Classify up to similarity all complex $3 \times 3$ matrices A such that A such that ${A^3} = 2{A^2} - A$.

Here is what I know: All matrices with complex entries have Jordan canonical forms. If matrices are similar, they have the same Jordan canonical form.

This is what I've tried: Let $g(t) = {t^3} - 2{t^2} + t$. Then $g(A) = 0$. So, $p(t)$, the minimal polynomial of $A$, can be: $t, t-1, {(t-1)^2}, t(t-1), t{(t-1)^2}$.

Now, I think that we classify all the matrices as similar to one of:

$\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$, $\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, $\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$, $\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$, $\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$, $\begin{bmatrix}1&1&0\\0&1&0\\0&0&0\end{bmatrix}$.

And my reasoning for that was just taking each one of the possible minimal polynomials, and finding the possible JCF's for it. But, I'm not very confident in this strategy at all. Is this the right way to go about the problem? If not, what would you suggest trying? If so, is my reasoning correct - can I just find the possible minimal polynomials, and say that each matrix we are looking for is similar to the JCF that corresponds to these minimal polynomials? Is there anything else I would need to do to prove that this answer is ok?


Yes; your answer and reasoning are completely correct, at least for this problem.

This approach is problematic, however, when you consider matrices of size larger that $3 \times 3$, where the characteristic and minimal polynomials do not completely determine a matrix up to similarity. For example, consider $$ \pmatrix{ 0&1\\ &0\\ &&0&1\\ &&&0 }, \pmatrix{ 0&1\\ &0\\ &&0\\ &&&0 } $$ In such a case, it is useful to note that the minimal polynomial tells you the size of the largest Jordan block associated with each eigenvalue, and nothing more.

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  • $\begingroup$ So, I've been discussing this problem with other people and I think that my answer above may be incorrect, because the minimal polynomial has to be $t{(t-1)^2}$, or $t(t-1)$, because the minimal polynomial and the characteristic polynomial must have the same zeros. Is this correct? $\endgroup$ – Gesa Nov 11 '14 at 20:14
  • $\begingroup$ In fact, we're not given any information about the characteristic polynomial in this problem, so your approach is fine. For example, the answer $A=0$ is certainly a solution. $\endgroup$ – Ben Grossmann Nov 11 '14 at 20:38

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