Proof of Spectral Theorem for a Self-Adjoint Operator in Finite Dimensions

Write (I$k$) for claim $k$ of Elementary Properties of an Orthonormal Family, (Q$k$) for claim $k$ of A Linear Subspace is a Vector Space and Inherits an Inner Product, (F$k$) for claim $k$ of Properties of Finite Sums of Vectors, (C$k$) for claim $k$ of The Orthogonal Complement of a Unit Vector, and (B$k$) for claim $k$ of Elementary Properties of a Self-Adjoint Operator. Sums of vectors are finite sums in a vector space.

The induction is over the dimension, quantified over all spaces at once. Let $A$ be the set of those $q\in\mathbb{N}$ with the following property: for every complex inner product space $X$ that is finite-dimensional with $X\ne\{0_{X}\}$ and $\dim X=q$, and every self-adjoint linear operator $R$ on $X$, there are an orthonormal basis $c\in X^{q}$ of $X$ and a tuple $\nu\in\mathbb{R}^{q}$ of real numbers with $R(c_{k})=\nu_{k}c_{k}$ for every $k\in[q]$. We show $A=\mathbb{N}$ by Principle of Induction for the Natural Numbers.

Base case. Let $\dim X=1$ and let $R$ be self-adjoint on $X$. By A Self-Adjoint Operator on a Finite-Dimensional Space has a Unit Eigenvector there are a unit vector $u\in X$ and a real number $\mu$ with $R(u)=\mu u$. Let $c\in X^{1}$ have component $c_{1}=u$; it is orthonormal, since its only component is a unit vector and $[1]$ contains no two distinct indices, hence linearly independent by (I3).

Let $\tilde{u}\in X^{1}$ be the $1$-tuple with component $u$ and let $W$ be the orthogonal complement of the span of $\tilde{u}$. By (C2), $W=\{0_{X}\}$, so (C1) gives, for every $x\in X$, $x=\langle u,x\rangle u+0_{X}=\langle u,x\rangle u$, which is $\sum_{k=1}^{1}\langle u,x\rangle c_{k}$ by (F1). Hence $c$ spans $X$ and is a basis, so an orthonormal basis. With $\nu\in\mathbb{R}^{1}$ given by $\nu_{1}=\mu$ we get $R(c_{1})=\nu_{1}c_{1}$, so $1\in A$.

Induction step. Let $q\in A$, let $X$ be a complex inner product space that is finite-dimensional with $X\ne\{0_{X}\}$ and $\dim X=q+1$, and let $R$ be a self-adjoint linear operator on $X$.

By A Self-Adjoint Operator on a Finite-Dimensional Space has a Unit Eigenvector there are a unit vector $u\in X$ and a real number $\mu$ with $R(u)=\mu u$; thus $u$ is an eigenvector of $R$ with eigenvalue $\mu$. Let $\tilde{u}\in X^{1}$ be the $1$-tuple with component $u$, let $U$ be the span of $\tilde{u}$, a linear subspace by claim 1 of The Span of a Finite Family is the Smallest Subspace Containing It, and let $W=U^{\perp}$, a linear subspace by The Orthogonal Complement of a Linear Subspace is a Linear Subspace.

By (C3) applied with $r=q$, the set $W$ is not $\{0_{X}\}$ and, with the inner product it inherits by (Q3), it has an orthonormal basis $f\in W^{q}$. In particular $W$ is a complex inner product space by (Q1) and (Q3), it is finite-dimensional because $f$ is a basis of it, and $\dim W=q$ by the definition of the dimension.

By (B3) we have $R(x)\in W$ for every $x\in W$, and by (B4) the map $R_{W}$ sending $x\in W$ to $R(x)$ is a self-adjoint linear operator on $W$. Since $q\in A$, there are an orthonormal basis $g\in W^{q}$ of $W$ and a tuple $\sigma\in\mathbb{R}^{q}$ of real numbers with $R(g_{k})=R_{W}(g_{k})=\sigma_{k}g_{k}$ for every $k\in[q]$.

Define $c\in X^{q+1}$ by $c_{k}=g_{k}$ for $k\in[q]$ and $c_{q+1}=u$, and $\nu\in\mathbb{R}^{q+1}$ by $\nu_{k}=\sigma_{k}$ for $k\in[q]$ and $\nu_{q+1}=\mu$.

$c$ is orthonormal. By (Q3) the inner product and the induced norm of $W$ are the restrictions of those of $X$, so every component of $g$ is a unit vector of $X$ and distinct components of $g$ are orthogonal in $X$; and $u$ is a unit vector. Let $i,j\in[q+1]$ be distinct. If both lie in $[q]$, then $c_{i}$ and $c_{j}$ are orthogonal. Otherwise exactly one of them equals $q+1$ by claim 5 of Properties of the Order on the Natural Numbers; and for $k\in[q]$ we have $g_{k}\in W$ and $u\in U$, so $\langle u,g_{k}\rangle=0$ by the definition of the orthogonal complement, while $\langle g_{k},u\rangle=\overline{0}=0$ by condition 1 of Complex Inner Product Space and claim 1 of Properties of Complex Conjugation and Modulus. Hence $c$ is orthonormal, and linearly independent by (I3).

$c$ spans $X$. Let $x\in X$. By (C1) we have $x=\langle u,x\rangle u+w$ with $w\in W$, and since $g$ spans $W$ there is a tuple $a$ of complex numbers on $[q]$ with $w=\sum_{k=1}^{q}a_{k}g_{k}$, the sum being the same formed in $W$ and in $X$ by (Q2). Let $a'$ be the tuple of complex numbers on $[q+1]$ with $a'_{k}=a_{k}$ for $k\in[q]$ and $a'_{q+1}=\langle u,x\rangle$. By the recursion and restriction parts of (F1),

the last equality by commutativity of addition in $X$. Hence $c$ is an orthonormal basis of $X$.

The eigenvalue equations. For $k\in[q]$ we have $R(c_{k})=R(g_{k})=\sigma_{k}g_{k}=\nu_{k}c_{k}$, and $R(c_{q+1})=R(u)=\mu u=\nu_{q+1}c_{q+1}$. Therefore $q+1\in A$.

By Principle of Induction for the Natural Numbers, $A=\mathbb{N}$. Applying the property of $A$ with $q=n$, $X=V$ and $R=T$ gives an orthonormal basis $e\in V^{n}$ and a tuple $\lambda\in\mathbb{R}^{n}$ of real numbers with $T(e_{k})=\lambda_{k}e_{k}$ for every $k\in[n]$. Each $e_{k}$ is a unit vector, hence $e_{k}\ne 0_{V}$ because $\langle e_{k},e_{k}\rangle=1\ne 0$ by (I1) while $\langle 0_{V},0_{V}\rangle=0$; so each $e_{k}$ is an eigenvector of $T$ with eigenvalue $\lambda_{k}$.