Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a $3 \times 3$ complex matrix with two distinct eigenvalues $e_1, e_2$. Write all possible Jordan canonical forms of $A$ and find the spectrum of $A$, in case(1) $A^{2}=A$ and case(2) $A^{2}=I$

So I know that the characteristic polynomial in $\mathbb{C}$ is always fully reducible, which implies that there are three eigenvalues for A, right? Then $e_3$ is either equal to $e_1$ or $e_2$. Then to find each $J_{i}$, would I find the kernel of each eigenvalue?

share|improve this question
    
Also, I think that for case 2, the spec is just the set {-1,1}. –  JackinAstoria Sep 27 '12 at 0:35
add comment

2 Answers 2

The Jordan Canonical Form of a $3 \times 3$ complex matrix is completely determined by its characteristic and minimal polynomials. You already know you have two distinct eigenvalues $e_1$ and $e_2$. Now if the third eigenvalue $e_3$ is different from those two, then your matrix is diagonalisable. If it is equal to $e_1$ or $e_2$, say $e_1$ for now your characteristic polynomial is $\chi(t) = (t - e_1)^2(t - e_2)$ while the minimal polynomial is either equal to the characteristic polynomial or is $(t-e_1)(t-e_2)$.

See this post here for an explanation of why in the $3 \times 3$ case the characteristic and minimal polynomials completely determine your jordan form.

share|improve this answer
    
thank you but some questions. So the matrix representation would be just two blocks with the first one just being a $1x1$ matrix and the second being a $2x2$? –  JackinAstoria Sep 27 '12 at 0:32
    
@JackinAstoria Well it depends on whether the characteristic polynomial is equal to the minimal polynomial or is not. See my post in the link. –  user38268 Sep 27 '12 at 0:38
    
I took a look and read it, but how can I determine the minimal polynomial, would I just assign it WLOG to either c1 or c2? –  JackinAstoria Sep 27 '12 at 0:42
    
@JackinAstoria By the way in case (1) and (2) I believe $A$ is always diagonalisable. –  user38268 Sep 27 '12 at 0:48
    
@BenjaLim - Indeed in both cases $A$ is diagonalisable as the minimal polynomial splits into distinct linear factors. –  Belgi Sep 27 '12 at 2:03
add comment

Hint 1: $A^{2}=A\implies A^{2}-A=0\implies A(A-I)=0\implies P(A)=0$ where $P(x)=x(x-1)$ so the minimal polynomial divides $P$ .

Hint 2: The roots of the minimal polynomials are all the roots of the characteristic polynomial and there are $2$ distinct eigenvalues.

share|improve this answer
    
Thanks, but I'm a bit confused. I think I remember, although possibly wrong, that eigenvalues for matrix A^2 are just the eigenvalues for matrix A, squared. So then in my OP, the spectrum in case 1, would that just be $c_{1}^{2}$ and $c_{2}^{2}$? Case 2? –  JackinAstoria Sep 27 '12 at 1:43
    
I say that from both hints you can deduce exactly what are the eigenvalues, can you see that ? –  Belgi Sep 27 '12 at 1:44
    
feeling dumb here, is it 0,1? –  JackinAstoria Sep 27 '12 at 1:47
    
Can you first tell what is the minimal polynomial by using hint 1 ? –  Belgi Sep 27 '12 at 1:47
    
The minimal one is 0? –  JackinAstoria Sep 27 '12 at 1:48
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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