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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?

Community
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2 Answers2

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  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 $\chi_A(X)=X^2(X-1)$, $\chi_B(X)=X(X-1)^2$.
Omran Kouba
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    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. – Carl Mummert May 28 '14 at 11:27
  • 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? – crazyGuy Nov 21 '20 at 16:39
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"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.

Frank Wan
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