Jordan carnonical form of a clinodiagonal matrix Let $A$ be a $n\times n$ matrix in the following form
\begin{pmatrix}
&&&a_1\\
&&\cdot&\\
&\cdot&&\\
a_n&&&
\end{pmatrix}
How can we find it Jordan carnonical form? It seems that $n=2$ is easy to calculate. For general $n$, it seems difficult.
 A: By an appropriate row/column permutation, your matrix is similar to a block diagonal matrix:
$$ \begin{pmatrix}
0 & a_n \\
a_1 & 0 \\
&& 0 & a_{n-1} \\
&& a_2 & 0 \\
&&&& \ddots \end{pmatrix} $$
From there you can easily Jordanize each block separately.
A: @Henning Makholm
A complement, that I give as an answer because it takes some room. 
Let us take the case where $n$ is even and all $a_k >0$ in order not to have to consider special cases (though not especialy difficult).
It is remarkable that this kind of matrices have very simple eigenvalues, and eigenvectors as well. 
The eigenvalues are immediately computed using the block decomposition of the similar matrix you have given  
$$\sqrt{a_1a_{n}}, \ -\sqrt{a_1a_{n}}, \ \sqrt{a_2a_{n-1}}, \ -\sqrt{a_2a_{n-1}} , \cdots .$$ 
In a surprising way, "the" associated eigenvectors are as well simple:
$$\begin{pmatrix}\sqrt{a_1}\\0\\0\\\cdots\\0\\0\\\sqrt{a_n}\end{pmatrix}
\begin{pmatrix}\sqrt{a_1}\\0\\0\\\cdots\\0\\0\\-\sqrt{a_n}\end{pmatrix}
\begin{pmatrix}0\\\sqrt{a_2}\\0\\\cdots\\0\\\sqrt{a_{n-1}}\\0\end{pmatrix}
\begin{pmatrix}0\\\sqrt{a_2}\\0\\\cdots\\0\\-\sqrt{a_{n-1}}\\0\end{pmatrix}
etc.$$
making a change of base matrix $P$ very simple.
