Involutory matrix diagonaliable

Question

On my exam today there's this question:

A is a real n by n matrix and it is its own inverse. Prove that A is diagonalizable.

It seems very few students solved it if any.

All I know is that it's eigenvalue has to be 1 or -1. But how do I know the dimension of the eigenspace is enough? I've searched through internet and the solutions I found is all about minimal polynomial which I haven't learnt. Is there any method using only properties of eigenvectors?

Answer 1

If you haven't covered minimal polynomials and related topics this was a hard question. Here's a simple proof, using nothing but the definitions. (The proof looks like magic - I don't see how anyone would think of it if they hadn't learned about minimal polynomials etc.)

First:

If $x$ is any vector then $x+Ax$ and $x-Ax$ are eigenvectors of $A$.

Proof: Say $z=x+Ax$. Then $$Az = A(x+Ax)=Ax+A^2x=Ax+x=z.$$Similarly if $z=x-Ax$ then $Az=-z$..

Hence:

Any vector is a linear combination of eigenvectors of $A$.

Proof: $x=\frac12(x+Ax)+\frac12(x-Ax)$.

Now say $E$ is the set of eigenvectors of $A$. We've shown that $E$ spans $\Bbb R^n$. Since any spanning set contains a basis, $E$ contains a basis for $\Bbb R^n$. That is, there is a basis consisting of eigenvectors, so $A$ is diagonalizable.