Does a square matrix satisfy only its characteristic equation?

Question

Say that we have a square matrix of order $N \times N$ and is NOT a diagonal matrix. Now can this matrix satisfy a polynomial of degree $N$ other than its own characteristic equation?

Answer 1

For sure, for example let $M$ be any $n \times n$ matrix such that $M \neq 0$ but $M^2 = 0$, for example we could take $M$ to be a matrix full of zeros, with a single 1 in the top-right spot. The characteristic equation of $M$ is $x^n$, but $M$ satisfies many more polynomials of degree $n$ as long as $n > 2$, for example it satisfies $M^2(M + 1) = 0$.

In general, this happens whenever the minimal polynomial of a matrix is a lower degree than the characteristic polynomial.

Answer 2

Yesx it can. Consider

$$M = \begin{pmatrix}1 & 1 & 0 \\ 0&1&0 \\ 0&0&1\end{pmatrix}.$$

Clearly, the characteristic polynomial of $M$ is a cubic. However, also note that $(M - I)^2 = O$.

Thus, $M$ satisfies any cubic polynomial of the form

$$(x - 1)^2(x - a).$$

Almost all values of $a \in \Bbb R$ will give you a cubic which is different from $M$'s characteristic polynomial