Prove there exist such that $A = [T]_\beta$ and $B = [T]_\gamma$

Question

Let $A$ and $B$ be similar $n \times n$ matrices. Prove that there exists an $n$-dimensional vector space $V$, a linear operator $T$ on $V$, and ordered bases $\beta$ and $\gamma$ for $V$ such that $A = [T]_\beta$ and $B = [T]_\gamma$.

Answer 1

The main thing here is how a change of basis acts on the representation of a linear transformation $T$.

There are few things that are often not clearly stated in textbook, so I will explicit them in a pedantic way for you just to be sure we are on the same page.

A linear transformation $T$ from a space to another space it's just a linear transformation, nothing else. Can be represented in various way, but since it is linear, when you choose a basis in vector space, you can usefully represent this linear transformation by a matrix. This means that once you choose a base on your vector space you can represent the linear transformation $T$ by a matrix $[T]$.

Now since you can pick different bases on the same vector space, namely $\beta$ and $\gamma$ in you example, then you might have different matrix representation $A = [T]_\beta$ and $B = [T]_\gamma$ of the same linear transformation $T$. I hope this thing were known to you, if they weren't you should look up again to it until they are perfectly clear.

But then everything is almost done. Because the main thing here is how does the matrix representation in the base $\beta$ change if I switch my base to the base $\gamma$?

It's really easy to show that if you have a matrix $P$ that alouds you to switch from the base $\beta$ and the base $\gamma$ then the matrix representation $A = [T]_\beta$ in the base $\beta$ and the matrix representation $B = [T]_\gamma$ in the base $\gamma$ are related by similarity i.e. $$A=P^{-1}BP$$

Now your statement becomes trivial: take a Vector space with canonical base and consider $T$ the linear operator represented by the matrix A. Then if the two matrix are similar, then there is a $P$ such that $A=P^{-1}BP$. Then the base you are looking is just the one expressed by the changing of coordinate expressed by $P$.