If $A$ is an involutory matrix, i.e. $A^2=I$, then is $A$ diagonalizable?
Is the matrix $A$ diagonalizable if $A^2=I$
17
$\begingroup$
linear-algebra
matrices
-
0@ArturoMagidin You said it slicker than me. I was thinking Maschke's Theorem. – 2012-01-17
1 Answers
29
Since $A^2=I$, then $A$ satisfies the polynomial $t^2-1 = (t-1)(t+1)$. Hence, the minimal polynomial of $A$ divides $(t-1)(t+1)$; so the minimal polynomial of $A$ splits and has distinct roots, so $A$ is diagonalizable.
As N.S. points out in the comments, the above fails if you are working in characteristic 2. There, the matrix $A=\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$ has minimal and characteristic polynomials $t^2+1 = (t+1)^2$, and it is not diagonalizable (the eigenspace of $1$ has dimension $1$). But if $1\neq -1$, you are set.
-
0@N.S. Ah, true. – 2012-01-17