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\begin{document}
\title{{\Large Semester 2026A, Exam A, Q1\vspace{-2em}}}

\maketitle
\begin{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 
\[
[T]_{\mathcal{C}}^{\mathcal{B}}\,[v]_{\mathcal{B}}=[T(v)]_{\mathcal{C}}
\]
\end{problem*}

\medskip{}

\rule[0.5ex]{1\columnwidth}{1pt}

\medskip{}

\begin{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

\[
=\left[a_{1}T(v_{1})+\dots a_{n}T(v_{n})\right]_{\mathcal{C}}\underset{\text{T is linear}}{=}\left[T(a_{1}v_{1}+\dots+a_{n}v_{n})\right]_{\mathcal{C}}=\left[T(v)\right]_{\mathcal{C}}
\]
\end{proof}
\begin{quote}
This is an exam problem I translated and solved. The original exam
was written by the Linear Algebra 1 course lecturers at HUJI.
\end{quote}

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