1
$\begingroup$

This is a question from Serre's Exercise book in Matrix theory. I don't even know how to start. Any help would be appreciated.

Assume that the characteristic of the field $k$ is not equal to 2. Given $M \in GL_n(k)$, show that the matrix \begin{align} \begin{pmatrix}0_n & M^{-1} \\ M & 0_n\end{pmatrix} \end{align}

is diagonalizable. Find its eigenvectors and eigenvalues. More generally, show that every involution ($A^2=I$) is diagonalizable.

$\endgroup$

2 Answers 2

1
$\begingroup$

If $v_i \in k^n$ are the eigenbasis of $M$, then $(v_i , Mv_i) \in k^{2n}$ is an eigenvector with eigenvalue 1, $(v_i, -Mv_i) \in k^{2n}$ is an eigenvector with eigenvalue -1. Linear independence follows from linear independent of $v_i$

Since $A^2-1 = 0$, the minimal polynomial is $x^2-1$, which has a differential of $2x$. Since the characteristic is not 2, and $0$ is not a root of $x^2-1$, it follows that there are no double roots.

$\endgroup$
1
$\begingroup$

Look at the minimal polynomial (a factor of $t^2 - 1$) which over a field of characteristic $\ne 2$ has no double roots.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .