What values of $a$ make matrix $A$ diagonalisable?

Question

I have the following question in an assignment paper.

Let $$A=\begin{bmatrix} 0 & a & 0\\ 1 & 0 & a\\ a & 1 & 0\end{bmatrix}$$ For what values of $a$ is $A$ diagonalisable?

Simply put, I don't know how to do it. In the $2 \times 2$ case we were asked, I completed the square of the characteristic polynomial and found that in all but $1$ choice of the unknown entry you got distinct eigenvalues and, therefore, distinct eigenvectors. At which point I just had to consider the one case for which I had eigenvalue of algebraic multiplicity $2$ and show that the geometric multiplicity of the eigenvector associated with it was $1$, I was done.

Any tips would be hugely appreciated, I've said it an assignment so reservation on full solution I understand but some hints would be amazing. Thank you.

Answer 1

Hint: the characteristic polynomial is $-\lambda(\lambda^2-a)-a(-\lambda-a^2)=-\lambda^3+2\lambda a+a^3=0$.

The discriminant is $-4(-2a)^3-27(a^3)^2=32a^3-27a^6$. So $a=0$ and $a=\frac {2\sqrt[3]4}3$, for $a\in \mathbb R$... In $\mathbb C$, there are $a=0$ (with multiplicity $3$, again), and $3$ nonzero solutions to $\Delta =0$. Ruling these out will insure diagonalizability; but in some of these cases the matrix might still be diagonalizable...

Answer 2

Taking $a^3 = \frac{32}{27} \; , $ this includes complex $a:$

$$ \frac{1}{864a} \left( \begin{array}{rrr} 40 & 36 a^2 & 24 a \\ -100 & -90 a^2 & 156 a \\ -9 a^2 & -24 a & 32 \end{array} \right) \left( \begin{array}{rrr} 0 & a & 0 \\ 1 & 0 & a \\ a & 1 & 0 \end{array} \right) \left( \begin{array}{rrr} 6 a & -12 a & 54 a^2 \\ 9 a^2 & 9 a^2 & -60 \\ 10 & 4 & 0 \end{array} \right) = \left( \begin{array}{ccc} \frac{3}{2} a^2 & 0 & 0 \\ 0 & - \frac{3}{4} a^2 & 1 \\ 0 & 0 & - \frac{3}{4} a^2 \end{array} \right) $$