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.

  • 1
    $\begingroup$ To answer the question in the title: yes, similar matrices (diagonalizable or not) have the same characteristic polynomial, therefore they have the same eigenvalues. $\endgroup$ – Lozenges Jun 8 '19 at 17:09
  • $\begingroup$ I think you mean “non-diagonalizable” instead of “non-diagonal.” The matrix is non-diagonal for any value of $k$. $\endgroup$ – amd Jun 8 '19 at 18:59
  • $\begingroup$ Regarding the underlying problem that you’re trying to solve, there’s a repeated eigenvalue, so you need to find a matrix with the same eigenvalues that is diagonalizable. $\endgroup$ – amd Jun 8 '19 at 19:08

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 $$p_A(\lambda) = \text{det}(A - \lambda I) = p_B(\lambda) = \text{det}(B - \lambda I)$$

  • $\begingroup$ Thank you for answering. As I understand, this is not valid in the other way around: matrices with the same characteristic are not (necessarly) similar, but they always are if they are diagonalizable. Right? $\endgroup$ – ungukla Jun 8 '19 at 17:34
  • $\begingroup$ 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. $\endgroup$ – paulinho Jun 8 '19 at 17:37

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