomial
$$\det([T - \lambda I_{V}]_{\mathcal{B}}) = \det([T]_{\mathcal{B}} - \lambda I_{n}) = 0$$
which are exactly the eigenvalues of $[T]_{\mathcal{B}}$.
Any comments on my solution? Any contribution is appreciated.