Transforming of a nilpotent matrix to a specific form

Question

I wish to build a question in the field of discrete state spaces representation (control theory). The canonical form has a very unique, but not singular, representation. I am focusing on the $A$ matrix of the representation of the form:

$$\begin{bmatrix} -a_1 & -a_2 & \cdots & -a_{N-1}& -a_N\\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 &0 \\ \end{bmatrix}$$

That is a subdiagonal matrix + a matrix of the same size where only the first row has specific values.

Other representations can be generated using a transformation matrix $T$ by $T^{-1}AT$. In my case, $A\in3\times3$.

I want to use a representation with a nilpotentic matrix $B$. Given, $A$ and $B$ which answer these definitions, is there a way to calculate $T$ which transition between them so that $B=T^{-1}AT$? I do not mind the values of $a_1,\ldots,a_N$ being dictated by $B$. All I care about is the form of the resulting $A$.

For example:

$$B=\begin{bmatrix} 2 & 2 & -2\\ 5 & 1 & -3\\ 1 & 5 & -3\\ \end{bmatrix}, A=\begin{bmatrix} -a_1 & -a_2 & -a_3\\ 1 & 0 & 0\\ 0 & 1 & 0\\ \end{bmatrix}$$

Is it even possible to transform between the two or am I missing a limitation here? If it is possible, is there a close solution for $T$?

Answer 1

It is impossible for any regular matrix $A$. Proof:

Let's suppose that there is a regular matrix $T$ for such $B = T^{-1}AT$ is nilpotent. Therefore there exists $n$ for such $B^n=0$. $$ 0 = B^n=(T^{-1}AT)^n=T^{-1}A^nT\\ T0T^{-1}=A^n\\ 0=A^n $$ However, $A$ is regular, so $A^n\neq0$. This is a contradiction.

Answer 2

This is an extended comment rather than a solution.

Your matrix $A$ is in what is called Rational Canonical Form. Note that the characteristic equation of $A$ is $X^n +a_1 X^{n-1} +\dots + a_n$.

If $B$ is nilpotent then the characteristic polynomial of $B$ is $X^n$.

Similar matrices have the same characteristic polynomial, so if you want $A$ to be in this shape, and want it to be similar to a nilpotent matrix $B$ then we must have that all the $a_j=0$.

Now if you start with a nilpotent $B$ then it is indeed similar to a matrix whose only non-zero elements lie on the subdiagonal, and are all either $1$ or $0$. There are algorithmic ways of finding an appropriate $T$: if you search for "finding a Jordan basis" you'll find them.

In the explicit case of the $3\times 3$ $B$ you give, you can do the following to find the columns of $T$: (i) find an eigenvector and call it $u_3$; (ii) find a solution of $Bx=u_3$ and call it $u_2$; (iii) find a solution of $Bx=u_2$ and call it $u_1$. (These steps just require us to solve linear equations.)