Let $X$ be the linear space of all $3\times 3$ matrices. Let
$N=\begin{bmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{bmatrix}$
And let $T:X\to X$ be the linear map $TA=AN-NA$.
Show that $T^k=0$ for some $k\geq 1$ and find the smallest such $k$. Then find the Jordan canonical form of $T$.
We can show that if
$A=\begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{bmatrix}$
$TA=\begin{bmatrix} -a_{21}&a_{11}-a_{22}&a_{12}-a_{23}\\ -a_{31}&a_{21}-a_{32}&a_{22}-a_{33}\\ 0&a_{31}&a_{32}\end{bmatrix}$
$T^2A=\begin{bmatrix} a_{31}&a_{32}-2a_{21}&a_{11}-2a_{22}+a_{32}\\ 0&-2a_{31}&a_{21}-2a_{32}\\ 0&0&a_{31}\end{bmatrix}$
$T^3A=\begin{bmatrix} 0&3a_{31}&-3a_{21}+2a_{32}\\ 0&0&-3a_{31}\\ 0&0&0\end{bmatrix}$
$T^4A=\begin{bmatrix} 0&0&6a_{31}\\ 0&0&0\\ 0&0&0\end{bmatrix}$
And then $T^5=0$.
You can also show that $T^5=0$ without calculating the components of each matrix, just with some matrix algebra and using that $N^3=0$. However, I am curious if there is a way to show that $k=5$ is the degree of nilpotency without using components.
What is the best method for deducing the JCF? From the above we have that the minimal polynomial for $T$ is $m_{T}(x)=x^5$, so all eigenvalues are $0$ and that the largest Jordan block is of size 5. How to find the other blocks? Is there a way to do it cleanly without using the components? This was on a preliminary exam, so I guess I'm just searching for ways to find answers using the least possible time.