= Change-of-basis matrix
{title2=$[z]_{\rm old}=P[z]_{\rm new}$}
Fix the convention: the columns of the <change-of-basis matrix> $P$ are the new <basis vectors> written in the old <basis>. Thus $P$ converts new <coordinate vectors> into old ones; its <matrix inverse> converts old <coordinate vectors> into new ones. For the preceding <basis>,
$$
P=\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad
P^{-1}=\frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad
P^{-1}\begin{pmatrix}2\\0\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}.
$$
Its <determinant> is $-2$, so $P$ is an <invertible matrix>. Checking which direction $P$ converts is essential: different books may call either $P$ or $P^{-1}$ the <change-of-basis matrix>.
For a <linear operator>, convert the input into the old <basis>, apply the old <matrix>, and convert the output back. Reading that sequence from right to left gives
$$
A_{\rm new}=P^{-1}A_{\rm old}P.
$$
For example, let the old <diagonal matrix> stretch horizontal components by $2$ and vertical components by $3$. Then
$$
A_{\rm old}=\begin{pmatrix}2&0\\0&3\end{pmatrix},\qquad
A_{\rm new}=\begin{pmatrix}2.5&-0.5\\-0.5&2.5\end{pmatrix},\qquad
A_{\rm new}\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}2\\2\end{pmatrix}.
$$
The output represents $2u+2v=(4,0)$, just as $A_{\rm old}(2,0)^T=(4,0)^T$. The <matrices> look different because their <bases> differ, but they encode the same <linear operator>. This relation is called <matrix similarity>; it preserves the underlying action.https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/46108efb43dcf33e989eb5e101fcf9b5_MIT18_06S10_L31.pdf[Strang, Lecture 31 transcript: coordinate conversion and similar matrices]{ref}
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