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