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

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 (?)

Bernard
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bm1125
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    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

2 Answers2

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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

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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.

xbh
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  • 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
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    @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
  • No problem. Glad to help! :) – xbh Aug 31 '18 at 11:37
  • $p(3)=0$ is not enough to ensure the matrix is not diagonalisable. You also have to compute the geometric multiplicity. – Bernard Aug 31 '18 at 11:43
  • Yeah, I said that at least these could be the desired cases, and the verification is necessary. – xbh Aug 31 '18 at 11:44
  • so after checking the only value I could find that make the matrix not diagonalizable is $\ a =4 $ (then I get two roots of $\ 2 $ with only one eigenvector) but when $\ a=5 $ I get two roots of $\ 3 $ and one root of $\ 2 $ and $\ p(3) =0 $ has two eigen vectors therefore $\ A$ is diagonalizable if a =5 . Did i get it right? – bm1125 Aug 31 '18 at 12:08
  • @bm1125 Sorry, I do not know. I didn't check them. I was lazy :D – xbh Aug 31 '18 at 12:10
  • Sure :) Just wanted to make sure my logic is right here. – bm1125 Aug 31 '18 at 12:13
  • Your logic seems fine to me. – xbh Aug 31 '18 at 12:16