How to calculate the Jordan canonical form of the following matrix?

Question

This is an exercise from my textbook.

"Calculate the Jordan normal form of the following matrix:

$$A=\begin{pmatrix} 1&2&3&4&\cdots&n\\ 0&1&2&3&\cdots&n-1\\ 0&0&1&2&\cdots&n-2\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&\cdots&1 \end{pmatrix}. $$ "

I know how to calculate the Jordan normal form of any given matrix, but this problem is a specific n-order square matrix, and there should be a special way to avoid brute force calculation. However, I don't know how to do it. Hope some body can help me solving this. Kindly help.

Answer 1

You problem is not very complicated since you already have a triangular matrix. Obviously $\chi_A = (X-1)^n$, prove that $\pi_A = \chi_A$.

By doing so you will have $A-I_n$ as a cyclic endomorphism since it is nilpotent of order $n$. Simply chose $x \neq 0$ and show that $B = ((A-I_n)^{n-1} x,(A-I_n)^{n-2} x,\dots , (A-I_n)x,x)$ forms a basis of $\mathbb{R}^n$.

Then write $A-In$ in the basis $B$ and deduce $A$ in the basis $B$, it is your Jordan form.

I suggest you study the proof of Jordan's decomposition or Frobenius' decomposition. Being with ease working with cyclic bases should help.

Answer 2

Note that all eigenvalues of $A$ are one and $A-I$ can be written in the form $B= \begin{bmatrix} 0 & C \\ 0 & 0\end{bmatrix}$, where $C$ has dimension $n-1$ and is easily checked to be nonsingular. Then $Bx = 0$ iff $x$ is a multiple of $e_1$. In particular, $\dim \ker (A-I) = 1$. Hence there is one Jordan block.