📂Matrix Algebra

Proof that if all eigenvalues of a symmetric real matrix are either 0 or 1, it is an idempotent matrix

Theorem

If all the eigenvalues of the symmetric matrix $A \in \mathbb{R}^{n \times n}$ are either $0$ or $1$, then $A$ is an idempotent matrix.

Explanation

This lemma is used in the proof of the equivalence conditions for the chi-square-ness of quadratic forms of normally distributed random vectors and the proof of Cochran’s theorem.

The converse does not hold.

Proof

Spectral Theory: If $A$ is a Hermitian matrix, then it can be unitarily diagonalized: $$ A = A^{\ast} \implies A = Q \Lambda Q^{\ast} $$

If the real matrix $A \in \mathbb{R}^{n \times n}$ is a symmetric matrix, then it is a Hermitian matrix and hence diagonalizable. Let $\Lambda$ be the diagonal matrix composed of the eigenvalues of $A$, and $Q = Q^{\ast} = Q^{T}$ be the unitary matrix, then we have $A = Q \Lambda Q^{T}$, yielding: $$ A^{2} = \left( Q \Lambda Q^{T} \right) \left( Q \Lambda Q^{T} \right) = Q \Lambda^{2} Q^{T} $$ However, since the diagonal matrix $\Lambda$ has only $0$ and $1$ as its principal diagonal elements, it follows that $\Lambda^{2} = \Lambda$, and thus $A$ is an idempotent matrix as shown below: $$ A^{2} = Q \Lambda^{2} Q^{T} = Q \Lambda Q^{T} = A $$

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