We have a matrix $A : 7x7$ with a minimal polynom: $$ m(t) = (t - \lambda)^3 $$
Also, we know that $A$ is not a Nilpotent matrix.
The rank of $A-\lambda I$ = 3
We need to find jordan forms for $A,A^2$.
For $A$ its ok:
We have: $$ m(t) = (t - \lambda)^3 $$
Therefore, the charactaristic polynomial: $$ p(t) = (t - \lambda)^7 $$
And $$ dim V_{\lambda} = 3 $$
Therefore: $$ G = diag\{J_{3}(\lambda), J_{2}(\lambda), \lambda, \lambda\} $$
NOW THE PROBLEM
Now i need to find jordan form for $A^2$.
The solution says: $$ P^{-1}A^2P = G^2 = diag\{J_{3}(\lambda)^2, J_{2}(\lambda)^2, \lambda^2, \lambda^2 \} = G' $$
And then he says that the charactaristic and minimal polynomial of $G'$ are: $$ p(t) = (t - \lambda^2)^7, m(t) = (t - \lambda^2)^3 $$
And because $A$ is not nilpotaent than $\lambda \neq 0$ and we conclude: $$ rank(G'-\lambda^2 I) = 3 $$
How did he conclude all of that? we dont know that?
Its like he conclude that $A^2$ has the same eigenvalues as $A$ just squared.
But why? how?
Thank you.
Hint: If $Ax=\lambda x$ then $A^2 x=A(Ax)=A(\lambda x)=\lambda Ax=\lambda^2 x$.
Updated: Clearly $(A^2-\lambda^2)^3=(A-\lambda)^3(A+\lambda)^3=0$. Now the multiplicity of $\lambda^2$ in the minimal polynomial determines the size of the largest Jordan block, which we know is of size $3$. It follows that $(t-\lambda^2)^3$ is the minimal polynomial of $A^2$.