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.

@R_Berger how is the determinant important? Do you mean the determinant of $A-\lambda I$? — Yanko, Mar 24 '18 at 18:03
Tip: the discriminant of a polynomial allows you to determine whether that polynomial has any multiple roots. In this case, taking the discriminant of the characteristic polynomial (as a function of $a$) will allow you to conclude diagonalizability for all but finitely many values of $a$. — Greg Martin, Mar 24 '18 at 18:13
I've been operating under the assumption $a\in \mathbb{C}$, I think this is wrong then. Thank you for your help — LM123, Mar 24 '18 at 18:42

score 4 · Accepted Answer · 2018-03-25T00:51:50.260

4

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...

edited Mar 25 '18 at 00:51

answered Mar 24 '18 at 19:00

1

For $a = 1$, when you plug in $\lambda = a$, you get a matrix of determinant $+1$, so $\lambda = a$ is not a root of the characteristic polynomial. (Or else I did the determinant wrong, a very real possibility.) – John Hughes Mar 24 '18 at 19:16
i get the characteristic polynomial as $-\lambda ^3+2a\lambda+a^3=0$ – LM123 Mar 24 '18 at 19:30
I think i lost a minus sign. – Mar 24 '18 at 19:31
either way, everyone seems to be working over the reals which largely solves my problem, the assignment document didn't specify so i, without thinking assumed it to be over the complex. – LM123 Mar 24 '18 at 19:35
How are you ruling out the two values of $a$ that produce repeated roots? Algebraic multiplicity $\gt1$ of an eigenvalue doesn’t automatically imply that the matrix isn’t diagonalizable. Although it’s obvious at a glance that the matrix with $a=0$ isn’t diagonalizable, I think that the other real value of $a$ that produces repeated roots needs further investigation. – amd Mar 24 '18 at 23:50
@amd I think you're right.. $\Delta \not =0$ would imply diagonalizability, but might not be necessary... – Mar 25 '18 at 00:16
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That aside, I like this approach. It narrows down the cases that need investigation very quickly. – amd Mar 25 '18 at 00:20

Will Jagy · Answer 2 · 2018-03-25T01:27:37.500

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

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

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

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