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

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?

• I think you mean “non-diagonalizable” instead of “non-diagonal.” The matrix is non-diagonal for any value of $k$. – amd Jun 8 '19 at 18:59
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 $$p_A(\lambda) = \text{det}(A - \lambda I) = p_B(\lambda) = \text{det}(B - \lambda I)$$
• Yes, correct! Take for example the zero matrix and the $2 \times 2$ matrix with all zeroes except one $1$ in the upper righthand corner. They have the same characteristic polynomial ($p(\lambda) = \lambda^2$) but cannot be similar, as they don't have the same rank. – paulinho Jun 8 '19 at 17:37