Do similar, non-diagonalizable matrices have the same eigenvalues?

Question

To better explain, I have the matrix \begin{bmatrix}3&2&0\\5&0&0\\k&b&-2\end{bmatrix} where k is chosen to make the matrix non-diagonal. I have to find, if possible, a matrix with the same eigenvalues which is not similar to this one, but I can't seem to find it.
Is it only this particular case, or in general non-diagonalizable matrices that are not similar have different eigenvalues?

Thanks in advance.

Answer 1

Lozenges is correct, and here is the proof of why. Let $A$ and $B$ be similar matrices, i.e. $A = S B S^{-1}$ for some invertible matrix $S$. The characteristic polynomial of $A$ is $$p_A(\lambda) = \text{det}(A - \lambda I) = \text{det}(S B S^{-1} - S (\lambda I) S^{-1})=\text{det}(S)\text{det}(B - \lambda I) \text{det}(S^{-1})$$ but $\text{det}(S)$ and $\text{det}(S^{-1})$ are inverses (this is a property of pairs of inverse matrices). This implies $$p_A(\lambda) = \text{det}(A - \lambda I) = p_B(\lambda) = \text{det}(B - \lambda I)$$