Showing that $A^2=I$ implies $A$ is diagonalizable

Question

Suppose $A$ is a square matrix such that $A^2=I$, and assume that $A\neq I$ and $A\neq -I$. I am trying to show that $A$ is diagonalizable based on the following hint: "Verify that $A(A+I)=A+I$ and $A(A-I)=-(A-I)$ and then look at nonzero columns of $A+I$ and $A-I$".

It is easy to see that $A(A+I)=A+I$ and $A(A-I)=-(A-I)$. However, I am not sure what is meant by looking at the columns since we don't have anything concrete. I notice however that the only possible eigenvalues for the matrix are $1$ and $-1$ and further that the above equalities imply that each vector $(A+I)v$ is an eigenvector corresponding to $1$ and that each vector $(A-I)v$ is an eigenvector corresponding to $-1$. But where do I go from here?

I understand that this question has been asked before. I want to know what might be meant by "look at the columns". In particular, nothing about decompositions or canonical forms appears in the book at this point. I am hoping for something elementary based off the hint given above.

Does this answer your question? $T^2=I$ implies $T$ is diagonalizable — Dietrich Burde, Sep 25 '20 at 13:22
@DietrichBurde No, I am specifically wondering what is meant by "look at the columns" — confusedmath, Sep 25 '20 at 13:25
Hint: Recall the definition of eigenvalue and eigenvector. What happens if we pack multiple eigenvectors together in a matrix in the definition. What do we get? — Stefan Octavian, Sep 25 '20 at 13:30
@StefanOctavian If we do that we get a matrix with eigenvectors. But I don't see why that means that matrix is invertible, or what it has to do with the columns of $A+I$ or $A-I$ — confusedmath, Sep 25 '20 at 14:25
$A^2 =I$ does not imply diagonalizeability over a field of characteristic 2, so something must be said about the field. — user8675309, Sep 25 '20 at 16:14
@confusedmath the point is that because $A(A - I) = -(A-I)$, we know that every non-zero column of $A - I$ is an eigenvector. — Ben Grossmann, Sep 25 '20 at 16:55

score 2 · Accepted Answer · answered Sep 25 '20 at 19:50

For simplicity, let's say you're working over $\mathbb{R}$.

As noted by Ben Grossman, the non-zero columns of $A + I$ and $A - I$ are eigenvectors.

[To see this, note (as you have) that $(A + I)v$ and $(A - I)v$ are eigenvectors and set $v = e_i $ ($i^{th}$ standard basis vector).]

Now, since $A^2 = I$, we know that Im $A = \mathbb{R}^n$. Also, since $(A+ I) + (A - I) = 2A$, the columns of $A + I$ and $A - I$ together form a spanning set for $\mathbb{R}^n$.

Therefore, we can take $n$ of these columns to find a basis for $\mathbb{R}^n$. Since these are $n$ linearly independent eigenvectors for $A$, $A$ must be diagonalizable.

Showing that $A^2=I$ implies $A$ is diagonalizable

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