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?
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.