2
$\begingroup$

This proves: Similar matrices have the same characteristic polynomial. (Lay P277 Theorem 4)

I prefer https://math.stackexchange.com/a/8407/53259, but this proves that they have the same eigenvalues.

Are they equivalent? What about in general, even for matrices which are NOT similar?

$\endgroup$
4
$\begingroup$
  1. If $A$ and $B$ have the same characteristic polynomial, then clearly the have the same eigenvalues, these are the zeros of the characteristic polynomial.
  2. The converse is generally not true: for example $$ A=\left[\matrix{1&0&0\cr 0&0&1\cr 0&0&0}\right],\quad B=\left[\matrix{1&1&0\cr 0&1&0\cr 0&0&0}\right] $$ we have $\sigma(A)=\sigma(B)=\{0,1\}$, but $\chi_A(X)=X^2(X-1)$, $\chi_B(X)=X(X-1)^2$.
$\endgroup$
2
  • 3
    $\begingroup$ Of course, if two matrices have the same eigenvalues with the same multiplicities (and we work over an algebraically closed field) then the matrices do have the same characteristic polynomial. $\endgroup$ May 28 '14 at 11:27
  • $\begingroup$ How did you get to the statement that if A and B have the same eigenvalues and multiplicities, then they have the same characteristic polynomial? $\endgroup$
    – crazyGuy
    Nov 21 '20 at 16:39
0
$\begingroup$

"What about in general, even for matrices which are NOT similar?" You can get matrices A and B, jordan matrices, with the same characteristic polynomial and same eigenvalues and the matrices ARE NOT EQUIVALENT Jordan matrices.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.