Canonic form of A matrix defined by a condition of powers equation I have to solve this old exam problem: A square matrix $A$ of order $4$ is not diagonalisable and satisfy this condition: $(A−I)^4 = 4(A−I)^2 = 9(A−I)^2$ List all the possible canonic form of $A$, the characteristic polynomial and the minimal polynomial. I already have the solution but I doesn't understand several part of it. The first thig that my prof do to solve it is write this: "By hypothesis we know that $A$ satisfies the following equations: $(t−1)^2(t+1)(t−3)=0$ $(t−1)^2(t+2)(t−4)=0$" First of all I don't get how he finds those equations. Can you help me?

After some days I successfully solved the problem:

First we have to breack the equivalence in two parts:

$(A-I)^4=4(A-I)^2$ and $(A-I)^4=9(A-I)^2$

So we know that the first equation is represented by the polynomial:

$(t-1)^4-4(t-1)^2=(t-1)^2((t-1)^2-4)=(t-1)^2(t^2-2t-3)=(t-1)^2(t+1)(t-3)=0$

And the second:

$(t-1)^4-9(t-1)^2=(t-1)^2((t-1)^2-9)=(t-1)^2(t^2-2t-8)=(t-1)^2(t+2)(t-4)=0$

Now the minimal polynomial must be in this case in the form $(t+n)^k$ with $k>1$ because $A$ is not diagonalisable. So the minimal polynomial is:

$m\substack{A}(t)=(t-1)^2$

And the possible canonic form are: $J\substack{1}=\begin{bmatrix}1&1&0&0\\\0&1&0&0\\\0&0&1&1\\\0&0&0&1\end{bmatrix}$ or $J\substack{2}=\begin{bmatrix}1&1&0&0\\\0&1&0&0\\\0&0&1&0\\\0&0&0&1\end{bmatrix}$

Now the only thing left is the characteristic polynomial that must be a 4th grade polynomial and because the only eigenvalues is $\lambda=1$ then

$p\substack{c}=(t-1)^4$.