Find the Jordan form of an anti-diagonal matrix

jordan-normal-formlinear algebramatrices

How to find the Jordan form of an anti-diagonal matrix?

$$\begin{bmatrix} &&&{}a_{1}\\ &&\ddots &\\ &a_{\text{} n-1}&&\\ a_{n}&&& \end{bmatrix}$$

It seems to find the characteristic polynomial and minimal polynomial, but how to find these?

Best Answer

The matrix in question is similar via a permutation matrix to $$ A=\begin{cases} \pmatrix{0&a_1\\ a_n&0}\oplus\pmatrix{0&a_2\\ a_{n-1}&0}\oplus\pmatrix{0&a_{(n-1)/2}\\ a_{(n+3)/2}&0}\oplus a_{(n+1)/2}&\text{when $n$ is odd},\\ \pmatrix{0&a_1\\ a_n&0}\oplus\pmatrix{0&a_2\\ a_{n-1}&0}\oplus\pmatrix{0&a_{n/2}\\ a_{n/2+1}&0}&\text{when $n$ is even}.\\ \end{cases} $$ The problem thus reduces to finding the Jordan form of $X=\pmatrix{0&a\\ b&0}$ over the algebraic closure of the underlying field.

  • If $X$ has two distinct eigenvalues $r_1$ and $r_2$, its Jordan form is $\pmatrix{r_1&0\\ 0&r_2}$.
  • If $X=0$, its Jordan form is obviously itself.
  • If $X\ne0$ has a double eigenvalue $r$ (e.g. when $a=0\ne b$ or when the field is $GF(2)$ and $ab=1$), since $X$ is not a scalar matrix, it must be non-diagonalisable. Hence its Jordan form is $\pmatrix{r&1\\ 0&r}$.
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