TITLE: Show matrix is diagonalizable
QUESTION [0 upvotes]: Let $A$ be an $n \times n$ matrix such that $A^2 = I$ and $A \neq I$. Show that $A$ is diagonalizable.

So far I tried multiplying both sides of the first equation by $P$ and by $P^{-1}$, but I don't know how to proceed from there.

REPLY [1 votes]: If your field's characteristic is not 2, then consider using basic tools of vector spaces (scaling and translating)     
$B := 2^{-1}(A+I)$
and $B^2 = 2^{-2}(A^2 + 2A + I) = 2^{-2}(I + 2A +I) = 2^{-2}\cdot 2 (A + I) = B $ 
so $B$ is idempotent.  Idempotent matrices are always diagonalizable, which implies $A$ is diagonalizable.  The standard argument for why uses a simple form of minimal polynomial argument -- I give it below.
(This approach makes it clear why things may break in a field of characteristic 2 -- i.e. because $2^{-1}$ doesn't exist).  I prefer leaning on idempotence since it is in some ways more important than involutions.
simple argument for diagonalizability of idempotent matrix
(assume $I \neq B \neq \mathbf 0$ as there is nothing to do in those cases.).  
$B^2 = B \longrightarrow B^2 - B = (B-I)(B-\mathbf 0)   = \mathbf 0$
consider $\dim \ker B = r$ then linear independent set $\big\{\mathbf v_1, ... ,\mathbf v_r\}$ in the kernel of $B$ and another linearly independent set $\{\mathbf v_{r+1},..., \mathbf v_n\}$ that generates the image of $B$ (rank-nullity)  
$\mathbf v_j \in \{\mathbf v_{r+1},..., \mathbf v_n\}\longrightarrow \big(B-I\big)\mathbf v_j = \mathbf 0$
so $\mathbf v_j$ is an eigenvector of $B$ with eigenvalue 1 which implies linear independence from those eigenvectors with eigenvalue 0 so
$\big\{\mathbf v_1, ... ,\mathbf v_r, \mathbf v_{r+1},..., \mathbf v_n\} $ is a linearly independent set which forms a basis for our n-dimensional vector space, i.e. $B$ has eigenvectors associated with eigenvalue 1 and 0 that form a basis so $B$ is diagonalizable.  

note: if OP is uncomfortable with minimal polynomial arguments and is working in $\mathbb R$ or $\mathbb C$, we could also prove idempotent matrices are diagonalizable by observing all eigenvalues must be 0 or 1, then writing out the Jordan blocks.  $P^{-1}B^kP = J^k = J$ thus $\big \Vert P^{-1}B^kP\big \Vert_F = \big \Vert J^k\big \Vert_F = \big \Vert J\big \Vert_F $ which implies the super diagonal associated with eigenvalue 1 is all zero (i.e. if this wasn't the case then we'd have the crude lower bound $k\leq \big \Vert J^k\big \Vert_F$ which implies $\big \Vert J^k\big \Vert_F \gt \big \Vert J\big \Vert_F$ for large enough $k$).  
As for eigenvalue 0, just repeat the argument above on $(I-B)$ which is idempotent.
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