A Self-Adjoint Operator on a Finite-Dimensional Space has a Unit Eigenvector

Let $V$ together with $\langle\cdot,\cdot\rangle$ be a \reftext{def:complex-inner-product-space-2026a}{complex inner product space} with \reftext{lem:vector-space-basic-identities-2026a}{zero vector} $0_{V}$, and suppose that $V$ is \reftext{def:finite-dimensional-vector-space-2026b}{finite-dimensional} and $V\ne\{0_{V}\}$. Let $T$ be a \reftext{def:linear-operator-2026a}{linear operator} on $V$ that is \reftext{def:self-adjoint-operator-2026b}{self-adjoint}.

Then there are a \reftext{def:unit-vector-2026a}{unit vector} $x_{0}\in V$ and a \reftext{def:real-numbers-c54-2026c}{real number} $\lambda$ such that $x_{0}$ is an \reftext{def:eigenvector-eigenvalue-2026a}{eigenvector of $T$ with eigenvalue $\lambda$}.