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

Write (O$k$) for claim $k$ of Elementary Arithmetic in an Ordered Field, (M$k$) for claim $k$ of Properties of Complex Conjugation and Modulus, (N$k$) for claim $k$ of The Induced Norm is a Norm, and Induces a Metric, (B$k$) for claim $k$ of Elementary Properties of a Self-Adjoint Operator, and (I$k$) for claim $k$ of Elementary Properties of an Orthonormal Family. Conditions on an inner product are numbered as in Complex Inner Product Space. Let $\lVert\cdot\rVert$ be the induced norm, a norm by (N2), let $d(u,v)=\lVert u-v\rVert$, a metric by (N3), and let $|z|$ be the modulus of a complex number $z$, which for a real number agrees with its absolute value by claim 8 of Properties of the Absolute Value in an Ordered Field. Inequalities between real numbers are those of the ordered field $\mathbb{R}$, and $a<b$ abbreviates the conjunction of $a\le b$ and $a\ne b$.

Step 0: three order facts. Let $a,b,c$ be real numbers.

(i) If $a\le b$ and $b<c$, then $a<c$. Indeed $a\le c$ by transitivity, and $a=c$ would give $c\le b$, which with $b\le c$ and antisymmetry forces $b=c$, contrary to $b<c$.

(ii) If $a<b$ and $0\le c$ with $c\ne 0$, then $ca<cb$. Indeed $ca\le cb$ by (O5); and $ca=cb$ would give $c(b-a)=cb-ca=0$, whence $b-a=c^{-1}(c(b-a))=0$ and $a=b$, contrary to $a\ne b$.

(iii) If $a\le b$ and $a'\le b'$, then $a+a'\le b+b'$: two applications of the first order axiom of Ordered Field give $a+a'\le b+a'$ and $b+a'\le b+b'$, and transitivity applies.

Step 1: the unit sphere and a bound for $T$. By claim 1 of Orthonormal Bases and Basis Size in a Finite-Dimensional Inner Product Space there are a natural number $n$ and an orthonormal basis $e\in V^{n}$ of $V$. Let

be the set of unit vectors. By claim 3 of Coordinate Isometry Determined by a Finite Orthonormal Basis, $S$ is nonempty and compact in $V$ equipped with the topology of the sets open in $(V,d)$.

By Every Linear Operator on a Space with a Finite Orthonormal Basis is Bounded the real number $C=\sum_{k=1}^{n}\lVert T(e_{k})\rVert$ satisfies $0\le C$ and $\lVert T(u)\rVert\le C\lVert u\rVert$ for every $u\in V$. Put $D=C+C$, so $0\le D$ by (O2).

Step 2: the Rayleigh quotient is continuous on $S$. Let $R_{T}$ be the Rayleigh quotient of $T$, a map from $S$ to $\mathbb{R}$.

For $y,z,v\in V$ we have $\langle y+z,v\rangle=\langle y,v\rangle+\langle z,v\rangle$, by conditions 1 and 2 together with the additivity of conjugation in (M1). Let $x,y\in S$. Since $(y-x)+x=y$ and $T(y)-T(x)=T(y-x)$ by linearity of $T$, conditions 2 and the identity just noted give

By the triangle inequality (M7), then Cauchy-Schwarz (N1), then the bound of step 1 together with (O5) and $\lVert x\rVert=\lVert y\rVert=1$, and finally step 0(iii),

the last equality using $\lVert y-x\rVert=d(y,x)=d(x,y)$ by the symmetry of a metric.

Now let $\varepsilon$ be a real number with $0\le\varepsilon$ and $\varepsilon\ne 0$. By (O1) and (O2), $0\le 1+D$; and $1+D\ne 0$, since $1=1+0\le 1+D$ by the first order axiom, so $1+D=0$ would give $1\le 0$ and hence $1=0$ by (O1) and antisymmetry. Put $\delta=\varepsilon(1+D)^{-1}$. By (O4) we have $0\le(1+D)^{-1}$ and $(1+D)^{-1}\ne 0$, so $0\le\delta$ by the second order axiom, and $\delta\ne 0$ because $\delta=0$ would give $\varepsilon=\delta(1+D)=0$.

