Idempotent Matrix

TL;DR

An idempotent matrix is unchanged when multiplied by itself.
Common examples are the identity matrix and projection matrices.
For idempotent matrices, rank equals trace; they are diagonalizable (per the source).

An idempotent matrix is a square matrix that satisfies the equation $A^2 = A,$ i.e., multiplying the matrix by itself yields the same matrix.

Idempotence means repeated multiplication by the matrix does not change the result beyond the first multiplication: A times A equals A.
The identity matrix is an idempotent matrix because it reproduces itself under multiplication.
Projection matrices that project vectors onto a subspace are idempotent because applying the projection twice has the same effect as applying it once.
The source lists several properties attributed to idempotent matrices: the rank equals the trace, an idempotent matrix’s inverse is described as idempotent (in the source), and idempotent matrices are described as diagonalizable.

The 2×2 identity matrix I is idempotent. The source shows:

I \cdot I = \begin{bmatrix}1 & 0\end{bmatrix} \cdot \begin{bmatrix}1 & 0\end{bmatrix} = \begin{bmatrix}11 + 00 & 10 + 00\end{bmatrix} = \begin{bmatrix}1 & 0\end{bmatrix} = I

The source gives the projection matrix P as:

P = \begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}0 & 1 & 0\end{bmatrix}

and shows that multiplying P by itself yields P:

P \cdot P = \begin{bmatrix}1 & 0 & 0\end{bmatrix} \cdot \begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}0 & 1 & 0\end{bmatrix} \begin{bmatrix}0 & 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}0 & 1 & 0\end{bmatrix} = P

The source lists these properties of idempotent matrices:

The rank of an idempotent matrix is always equal to its trace. The source gives examples: the identity matrix has rank 2 because it has 2 non-zero elements on the main diagonal, and the projection matrix has rank 2 for the same reason.
The source states that the inverse of an idempotent matrix is also an idempotent matrix and illustrates this by writing the equation $P \cdot P^{-1} = I$ with the matrices shown as: $\begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}a & b & c\end{bmatrix} \begin{bmatrix}0 & 1 & 0\end{bmatrix} \begin{bmatrix}d & e & f\end{bmatrix} = \begin{bmatrix}1 & 0\end{bmatrix} \begin{bmatrix}g & h\end{bmatrix} \begin{bmatrix}0 & 1\end{bmatrix}$ and then gives the inverse (as presented in the source): $P^{-1} = \begin{bmatrix}a & b & c\end{bmatrix} \begin{bmatrix}d & e & f\end{bmatrix} \begin{bmatrix}g & h\end{bmatrix}$
The source asserts that idempotent matrices are always diagonalizable and provides the decomposition for P as $P = D \cdot V$ with $D = \begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}0 & 1 & 0\end{bmatrix} \quad\text{and}\quad V = \begin{bmatrix}1 & 0 & 0\end{bmatrix} \begin{bmatrix}0 & 1 & 0\end{bmatrix}.$