Find all matrices that commute with a Jordan block with $0$ diagonal

Question

Question: Find all $n\times n$ matrices that commutes with $$T=\left(\begin{matrix}0&1&&\\&0&\ddots&\\&&\ddots&1\\&&&0\end{matrix}\right)$$

You are probably wondering why I am answering my own question. Well, as I was typing my question I accidentally solved it...but I didn't want to waste all this typing, and I didn't find this one on MSE so I decided to post it anyway. This could be a duplicate, if so, welcome to point it out.

Answer 1

For any matrix $U,$ we can calculate any polynomial $$ W = a_0 I + a_1 U + a_2 U^2 + ... + a_{n-1} U^{n-1} $$ and conclude that $$ WU = UW $$ Also, there is no reason to worry about exponents $n$ or higher, because of Cayley-Hamilton.

If every eigenvalue occurs in just a single Jordan block, which is equivalent to the minimal polynomial being equal to the characteristic polynomial, then the only matrices that commute with $U$ are themselves polynomials in $U$

Answer 2

Suppose $S=(s_{ij})$ commutes with it, then

$$\left(\begin{matrix}s_{11}&s_{12}&\cdots&s_{1n}\\s_{21}&s_{22}&\cdots&s_{2n}\\&&\ddots&\\s_{n1}&s_{n2}&&s_{nn}\end{matrix}\right)\left(\begin{matrix}0&1&&\\&0&\ddots&\\&&\ddots&1\\&&&0\end{matrix}\right)=\left(\begin{matrix}0&1&&\\&0&\ddots&\\&&\ddots&1\\&&&0\end{matrix}\right)\left(\begin{matrix}s_{11}&s_{12}&\cdots&s_{1n}\\s_{21}&s_{22}&\cdots&s_{2n}\\&&\ddots&\\s_{n1}&s_{n2}&&s_{nn}\end{matrix}\right)$$

$$\implies\left(\begin{matrix}0&s_{11}&\cdots&s_{1,n-1}\\0&s_{21}&\cdots&s_{2,n-1}\\\vdots&\vdots&&\vdots\\0&s_{n1}&\cdots&s_{n,n-1}\end{matrix}\right)=\left(\begin{matrix}s_{21}&s_{22}&\cdots&s_{2n}\\\vdots&\vdots&&\vdots\\s_{n1}&s_{n2}&\cdots&s_{nn}\\0&0&\cdots&0\end{matrix}\right)$$

Hence $s_{ij}=0$ if $j=1,i\neq1$ or if $i=n,j\neq n$. And $s_{ij}=s_{i+1,j+1}$ for $1\leq i,j\leq n-1$. This means $S$ is of the form

$$\left(\begin{matrix}a_1&a_2&\cdots&a_n\\0&a_1&\ddots&\vdots\\&&\ddots&a_2\\0&&0&a_1\end{matrix}\right)$$

And it is easier to verify all matrices of this form commutes with the Jordan block above.

As a corollary, the set of all matrices having the same form as $S$ above is closed under multiplication.