## Rank and trace are equal for a real symmetric idempotent matrix

Proposition. Let \$latex mathbf{X} in mathbb{R}^{n times n}\$ be a matrix that is symmetric (\$latex mathbf{X}^top = mathbf{X}\$) and idempotent (\$latex mathbf{X}^2 = mathbf{X}\$). Then the rank of \$latex mathbf{X}\$ is equal to the trace of \$latex mathbf{X}\$. In fact, they are both equal to the sum of the eigenvalues of \$latex mathbf{X}\$.

The proof is relatively straightforward. Since \$latex mathbf{X}\$ is real and symmetric, it is orthogonally diagonalizable, i.e. there is an orthogonal matrix \$latex mathbf{U}\$ (\$latex mathbf{U}^top mathbf{U} = mathbf{I}\$) and a diagonal matrix \$latex mathbf{D}\$ such that \$latex mathbf{D} = mathbf{UXU}^top\$ (see here for proof).

Since \$latex mathbf{X}\$ is idempotent,

\$latex begin{aligned} mathbf{X}^2 &= mathbf{X},
mathbf{U}^top mathbf{D}^2 mathbf{U} &= mathbf{U}^T mathbf{DU},
mathbf{D}^2 &= mathbf{D}. end{aligned}\$

Since \$latex mathbf{D}\$ is a diagonal matrix, it implies that the entries on the diagonal must be zeros or ones. Thus, the number of ones on the diagonal (which is \$latex text{rank}(mathbf{D})…

View original post 16 more words