Find the Jordan canonical form of this matrix

Question

I’m working on my linear algebra assignment and now struggling to solve this problem, say, on the complex field, find the Jordan canonical form of $$A=\begin{bmatrix}\quad &\quad &\quad & a_1\\ \quad & \quad & a_2 & \quad \\ \quad & \dots & \quad & \quad \\ a_n &\quad & \quad & \quad \\ \end{bmatrix}.$$ Usually I’ll work with characteristic polynomial in order to obtain the eigenvalues when being asked to find JCF. However in this problem, at least I believe, finding characteristic polynomial can be a really tough task. See $\lambda I- A$ does not have a friendly look and there is no way to get the determinant and find all the roots.

I do believe the problem is rather difficult and I really do not have any direction in my mind. I would be very appreciated if anyone could offer some help.

Answer 1

I will just indicate a possible approach:

Take a look first at the case $n=2$. Then you have the matrix $\begin{pmatrix} 0 & a_1\\ a_2 & 0 \end{pmatrix} $, you can find the JCF of this matrix by simply dividing by dividing the second column by $a_1$. Namely you have $P=\begin{pmatrix}1 & 0\\ 0 & a_1\end{pmatrix}$, and

$\begin{pmatrix}1 & 0\\ 0 & a_1\end{pmatrix}\cdot \begin{pmatrix} 0 & a_1\\ a_2 & 0 \end{pmatrix}\cdot \begin{pmatrix}1 & 0\\ 0 & a_1\end{pmatrix}^{-1}=\begin{pmatrix}0 & 1\\ a_1a_2 & 0\end{pmatrix}$.

Now you can look at the case $n=3$ and use your knowledge of the case $n=2$ to solve this. The matrix is

$\begin{pmatrix} 0 & 0 & a_1\\ 0 & a_2 & 0\\ a_3& 0 & 0 \end{pmatrix}. $

Its trace is $a_2$ so you know you want to move the central element to a diagonal position, and then we will have a $2\times 2$ Jordan block with $0$ diagonal. Necessarily it will involve $a_1$ and $a_3$ and you can show that the JCF of this matrix will be $ \begin{pmatrix} a_2 & 0 & 0\\ 0 & 0 & 1\\ 0& a_1a_3 & 0 \end{pmatrix}. $

With some work you can show that the Jordan blocks will be $\begin{pmatrix}0 & 1\\ a_ia_{n-i+1} & 0\end{pmatrix}$ for all $i$ with the first diagonal element $a_{\frac{n+1}{2}}$ when $n$ is odd.

EDIT: If you're working over an algebraically closed field like $\mathbb{C}$ you can further simplify your Jordan blocks. Can you see what the answer should be then?

Answer 2

The matrix is reducible into two-dimensional subspaces $e_i\leftrightarrow e_{n-i}$. By a permutation of the basis elements, $e_1\mapsto e_1$, $e_n\mapsto e_2$, etc., it takes the following form $$\begin{pmatrix}B_1&&\\&B_2&\\&&\ddots\end{pmatrix},\qquad B_1=\begin{pmatrix}0&a_1\\a_n&0\end{pmatrix},B_2=\begin{pmatrix}0&a_2\\a_{n-1}&0\end{pmatrix},\ldots$$

Hence the Jordan form consists of $2\times2$ blocks with eigenvalues $\pm\sqrt{a_ia_{n-i+1}}$, one for each $B_i$; except that when $n$ is odd, one of the blocks is $1\times1$ $[a_{(n+1)/2}]$.

Thus the Jordan matrix is $$\begin{pmatrix}J_1&&\\&J_2&\\&&\ddots\end{pmatrix},\qquad J_i=\begin{pmatrix}\sqrt{a_ia_{n-i+1}}&0\\0&-\sqrt{a_ia_{n-i+1}}\end{pmatrix}, J_{(n+1)/2}=a_{(n+1)/2}\ \textrm{if $n$ odd}$$

Edit: Of course, the Jordan form is not unique, the $J_i$ can be ordered differently. Also, if $a_i\ne0$ but $a_{n-i+1}=0$ (or the other way round), then the above $J_i$ should be replaced by $J_i=\begin{pmatrix}0&1\\0&0\end{pmatrix}$.

Answer 3

$\newcommand{\diag}{\mathrm{diag}}$ If $n = 2k\,(k \geq 1)$ is even, denote $b_i = a_{i}a_{2k - i + 1}, i = 1, 2, \ldots, k$. Let $c_1, \ldots, c_t$ be all distinct values of $b_1, \ldots, b_k$, and $e_1, \ldots, e_t$ be their multiplicities so that $e_1 + \cdots + e_t = k$. If $0 \notin \{c_1, \ldots, c_t\}$, then it is easy to verify that $\lambda I - A$ is equivalent to \begin{equation*} \diag(\underbrace{1, \ldots, 1}_{k \text{ terms}}, \underbrace{\lambda^2 - c_1, \ldots, \lambda^2 - c_1}_{e_1 \text{ terms}}, \ldots, \underbrace{\lambda^2 - c_t, \ldots, \lambda^2 - c_t}_{e_t \text{ terms}}). \end{equation*} If $0 \in \{c_1, \ldots, c_t\}$, without loss of generality, assume $c_1 = 0$, then let $f_1$ denote the number of terms such that $a_{i} = a_{2k - i + 1} = 0 $ so that $e_1 - f_1$ is the number of terms such that one of $\{a_{i}, a_{2k - i + 1}\}$ is not zero and the remaining one is zero. Note $f_1$ may equal to $0$ or $e_1 $. In this case, it can be shown that $\lambda I - A$ is equivalent to \begin{equation*} \diag(\underbrace{1, \ldots, 1}_{k - f_1 \text{ terms}}, \underbrace{\lambda, \ldots, \lambda}_{2f_1 \text{ terms}}, \underbrace{\lambda^2, \ldots, \lambda^2}_{e_1 - f_1 \text{ terms}}, \underbrace{\lambda^2 - c_2, \ldots, \lambda^2 - c_2}_{e_2 \text{ terms}}, \ldots, \underbrace{\lambda^2 - c_t, \ldots, \lambda^2 - c_t}_{e_t \text{ terms}}). \end{equation*}

For each of the above two cases, it is straightforward to verify that the group of elementary divisors is \begin{equation*} \underbrace{\lambda \pm \sqrt{c_1}, \ldots, \lambda \pm \sqrt{c_1}}_{e_1 \text{ terms}}, \ldots, \underbrace{\lambda \pm \sqrt{c_t}, \ldots, \lambda \pm \sqrt{c_t}}_{e_t \text{ terms}}. \end{equation*} Therefore, the Jordan form of $A$ is \begin{equation*} \diag(\underbrace{\sqrt{c_1}, \ldots, \sqrt{c_1}}_{e_1 \text{ terms}}, \underbrace{-\sqrt{c_1}, \ldots, -\sqrt{c_1}}_{e_1 \text{ terms}},\ldots, \underbrace{\sqrt{c_t}, \ldots, \sqrt{c_t}}_{e_t \text{ terms}}, \underbrace{-\sqrt{c_t}, \ldots, -\sqrt{c_t}}_{e_t \text{ terms}}). \end{equation*}

The odd $n$ case can be treated analogously.