Let $T$ be a linear operator on a finite-dimensional vector space $V$. We define the determinant of $T$, denoted $\det(T)$, as follows: choose any ordered basis $\mathcal{B}$ for $V$, and define $\det(T) = \det([T]_{\mathcal{B}})$.
(a) Prove that if $\mathcal{B}$ and $\mathcal{B}'$ are two ordered bases for $V$, then $\det([T]_{\mathcal{B}}) = \det([T]_{\mathcal{B}'})$.
(b) Prove that $T$ is invertible if and only if $\det(T)\neq 0$.
(c) Prove that if $T$ is invertible, then $\det(T^{-1}) = [\det(T)]^{-1}$.
(d) Prove that if $U$ is also a linear operator on $V$, then $\det(TU)= \det(T)\det(U)$.
(e) Prove that $\det(T-\lambda I_{V}) = \det([T]_{\mathcal{B}} - \lambda I)$ for any scalar $\lambda$ and any ordered basis $\mathcal{B}$ for $V$.
MY ATTEMPT
(a) It suffices to know that $[T]_{\mathcal{B}}$ and $[T]_{\mathcal{B}'}$ are similar. More precisely, $[T]_{B} = [I]_{\mathcal{B}'}^{\mathcal{B}}[T]_{\mathcal{B}'}[I]_{\mathcal{B}}^{\mathcal{B}'}$, we conclude that \begin{align*} \det([T]_{\mathcal{B}}) = \det([I]_{\mathcal{B}'}^{\mathcal{B}}[T]_{\mathcal{B}'}[I]_{\mathcal{B}}^{\mathcal{B}'}) = \det([I]_{\mathcal{B}'}^{\mathcal{B}})\det([T]_{\mathcal{B}'})\det([I]_{\mathcal{B}}^{\mathcal{B}'}) = \det([T]_{\mathcal{B}'}) \end{align*}
(b) It results from the fact that $T$ is invertible iff $[T]_{\mathcal{B}}$ is invertible iff $\det([T]_{\mathcal{B}})\neq 0$.
(c) If $T$ is invertible, then $T^{-1}T = TT^{-1} = I_{V}$. Moreover, we do also have \begin{align*} [T^{-1}T]_{\mathcal{B}} = [T^{-1}]_{\mathcal{B}}[T]_{\mathcal{B}} = I & \Rightarrow \det([T^{-1}]_{\mathcal{B}})\det([T]_{\mathcal{B}}) = 1\\\\ & \Rightarrow \det([T^{-1}]_{\mathcal{B}}) = [\det([T]_{\mathcal{B}})]^{-1}\\\\ & \Rightarrow \det(T^{-1}) = [\det(T)]^{-1} \end{align*}
(d) Similarly, we do also have that \begin{align*} \det(TU) = \det([TU]_{\mathcal{B}}) = \det([T]_{\mathcal{B}}[U]_{\mathcal{B}}) = \det([T]_{\mathcal{B}})\det([U]_{\mathcal{B}}) = \det(T)\det(U) \end{align*}
(e) Finally, we conclude that \begin{align*} \det(T - \lambda I_{V}) = \det([T - \lambda I_{V}]_{\mathcal{B}}) = \det([T]_{\mathcal{B}} - \lambda I) \end{align*}
I am mainly concerned with the wording of my proofs. Any contribution is appreciated.
The proof looks good! It's elegantly worded, highlighting the important steps, without being unnecessarily wordy. There's a good balance between mathematical exposition and use of symbols, and the symbols and exposition fit together in such a way that one can read the entire proof (symbols and words) as sentences that are grammatically sensible (which is not a particularly common skill in mathematics students at your level).
On the topic of grammar, I would suggest that you put punctuation at the end of your centre-aligned chunks of mathematics. You should consider them as part of the sentence preceding them, and you should end the sentence with the appropriate punctuation mark (here, some periods would do). This is, of course, a very minor note!
Another thing you could consider doing is citing the results you're using. You can refer to the lectures, to the textbook, or wherever in the course you've discovered these results. Things like $\det(AB) = \det(A)\det(B)$ will have been either proven or at least explicitly stated and assumed in the course somewhere, and it's a good habit to get into to refer the reader to these results when they are first used. I wouldn't say that this change is necessary though.
In terms of mathematical content, I would suggest adding a little bit to part (a). You talk about $[T]_{\mathcal{B}}$ and $[T]_{\mathcal{B}'}$ being similar, as $[T]_{B} = [I]_{\mathcal{B}'}^{\mathcal{B}}[T]_{\mathcal{B}'}[I]_{\mathcal{B}}^{\mathcal{B}'}$. It strikes me as an omission here not to note that $[I]_{\mathcal{B}'}^{\mathcal{B}}$ and $[I]_{\mathcal{B}}^{\mathcal{B}'}$ are inverses. It seems like a point worth including, since you'll be using another result: $\left([A]_{\mathcal{B}}^{\mathcal{B}'}\right)^{-1} = [A^{-1}]_{\mathcal{B}'}^{\mathcal{B}}$, and the fact that $I^{-1} = I$.
Hope that helps!