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?

  • $\begingroup$ Also, I think that for case 2, the spec is just the set {-1,1}. $\endgroup$ – JackinAstoria Sep 27 '12 at 0:35

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.

  • $\begingroup$ 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 $2x2$? $\endgroup$ – JackinAstoria Sep 27 '12 at 0:32
  • $\begingroup$ @JackinAstoria Well it depends on whether the characteristic polynomial is equal to the minimal polynomial or is not. See my post in the link. $\endgroup$ – user38268 Sep 27 '12 at 0:38
  • $\begingroup$ 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? $\endgroup$ – JackinAstoria Sep 27 '12 at 0:42
  • $\begingroup$ @JackinAstoria By the way in case (1) and (2) I believe $A$ is always diagonalisable. $\endgroup$ – user38268 Sep 27 '12 at 0:48
  • $\begingroup$ @BenjaLim - Indeed in both cases $A$ is diagonalisable as the minimal polynomial splits into distinct linear factors. $\endgroup$ – Belgi Sep 27 '12 at 2:03

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.

  • $\begingroup$ 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? $\endgroup$ – JackinAstoria Sep 27 '12 at 1:43
  • $\begingroup$ I say that from both hints you can deduce exactly what are the eigenvalues, can you see that ? $\endgroup$ – Belgi Sep 27 '12 at 1:44
  • $\begingroup$ feeling dumb here, is it 0,1? $\endgroup$ – JackinAstoria Sep 27 '12 at 1:47
  • $\begingroup$ Can you first tell what is the minimal polynomial by using hint 1 ? $\endgroup$ – Belgi Sep 27 '12 at 1:47
  • $\begingroup$ The minimal one is 0? $\endgroup$ – JackinAstoria Sep 27 '12 at 1:48

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