if A diagonalizable then show that $a=0$

Question

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?

Answer 1

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 $a=0$.

Answer 2

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 $a=0$.