True/false questions about minimal and characteristic polynomials of a matrix

Question

We have the matrix $A= \begin{pmatrix} 0 &2 &2 \\ 2& 0 &2 \\ 2& 2 & 0 \end{pmatrix}$, then one of the following is true:

1. $f_A(x)=m_A(x) $

2. The matrix $C=A^4-4A^2+5I$ isn't diagnolizable over $\mathbb R$

3. For the matrix $B=A^5+6A^3+A^2-I$ we have $f_B(x)\neq m_B(x)$

4. For the matrix $B=A^5+6A^3+A^2-I$ we have $f_B(x)$ is a simple polynomial over $\mathbb C$

EDIT: I get that $f_A(x) = x(2+x)(2-x)$ and it looks like I can't zero the matrix with any smaller combination than $f_A$, so it's maybe true.

For 2 I get that eigenvalues of $A$ aren't zeroing $x^4-4x^2=x^2(x^2-4)=-5$, I'm not sure what to make of it though.

For 3, we have $f_B(x) = x^2(x^3+6x+1)-1$ (?) that can't be a minimal polynomial so that's maybe true.

For 4, I don't think it's true for the same reason as 3.

Notes: $f_A(x)$ is the characteristic polynomial of $A$, $m_A(x)$ is the minimal polynomial of A, a "simple" polynomial: $x(x+1)$, not "simple" poly': $x^2(x+3)$

Answer 1

The matrix $A$ is real and symmetric, so it must be diagonalisable. Therefore the minimal polynomial (supposedly called $m_A$) has simple roots.

The matrix $A+2I$ has rank $1$ and trace $6$, so its characteristic polynomial is $X^2(X-6)$, and the characteristic polynomial of $A$ (supposedly called $f_A$) is obtained from it by substituting $X+2$ for $X$: $f_A=(X+2)^2(X-4)$ (equivalently, you can see directly that $\lambda=-2$ is an eigenvalue with geometric multiplicity$~2$, and the remaining eigenvalue must be $4$ to make their sum equal to $0$, the trace of$~A$). Since this has a double root, point 1 must fail. In fact $m_A=(X+2)(X-4)$ is now forced.

Any polynomial of a diagonalisable matrix is diagonalisable (on the same basis of eigenvectors), so point 2 must be false as well.

In fact, there is a eigenspace for $A$ of dimension$~2$ (for eigenvalue $\lambda=-2$ of $A$), and this is contained in the eigenspace of any polynomial $P[A]$ of$~A$ (for the eigenvalue $\lambda=P[-2]$ of $P[A]$), so $f_{P[A]}$ always has a multiple root, and can never be equal to $m_{P[A]}$ (for the same reason as for $f_A$ and $m_A$ above). So point 3 is true and point 4 is false, independently of the precise polynomial $P=X^5+6X^3+X^2-1$ used there.