Proof of The Orthogonal Complement of a Unit Vector

Write (I$k$) for claim $k$ of Elementary Properties of an Orthonormal Family, (P$k$) for claim $k$ of The Span of a Finite Family is the Smallest Subspace Containing It, (F$k$) for claim $k$ of Properties of Finite Sums of Vectors, (Q$k$) for claim $k$ of A Linear Subspace is a Vector Space and Inherits an Inner Product, and (V$k$) for claim $k$ of Elementary Identities in a Vector Space. Conditions on an inner product are numbered as in Complex Inner Product Space, and the axioms of a vector space are used freely. Sums of vectors are finite sums.

The $1$-tuple $\tilde{u}$ is orthonormal, since its only component is a unit vector and $[1]$ contains no two distinct indices. By (P1), $u\in U$, and by the definition of the span together with (F1) the elements of $U$ are exactly the vectors $cu$ with $c$ a complex number.

We also record that $\langle y,0_{V}\rangle=\langle y,0\cdot 0_{V}\rangle=0\,\langle y,0_{V}\rangle=0$ for every $y\in V$, by (V3) and condition 3.

A vector lies in $W$ exactly when $\langle u,\cdot\rangle$ kills it. If $y\in W$ then $\langle u,y\rangle=0$ because $u\in U$. Conversely suppose $\langle u,y\rangle=0$ and let $cu\in U$. By conditions 1 and 3 and claim 1 of Properties of Complex Conjugation and Modulus,

so $y\in W$.

Claim 1. Existence. Let $x\in V$. Apply (I4) to the orthonormal tuple $\tilde{u}$ and the vector $x$: with $p=\sum_{k=1}^{1}\langle u,x\rangle u=\langle u,x\rangle u$ by (F1), and $w=x-p$, one gets $x=w+p$ and $\langle u,w\rangle=0$, so $w\in W$ by the criterion above and $x=\langle u,x\rangle u+w$.

Uniqueness. Suppose $\lambda u+w=\lambda' u+w'$ with $w,w'\in W$ and $\lambda,\lambda'$ complex. Applying $\langle u,\cdot\rangle$ and using conditions 2 and 3 together with $\langle u,w\rangle=\langle u,w'\rangle=0$,

By (I1) applied to $\tilde{u}$ with the $1$-tuple of coefficients $1$ we have $\langle u,u\rangle=1$, so $\lambda=\lambda'$ and hence $w=w'$. In particular the pair produced above is the only one, and its first entry is $\langle u,x\rangle$.

Claim 2. Let $n=1$. By the definition of the dimension there is a basis $b\in V^{1}$ of $V$, so by spanning and (F1) every $x\in V$ satisfies $x=cb_{1}$ for some complex $c$. Also $u\ne 0_{V}$: otherwise $\langle u,u\rangle=\langle 0_{V},0_{V}\rangle=0$ by the record above, contradicting $\langle u,u\rangle=1$ and $1\ne 0$. Write $u=cb_{1}$; then $c\ne 0$, since otherwise $u=0\,b_{1}=0_{V}$ by (V3). Hence $b_{1}=c^{-1}u\in U$, so (P3) gives $V=\operatorname{span}(b)\subseteq U$ and therefore $U=V$.

Now let $x\in W$. Since $x\in U$, the definition of the orthogonal complement gives $\langle x,x\rangle=0$, so $x=0_{V}$ by condition 4. Thus $W=\{0_{V}\}$.

Claim 3. Let $n=r+1$.

$W\ne\{0_{V}\}$. If $W=\{0_{V}\}$, then claim 1 gives $x=\langle u,x\rangle u$ for every $x\in V$, so $\tilde{u}$ spans $V$; being orthonormal it is linearly independent by (I3), hence a basis of $V$ of length $1$. By claim 2 of Orthonormal Bases and Basis Size in a Finite-Dimensional Inner Product Space and the definition of the dimension, $n=1$. But $n=r+1$ is a successor and $1$ is not a successor, by Natural Numbers, a contradiction.

A spanning tuple of $W$. By claim 1 of Orthonormal Bases and Basis Size in a Finite-Dimensional Inner Product Space and the definition of the dimension there is an orthonormal basis $a\in V^{n}$ of $V$. Let $P$ assign to each $x\in V$ the vector $P(x)=x-\langle u,x\rangle u$, which lies in $W$ by claim 1. Conditions 2 and 3, together with the vector space axioms, make $P$ a linear map from $V$ to $V$. Let $g\in W^{n}$ be the tuple with $g_{k}=P(a_{k})$.

By (Q1) the set $W$ is a vector space under the operations of $V$, and by (Q2) finite sums formed in $W$ agree with those formed in $V$. Let $y\in W$. Since $a$ spans $V$ there is a tuple $c$ of complex numbers with $y=\sum_{k=1}^{n}c_{k}a_{k}$, so by (F4) and the linearity of $P$,

On the other hand $\langle u,y\rangle=0$, so $P(y)=y-0\,u=y$ by (V3). Hence $g$ spans the vector space $W$.

An orthonormal basis of $W$. Since $W\ne\{0_{V}\}$, A Finite Spanning Family Contains a Basis applied to $W$ and $g$ gives a natural number $s$ and a basis $h\in W^{s}$ of $W$. By (Q3) the set $W$ with the restricted inner product is a complex inner product space, so Gram-Schmidt Orthonormalisation applied to the linearly independent tuple $h$ gives an orthonormal $f\in W^{s}$ whose span in $W$ equals that of $h$. As $h$ spans $W$, so does $f$; and $f$ is linearly independent by (I3). Hence $f$ is an orthonormal basis of $W$.

Adjoining $u$. Let $e\in V^{s+1}$ have $e_{k}=f_{k}$ for $k\in[s]$ and $e_{s+1}=u$. By (Q3) the inner product of $W$ is the restriction of that of $V$, so every component of $f$ is a unit vector of $V$ and distinct components of $f$ are orthogonal in $V$; and $u$ is a unit vector. Let $i,j\in[s+1]$ be distinct. If both lie in $[s]$, then $e_{i}$ and $e_{j}$ are orthogonal. Otherwise exactly one of them equals $s+1$, by claim 5 of Properties of the Order on the Natural Numbers; and for $k\in[s]$ we have $f_{k}\in W$, so $\langle u,f_{k}\rangle=0$, while $\langle f_{k},u\rangle=\overline{0}=0$ by condition 1 and claim 1 of Properties of Complex Conjugation and Modulus. Hence $e$ is orthonormal, and linearly independent by (I3).

Let $x\in V$. By claim 1, $x=\langle u,x\rangle u+w$ with $w\in W$, and $w=\sum_{k=1}^{s}d_{k}f_{k}$ for some tuple $d$ of complex numbers, the sum being the same in $W$ and in $V$ by (Q2). Let $c$ be the tuple of complex numbers on $[s+1]$ with $c_{k}=d_{k}$ for $k\in[s]$ and $c_{s+1}=\langle u,x\rangle$. By the recursion and restriction parts of (F1),

the last equality by commutativity of addition in $V$. So $e$ spans $V$ and is therefore an orthonormal basis of $V$.

The size. Both $a\in V^{n}$ and $e\in V^{s+1}$ are bases of $V$, so $s+1=n=r+1$ by claim 2 of Orthonormal Bases and Basis Size in a Finite-Dimensional Inner Product Space. The successor map is injective by Natural Numbers, so $s=r$. Thus $f\in W^{r}$ and $e\in V^{r+1}$ are as asserted.