1
$\begingroup$

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

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 (?)

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – xbh Aug 31 '18 at 11:11
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – bm1125 Aug 31 '18 at 11:29
  • 1
    $\begingroup$ @bm1125 You got it right! $\endgroup$ – xbh Aug 31 '18 at 11:30
  • $\begingroup$ 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 $ ? $\endgroup$ – bm1125 Aug 31 '18 at 11:33
  • $\begingroup$ @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. $\endgroup$ – xbh Aug 31 '18 at 11:35
  • $\begingroup$ ohh! thanks you! :) $\endgroup$ – bm1125 Aug 31 '18 at 11:36
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $\Delta \geq 0$ is diagonalizability in $\mathbb{R}$, no? @giannispapav $\endgroup$ – PackSciences Aug 31 '18 at 11:15
  • $\begingroup$ @PackSciences sorry I don't understand what you are asking $\endgroup$ – giannispapav Aug 31 '18 at 11:17
  • $\begingroup$ The criteria $\Delta \geq 0$ you've written is only for diagonalizability in $\mathbb{R}$, not $\mathbb{C}$, right? @giannispapav $\endgroup$ – PackSciences Aug 31 '18 at 11:18
  • $\begingroup$ @PackSciences I didn't use any criteria. Just pointed out that $\Delta\geq0$ $\endgroup$ – giannispapav Aug 31 '18 at 11:19
  • 1
    $\begingroup$ Oh ok, I am sorry. $\endgroup$ – PackSciences Aug 31 '18 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.