if A diagonalizable then show that $a=0$ Let $A= \begin{pmatrix}
     2 & 0 & 0\\
     a & 2 & 0 \\
b & c & -1\\
        \end{pmatrix}$
if A diagonalizable then show $a=0$.
$P_A(x)=|xI-A|=(x-2)^2(x+1)=x^3-3x^2+4$
since A diagonalizable there is a P matrix such as $A=P^{-1}DP$
$D= \begin{pmatrix}
     2 & 0 & 0\\
     0 & 2 & 0 \\
0 & 0 & -1\\
        \end{pmatrix}$  
from cayley hamil. $A^3-3A^2+4I=0$ how do we continue?

A matrix is diagonalisable if and only if its minimal polynomial is the product of distinct factors. Since the characteristic polynomial is $(x-2)^2(x+1)$, we see that $A$ is diagonalisable if and only if the polynomial $(x-2)(x+1)$ is the minimal polynomial.
Calculating this we get 
$$\begin{pmatrix} 
0&0&0 \\
a&0&0\\
b&c&-3\\
\end{pmatrix}
\begin{pmatrix} 
3&0&0 \\
a&3&0\\
b&c&0\\
\end{pmatrix}
=
\begin{pmatrix} 
0&0&0 \\
3a&0&0\\
ac&0&0\\
\end{pmatrix}
$$
And this is the zero matrix of and only if $a=0$.
