linear algebra - Same characteristic polynomial $iff$ same eigenvalues?

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?

Answer

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 $\chi_A(X)=X^2(X-1)$, $\chi_B(X)=X(X-1)^2$.