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.

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1
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