<s>[INST] 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?
 [/INST] Check whether there are 2 different eigenvectors to $2$. An $2$-eigenvector $v$ satisfies 
$$
\begin{pmatrix}
0 & 0 & 0\\
a & 0 & 0 \\
b & c & -3
\end{pmatrix}
\begin{pmatrix}
v_1 \\ v_2 \\ v_3
\end{pmatrix}=0.
$$
This has two independent solutions, iff the above matrix has rank 1, which is iff $a=0$.
</s>