Finding values of $\ a$ so $\ A$ will not be diagonalizable

$\ A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & a & a-2 \\ 0 & -2 & 0 \end{bmatrix} \\ a \in \mathbb R$

I need to find for which $\ a$ values $\ A$ will not be diagonalizable $\ A$

I was thinking trying the elimination way so finding values which $\ A$ can be diagonalize first.

so the characteristic polynomial of $\ A$ is $\ p(t) = (\lambda-3)(\lambda^2-a(\lambda-2)-4)$

But then after trying many numbers of $\ a$ , $\ (0,1,2,-1,)$ I see that it is wrong because there are too many possible values for $\ a$ to make the matrix diagonalizable. So maybe trying to figure out which values of a will give me less eigenvalues than needed (?)

• If you get two distinct real roots and neither of them equals $3$, then the matrix is diagonalizable. So at least you need one of the roots be $3$. To check the value $a$, you need to apply the usual diagonalization process to get rid of the one admits diagonalization. – xbh Aug 31 '18 at 11:11

As I commented, possible cases are: $p(3) = 0$ or $p$ has two equal roots, where $p(x) = x^2 - a(x-2) - 4$.

$p(3) = 0$ yields $a = 5$; $\varDelta = 0$ yields $a^2 + 16 - 8a = 0$, i.e. $a = 4$. Now check these cases by determining eigenspaces. I will let you take it from here.

• Thanks! you said that if we get another two distinct roots that neither of them 3 the matrix is diagonalizable. If I understand correctly that is because it is a 3x3 matrix with three distinct eigenvalues (and each have at least one corresponding eigenvector) it means that i must be diagonalizable ? Yet it could have three same roots for the polynomial and still be diagnoalizable , am I correct? – bm1125 Aug 31 '18 at 11:29
• @bm1125 You got it right! – xbh Aug 31 '18 at 11:30
• Thank you. I have really dumb question.. I'm really not sure what does it mean $\ \Delta = 0$ and how did you get from there to $\ a^2 + 16 - 8a =0$ ? – bm1125 Aug 31 '18 at 11:33
• @bm1125 Oh that is not dumb at all. My bad. $\varDelta$ is the discriminant of $p(x)$ as taught before college: the discriminant of $ax^2 + bx + c =0$ is $\varDelta = b^2 - 4ac$. And $\varDelta = 0$ indicates that the equation has two equal roots. – xbh Aug 31 '18 at 11:35
• ohh! thanks you! :) – bm1125 Aug 31 '18 at 11:36

Hint

for $\lambda^2-a\lambda+2a-4=0$ it is $\Delta=(a-4)^2\geq0$

So for $a\not=4$, $(\lambda-3)(\lambda^2-a\lambda+2a-4)=(\lambda-3)(\lambda-2)(\lambda-a+2)$

So if $a\not=4$ and $a\not=5$ $p(\lambda)$ is a product of distinct monic factors

• $\Delta \geq 0$ is diagonalizability in $\mathbb{R}$, no? @giannispapav – PackSciences Aug 31 '18 at 11:15
• @PackSciences sorry I don't understand what you are asking – giannispapav Aug 31 '18 at 11:17
• The criteria $\Delta \geq 0$ you've written is only for diagonalizability in $\mathbb{R}$, not $\mathbb{C}$, right? @giannispapav – PackSciences Aug 31 '18 at 11:18
• @PackSciences I didn't use any criteria. Just pointed out that $\Delta\geq0$ – giannispapav Aug 31 '18 at 11:19
• Oh ok, I am sorry. – PackSciences Aug 31 '18 at 11:20