If the characteristic polynomial of $A,B$ is the same does it mean that $A,B$ are similar?
So I read that it's true only if $A,B$ are diagonalizable, but why? if the characteristic polynomial is the same, then $A,B$ have exactly the same eigenvalues and the same amount of each eigenvalue, so they both have the same similar diagonal matrix. What am I missing?
User John Brevik gives as a hint:
How could you have come up with this hint yourself? Well, as you say, the characteristic polynomial only cares about eigenvalues, so you need to find two matrices which are not similar but which have the same eigenvalues. If we restrict to the 2x2 case, that's equivalent to insisting that they have the same trace and determinant.
Now, you've already identified that at least one of the matrices needs not to be diagonalisable. What's the least work we could do to ensure our matrices aren't similar? If one is diagonal and one is not diagonalisable, they're instantly not similar.
OK, what's the simplest possible thing we could take for the diagonal matrix? Either the zero matrix or the identity should spring to mind. And what about our non-diagonalisable matrix? Well, it's a theorem that every complex matrix is upper-triangulable, so we can WLOG that the bottom-left element is 0; the simplest possible thing for the top-right element is 1. This results in a matrix which is in Jordan normal form, so is not diagonalisable.
The resulting pairs of matrices are either $$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ or $$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} ,\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$
both of which work as counterexamples.