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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?

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    Also, I think that for case 2, the spec is just the set {-1,1}.2012-09-27

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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.

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    @BenjaLim - Indeed in both cases $A$ is diagonalisable as the minimal polynomial splits into distinct linear factors.2012-09-27
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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.

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    Okay,$1$has to be a root as that solves that polynomial, but so does$0$but you're saying 0 can't be.2012-09-27