3
$\begingroup$

Given: Matrix $A$ with characteristic polynomial $p(x) = (x+3)^2(x-1)(x-5)$

Also given: $\rho(A+2I) + \rho(A+3I) + \rho(A-5I) = 9$ (btw $\rho$ means rank of the matrix)

Prove: $A$ is diagonalizable.

I tried by saying first the the eigenvalues are $-3,1,5.$

Then, I know that their algebraic multiplicity of $-3$ is $2$, of $1$ is $1$, and of $5$ is $1$.

Now I need only to prove that the geometric multiplicity of $-3 $ is $2$ to show that $A$ is diagonalizable.

How can I prove it by using $\rho(A+2I) + \rho(A+3I) + \rho(A-5I) = 9$ ?

$\endgroup$
  • 2
    $\begingroup$ You know the rank of $A+2I$ and of $A-5I$. From that, compute that the rank of $A+3I$ is $2$. $\endgroup$ – Daniel Fischer Jun 14 '14 at 14:28
  • $\begingroup$ I know that A's rank has to be 4 because the sum of the algebric multiplexing is 4. and also that A is singular. $\endgroup$ – Ilan Aizelman WS Jun 14 '14 at 14:33
1
$\begingroup$

Hint: You know $$\rho(A-5I)=4-1=3$$ (why ?) and that $$\rho(A+2I)=4$$ (why ?)

What do you conclude from that, given the work you already did ?

$\endgroup$
  • $\begingroup$ Is it true that $\rho(A-5I)$ = 1 ? If yes, is it because its algebric multiplexing is 1? $\endgroup$ – Ilan Aizelman WS Jun 14 '14 at 14:35
  • 1
    $\begingroup$ @IlanAizelmanWS - Yes. This would imply that the geometric multiplexing is $\leq 1$. But you also know that it is at least $1$ since it is an eigenvalue and so the geometric multiplexing is $1$ $\endgroup$ – Belgi Jun 14 '14 at 14:36
  • $\begingroup$ And another last thing, How did you know that "n" is 4? I mean, Is it because the sum of the algebric multiplexing is 4? (2+1+1)? $\endgroup$ – Ilan Aizelman WS Jun 14 '14 at 14:40
  • $\begingroup$ @IlanAizelmanWS - By the way, do you study for the "Algebra aleph" exam ? $\endgroup$ – Belgi Jun 14 '14 at 14:40
  • 1
    $\begingroup$ @IlanAizelmanWS - oh yes, your'e right $\endgroup$ – Belgi Jun 14 '14 at 14:42

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.