Let $A$ be the complex $3 × 3$ matrix
$A = \begin{bmatrix}2&0&0 \\ x&2 & 0 \\y&z&-1\end{bmatrix}$
Find all triples $(x, y, z)$ for which the characteristic and minimal polynomials of $A $ are different.
My attempts : the Characteristic polynomial of $A = (\lambda +1)(\lambda -2)^2$
Case $1$: if $x=y= z \neq 0$ ,then minimial polynomial of $A = (\lambda +1)(\lambda -2)^2$
Case $2$ : if $x=y= z =0$,then minimial polynomial of $A = (\lambda +1)(\lambda -2)$
Here im confused that how can i find all triples $(x, y, z)$ for which the characteristic and minimal polynomials of $A $ are different.
The minimal polynomial is either $(\lambda+1)(\lambda-2)$ or $(\lambda+1)(\lambda-2)^2.$
We can try to compute $(A+I)(A-2I)$ and see whether we obtain $0$.
$$A+I=\begin{bmatrix}3&0&0 \\ x&3 & 0 \\y&z&0\end{bmatrix}, A-2I=\begin{bmatrix}0&0&0 \\ x&0 & 0 \\y&z&-3\end{bmatrix} $$
$$(A+I)(A-2I)=\begin{bmatrix} 0 & 0 & 0 \\ 3x & 0 & 0\\ xz & 0 & 0\end{bmatrix} $$
Can read off the required condition from the matrix above?
Edit:
We obtain $3x=0$ and $xz=0$.
which means $x=0$ and ($x=0$ or $z=0$)
which is equivalent to $x=0$.