Diagonalizability of an involution over a Field Suppose $A$ is a matrix over a field $F$ of characteristic $\neq 2$, with $A\cdot A=I$. Show that $A$ is similar to a diagonal matrix whose entries are either $1$ or $-1$.
 A: Addendum: in his answer, Marc van Leeuwen rightly observes that this does not work when $1 = -1$. In my hints below I was tacitly (but incorrectly) assuming $1 \ne -1$, that is, the field has characteristic $\ne 2$.
Hint 1

Show that $A$ is a root of $x^{2} - 1$.

Hint 2

Show that the minimal polynomial of $A$ is a divisor of $x^{2} - 1$.

Hint 3

Since the roots of the minimal polynomial are distinct, $A$ is diagonalizable. (This is where I need $1 \ne -1$.)

Hint 4

The eigenvalues are the roots of the minimal polynomial, and they can only be $1, -1$, as $x^{2} - 1 = (x - 1)(x + 1)$.

A: This is false in this generality, since it fails for fields of characteristic$~2$. There the only diagonal matrix whose entries are $\pm1$ is the identity matrix, and it is not similar to anything else; nevertheless there are plenty non-identity solutions to $A^2=I$ in characteristic$~2$, like $A=(\begin{smallmatrix}1&1\\0&1\end{smallmatrix})$.
Assuming $\operatorname{char}F\neq2$ however the result holds. $A$ is annihilated by substitution into the polynomial $X^2-1$ which splits as $(X+1)(X-1)$, with simple roots in this case. It is a theorem that such matrices are always diagonalisable, and the eigenvalues must of course be roots of that polynomial, so $+1$ or $-1$.
The fact that any square matrix$~A$ annihilated by substitution into a polynomial$~P$ that splits into distinct factors of degree$~1$ over $K$ is diagonalisable over$~K$ can be proved in various ways. If you already know the closely related fact that the minimal polynomial of$~A$ has such a factorisation if and only if $A~$is diagonalisable over$~K$, then the statement follows from that fact because the minimal polynomial has to divide$~P$, so it is a product of a subset of the factors (all distinct) of$~P$. The "if" part of the latter statement is clear by considering the diagonal form; the "only if" part can for instance be proved by induction on the number of factors (or the degree, which is the same) of the minimal polynomial. For no factors at all one is in dimension$~0$ and there is nothing to prove. Otherwise let $X-\lambda$ be one of the factors of the minimal polynomial$~\mu$, and $\mu'=\mu/(X-\lambda)$ the product of the remaining factors. Then $\mu'$ is the minimal polynomial of the restriction of (the endomorphism defined by) $A$ to the subspace $W=\operatorname{Im}(A-\lambda I)$. By induction this restriction is diagonalisable, so $W$ is a direct sum of eigenspaces of (the restriction of) $A$, all for eigenvalues distinct from$~\lambda$ (since $\lambda$ is not an root of $\mu'$ by hypothesis of distinct factors). As a sum of eigenspaces for distinct eigenvalues, the sum $W+\ker(A-\lambda I)$ is direct. Its dimension is that of the whole space by the rank-nullity theorem, so the whole space is a (direct) sum of eigenspaces and $A$ is diagonalisable.
Another, maybe simpler, proof is based on the general fact that if a space is annihilated by a polynomial $P[A]$ of a matrix$~A$, and if $P$ decomposes into a number of factors$~P_i$ that are pairwise relatively prime, then the space is the direct sum of the subspaces $\ker(P_i[A])$. (This can be proved by induction on the number of factors, using Bezout coefficients for the basic case of $2$ factors.) Apply this to the degree$~1$ factors of the given polynomial$~P$, and observe that the subspaces $\ker(P_i[A])$ are eigenspaces of$~A$ (or of dimension$~0$).
