TITLE: Find all 4x4 A matrices so that $A^4=A^6$
QUESTION [5 upvotes]: Find all 4x4 A matrices so that $A^4=A^6$.
I think the method has to do something with eigenvalues, eigenvectors etc'... 
Thanks in advance for any assistance!

REPLY [2 votes]: Hint:
We can write $A^6 - A^4 = 0$, which is to say that $A$ "satisfies" the polynomial $x^6 - x^4 = 0$.  We then know (by a theorem that is probably in your textbook) that the minimal polynomial of $A$ divides $x^6 - x^4 = x^4(x-1)(x+1)$.  
What does this tell us about the Jordan-canonical form of $A$?
Connection between minimal polynomials and J-C form:

$q_A(x) = (x-\lambda_1)^{m_1}\cdots(x-\lambda_k)^{m_k}$, where $\lambda_1,\dots,\lambda_k$ are the eigenvalues of $A$, and $m_k$ is the length of the longest Jordan block associated with $\lambda_k$.
$m_k \geq 1$ for every eigenvalue of $A$. $A$ is diagonalizable if and only if $m_k=1$ for all $k$.