Are matrices only similar to one other matrix?

Question

If I have a matrix, A, that represents a linear transformation in terms of fixed bases for the source and the target. And then I consider another set of bases for the source and the target which would have another associated matrix, B, and it just so happened to be that A and B are similar - would it be that B is the only matrix that is similar to A?

Like I understand that you can have many more matrices that are equivalent to A simply just by considering different pairs of bases, but would you be able to find another one that is similar to A? If you could, would it also be similar to B?

Answer 1

Here is an illustration of a counterexample for particular $2\times 2$ matrices.

Let $A=\begin{pmatrix}w&x\\y&z\end{pmatrix}$, and consider $P=\begin{pmatrix}1&b\\0&1\end{pmatrix}$. Note $P^{-1}=\begin{pmatrix}1&-b\\0&1\end{pmatrix}$, and

$$B= P^{-1}AP=\begin{pmatrix}w-by&x-bz\\y&z\end{pmatrix}\begin{pmatrix}1&b\\0&1\end{pmatrix}$$ $$=\begin{pmatrix}w-by&bw+x-b^2y- bz\\ y&by+z\end{pmatrix}$$ is similar to $A$. It should be clear that by choosing different values of $b$, you get different matrices $B$ (each of which is similar to $A$) unless $A$ is very special. The only time you have $A=B$ for every choice of $b$ is when $y=0$ and $w=z$, i.e., when $A=\begin{pmatrix}w&x\\0&w\end{pmatrix}$.

Answer 2

If $A$ and $B$ are matrix representations of a transformation $T$ under bases $\mathcal{A}$ and $\mathcal{B}$, then $A$ is similar to $B$ (so $A = MBM^{-1}$). You can find the similarity matrix $M$ by taking the appropriate change-of-basis matrix.

Whether or not you can find two $\textbf{different}$ matrices $A$ and $B$ depends on the transformation: for example the zero transformation only has one matrix, no matter the basis.