Matrix roots of the characteristic equation

Question

Let A be a matrix of $n \times n$ dimensions and $p( \lambda)= \det (A- \lambda I)$.

Then $p(A)=0$ by Caylee-Hamilton.

Are there any other matrices that satisfy the characteristic equation of A?

Answer 1

Amongst diagonalizable matrices, the other ones are those with a subset of eigenvalues of $A$ (not even necessarily the same multiplicity). Suppose $B$ is diagonalizable. Then there is an invertible matrix $E$ and diagonal matrix $\Lambda$ s.t. $$B = E\Lambda E^{-1}$$ Then for any polynomial we have that $$p(B) = Ep(\Lambda) E^{-1}$$ If $p$ is the characteristic polynomial of $A$ then we know that $p(\Lambda)$ is identically zero iff every eigenvalue of $B$ is an eigenvalue of $A$.

Answer 2

Correction (I wrote too fast). If $B\in M_n(K)$ is similar to $A$, then $p(B)=0$. Yet the converse is false as the following instance shows: take $A=diag(I_p,0_{n-p})$ and $B=I_n$; Then $p(\lambda)=(\lambda-1)^p\lambda^{n-p}$ and $p(B)=0$.