Let $y\in S$ with $d(x,y)<\delta$. By (O1) and the first order axiom, $D\le 1+D$, so (O5) with the nonnegative factor $d(x,y)$ gives $D\,d(x,y)\le(1+D)d(x,y)$; and step 0(ii) gives $(1+D)d(x,y)<(1+D)\delta=\varepsilon$. Combining with the displayed bound and step 0(i) twice,

This is exactly the continuity hypothesis of Extreme Value Theorem on a Compact Subset of a Metric Space.

Step 3: a maximiser. By Extreme Value Theorem on a Compact Subset of a Metric Space applied to the metric space $(V,d)$, the nonempty compact set $S$ and the map $R_{T}$, there is $x_{0}\in S$ with

Put $\lambda=R_{T}(x_{0})$, a real number by (B1).

Step 4: the operator $\lambda\,\mathrm{id}-T$. Let $A$ send $x\in V$ to $A(x)=\lambda x-T(x)$. It is a linear operator on $V$, since $T$ is and by the axioms of a vector space. Since $\lambda$ is real, $\overline{\lambda}=\lambda$ by (M1), so for $u,v\in V$, using the identity of step 2, conditions 1 and 3, and self-adjointness of $T$,

so $A$ is self-adjoint.

We check that $A$ is positive semi-definite. For $x\in V$, condition 3 gives $\langle x,A(x)\rangle=\lambda\langle x,x\rangle-\langle x,T(x)\rangle$, a real number by condition 4 and (B1).

If $x=0_{V}$, then $T(0_{V})=0_{V}$ by linearity and $\lambda 0_{V}=0_{V}$ by claim 4 of Elementary Identities in a Vector Space, so $A(0_{V})=0_{V}$; and $\langle 0_{V},0_{V}\rangle=\langle 0_{V},0\cdot 0_{V}\rangle=0\,\langle 0_{V},0_{V}\rangle=0$ by claim 3 of that lemma and condition 3. So $\langle x,A(x)\rangle=0$.

If $x\ne 0_{V}$, put $t=\lVert x\rVert$; by the positivity of a norm $t$ is real with $0\le t$ and $t\ne 0$. By (O4), $0\le t^{-1}$ and $t^{-1}\ne 0$, so by claim 8 of Properties of the Absolute Value in an Ordered Field and absolute homogeneity of the norm, $\lVert t^{-1}x\rVert=|t^{-1}|\,t=t^{-1}t=1$; thus $y=t^{-1}x$ lies in $S$, and $ty=x$. Applying (I1) to the orthonormal $1$-tuple with component $y$ and the coefficient $1$ gives $\langle y,y\rangle=1$, so

which satisfies $0\le\lambda-R_{T}(y)$ by step 3 and (O3). Using the identity of step 2, conditions 1 and 3, $\overline{t}=t$ from (M1), and linearity of $A$,

The second order axiom gives $0\le t\,t$ and then $0\le\langle x,A(x)\rangle$.

Step 5: conclusion. By (I1) again, $\langle x_{0},x_{0}\rangle=1$, so $\langle x_{0},A(x_{0})\rangle=\lambda-R_{T}(x_{0})=0$. By claim 2 of Cauchy-Schwarz Inequality for a Positive Semi-Definite Self-Adjoint Operator, $A(x_{0})=0_{V}$, that is, $T(x_{0})=\lambda x_{0}$. Finally $x_{0}\ne 0_{V}$, since $\langle x_{0},x_{0}\rangle=1\ne 0=\langle 0_{V},0_{V}\rangle$. Hence $x_{0}$ is a unit vector that is an eigenvector of $T$ with the real eigenvalue $\lambda$.