Suppose that $M$ is a $5 \times 5$ matrix with real entries and $p(x)= \text{det}( xI - M)$, then which of the following is true?
- $p(0)= \text{det}(M)$
- Every eigenvalue of $M$ is real if $p(1)+ p(2) = 0 = p(2)+ p(3).$
- $M^{-1}$ is necessarily a polynomial in $M$ of degree $4$ if $M$ is invertible.
- $M$ is not invertible if $M^2- 2M= 0.$
Option $1$ is false for $M= I_{5}$ , since then $p(0)= -1$ but $\text{det}(M)= \text{det} (I_{5})= 1.$ Also for $M= 2I$, $4$ is false. $3$ is true by Cayley- Hamilton theorem, but I'm unable to construct a counter- example for $2$. Any suggestions for that?
4 is $\color{red}{\text{false}}$, since if $M(M-2)=0$, then $x(x-2)$ is the minimal polynomial of $M$ and since the characteristic polynomial must be a multiple of $x(x-2)$ and therefore at least one of the eigenvalues is zero which means $M$ is not invertible ($M$ is singuar), $\color{red}{\text{unless }M=2I}$ and that's why this statement is false, the only exception is $M=2I$.
2 is $\color{red}{\text{false}}$, since $p(x)=x^5+bx^4+cx^3+dx^2+ex+f$ is a monic polynomial of degree 5. with the given conditions we get two equations
$$\begin{cases}17a+9b+5c+3d+2e+33=0 \\ 97a+35b+13c+5d+2e+275=0\end{cases}$$
by solving for $(d,e)$ and substituting in the characteristic polynomial $p(x)$ we get $$p(x)=x^5+ax^4+bx^3+cx^2+(-4c-13b-40a-121)x+(\frac{7}{2}c+15b+\frac{103}{2}a+165)$$ This equation is very general and by a simple substitution of $a=b=c=0$ we get the polynomial $p(x)=x^5-121x+165$ which has two complex roots, so in general the second statement is false.