finding Jordan canonform of $A^3$ given the minimal polynomial of $A$

Question

I'm having trouble with the following question: we were given the following minimal polynomial of a matrix $A\in \mathbb {C}^{4 \times 4}: X^4-2X^3+X^2$. Now we are asked to find the Jordan canonical form of $A$ as well as $A^3.$ The case for $A$ is pretty easy, but I can't see how to derive the case for $A^3$ from it. edit: the minimal polynomial can be split into: $X^2(X-1)^2$. From what we have learned the Jordan form will then be: $\begin{pmatrix} 0&0&0&0\\ 1&0&0&0\\ 0&0&1&0\\ 0&0&1&1\\ \end{pmatrix}$. We can know that matrix $A$ can be written in the form $A=P^{-1}JP,$ with $J$ the Jordan form. Then $A^3=P^{-1}J^3P,$ with $J^3=\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&4&1\\ \end{pmatrix}.$ The minimal polynomial of $A^3$ will be: $(X-1)^2.$ Since $(A^3-I)^2=(P^{-1}J^3P-P^{-1}IP)^2=P^{-1}(J^3-I)^2P=0.$ From here on out I am not sure how to proceed. I think since $\dim(Ker(J^3-I))=3$ this means the invariant system is: $\{2,1,1\}.$ From which would follow that the Jordanform is: $\begin{pmatrix} 1&0&0&0\\ 1&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{pmatrix}.$ But on the last part I am not so sure about my reasoning.

Answer 1

Four mistakes (the first one is benign):

• $J^3=\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&\color{red}3&1\\ \end{pmatrix}$ (not $\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&4&1\\ \end{pmatrix}$).
• $(J^3-I)^2=\begin{pmatrix} I_2&0\\0&0\\ \end{pmatrix}\color{red}\ne0$;
• the minimal polynomial of $A^3$, i.e. of $J^3$, is not $(X-1)^2$ but $\color{red}X(X-1)^2$;
• $\dim\ker(J^3-I)=\color{red}1$, not $3$.

Now, directly (via a rescaling of the fourth basis vector), $J^3$ is similar to $\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&1&1\\ \end{pmatrix}$, which is therefore the Jordan form of $A^3$. Note that the Jordan form of $A^n$ is the same for every $n\ge2$.