📂Matrix Algebra

Proof that the eigenvalues of an idempotent matrix are either 0 or 1

Theorem

The eigenvalues of an idempotent matrix are only $0$ or $1$.

Explanation

This lemma is used in the proof of the equivalence condition for the chi-squared property of a quadratic form of a normally distributed random vector.

For the converse of this theorem to hold, the given idempotent matrix must be a real symmetric matrix.

Proof ¹

Let $A$ be an idempotent matrix, in other words, $A^{2} = A$. Assume that $\lambda$ and $x$ are an eigenvalue and eigenvector of $A$, respectively, then $$A^{2} x = A \lambda x = \lambda A x = \lambda^{2} x$$ and $Ax = \lambda x$, since $A^{2} x = A x$, we obtain $\lambda^{2} x = \lambda x$. Since $x$ is an eigenvector, it is not the zero vector, and from $\lambda^{2} x - \lambda x = \mathbf{0}$, we get $\lambda^{2} - \lambda = 0$.

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