Problem. Let \(V,W\) be vector spaces, let \(T\colon V\to W\) be a linear map, let \(\mathcal{B},\mathcal{C}\) be bases of \(V\) and \(W\) respectively, and let \(v\in V\). Prove that \[[T]_{\mathcal{C}}^{\mathcal{B}}\,[v]_{\mathcal{B}}=[T(v)]_{\mathcal{C}}\]
Proof. Let us mark \(\mathcal{B}=\left(v_{1},\dots,v_{n}\right)\), \(\mathcal{C}=\left(w_{1},\dots,w_{m}\right)\) and \(v=a_{1}v_{1}+\dots+a_{n}v_{n}\).
\[[T]_{\mathcal{C}}^{\mathcal{B}}\,[v]_{\mathcal{B}}=\begin{pmatrix}| & & |\\ \left[T(v_{1})\right]_{\mathcal{C}} & \dots & \left[T(v_{n})\right]_{\mathcal{C}}\\ | & & | \end{pmatrix}\,\begin{pmatrix}a_{1}\\ \vdots\\ a_{n} \end{pmatrix}=a_{1}\left[T(v_{1})\right]_{\mathcal{C}}+\dots+a_{n}\left[T(v_{n})\right]_{\mathcal{C}}\]
Let us recall that \(\left[C_{1}\right]_{\mathcal{D}}+\left[C_{2}\right]_{\mathcal{D}}=\left[C_{1}+C_{2}\right]_{\mathcal{D}}\) and that \(a_{1}\left[C_{1}\right]_{\mathcal{D}}=\left[a_{1}C_{1}\right]_{\mathcal{D}}\). Thus
\[=\left[a_{1}T(v_{1})+\dots a_{n}T(v_{n})\right]_{\mathcal{C}}\underset{\text{T is linear}}{=}\left[T(a_{1}v_{1}+\dots+a_{n}v_{n})\right]_{\mathcal{C}}=\left[T(v)\right]_{\mathcal{C}}\] ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.