Let $A$ be an $n\times n$ idempotent complex matrix.
Then prove that $A$ is diagonalizable.

Definition.

An $n\times n$ matrix $A$ is said to be idempotent if $A^2=A$.
It is also called projective matrix.

Proof.

In general, an $n \times n$ matrix $B$ is diagonalizable if there are $n$ linearly independent eigenvectors. So if eigenvectors of $B$ span $\R^n$, then $B$ is diagonalizable.

We prove that $\R^n$ is spanned by eigenspaces. Every vector $\mathbf{v}\in \R^n$ can be expresses as
\[\mathbf{v}=(\mathbf{v}-A\mathbf{v})+A\mathbf{v}=\mathbf{v}_0+\mathbf{v}_1,\] where we put $\mathbf{v}_0=\mathbf{v}-A\mathbf{v}$ and $\mathbf{v}_1=A\mathbf{v}$.

We claim that $\mathbf{v}_0$ and $\mathbf{v}_1$ are elements in the eigenspaces corresponding to (possible) eigenvalues $0$ and $1$, respectively.
To see this, we compute
\begin{align*}
A\mathbf{v}_0&=A(\mathbf{v}-A\mathbf{v})\\
&=A\mathbf{v}-A^2\mathbf{v}\\
&=A\mathbf{v}-A\mathbf{v} && \text{since $A$ is idempotent}\\
&=O=0\mathbf{v}_0.
\end{align*}

Thus, we have $A\mathbf{v}_0=0\mathbf{v}_0$, and this means that $\mathbf{v}_0$ is a vector in the eigenspace corresponding to the eigenvalue $0$.
(If $0$ is not an eigenvalue of $A$, then $\mathbf{v}_0=\mathbf{0}$.)

We also have
\begin{align*}
A\mathbf{v}_1=A(A\mathbf{v})=A^2\mathbf{v}=A\mathbf{v}=\mathbf{v}_1,
\end{align*}
where the third equality holds as $A$ is idempotent.
This implies that $\mathbf{v}_1$ is a vector in the eigenspace corresponding to eigenvalue $1$. (If $1$ is not an eigenvalue of $A$, then $\mathbf{v}_1=\mathbf{0}$.)

It follows that every vector $\mathbf{v}\in \R^n$ is a sum of eigenvectors (or the zero vector).
That is, $\R^n$ is spanned by eigenvectors.

By the general fact mentioned at the beginning of the proof, we conclude that the idempotent matrix $A$ is diagonalizable.

Different Proofs

Three other different proofs of the fact that every idempotent matrix is diagonalizable are given in the post “Idempotent Matrices are Diagonalizable“.

The post Idempotent (Projective) Matrices are Diagonalizable first appeared on Problems in Mathematics.