Proof of Rules for Orthogonal Matrices: Inverse, Products, Isometry, Columns, Determinant, Orthonormal Extension and Expansion

Each result cited is universally quantified over the data in its own statement. We work in Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices. By Orthogonal Real Matrices §membership, a matrix $P\in\mathcal{M}_{n}(\mathbb{R})$ lies in $\mathrm{O}(n)$ if and only if $P^{\top}P=I_{n}$; in particular $Q^{\top}Q=I_{n}$ and $R^{\top}R=I_{n}$.

Clause inverse. Since $Q^{\top}Q=I_{n}$, Rules for Matrix Inverses §one-sided with $Q$ in place of $A$ and $Q^{\top}$ in place of $C$ shows that $Q$ is invertible and $Q^{-1}=Q^{\top}$. By Inverse Matrix and Invertible Real Square Matrix §invertible, $Q^{-1}$ is an inverse of $Q$, so $QQ^{-1}=I_{n}$ by Inverse Matrix and Invertible Real Square Matrix §inverse; that is, $QQ^{\top}=I_{n}$. Finally $(Q^{\top})^{\top}=Q$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §transpose, so $(Q^{\top})^{\top}Q^{\top}=QQ^{\top}=I_{n}$ and $Q^{\top}\in\mathrm{O}(n)$.

Clause product. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §transpose and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §identity, $I_{n}^{\top}I_{n}=I_{n}I_{n}=I_{n}$, so $I_{n}\in\mathrm{O}(n)$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §transpose, $(QR)^{\top}=R^{\top}Q^{\top}$; then by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §associative, applied twice in the second equality, and by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §identity,

$$(QR)^{\top}(QR)=\bigl(R^{\top}Q^{\top}\bigr)(QR)=R^{\top}\bigl((Q^{\top}Q)R\bigr)=R^{\top}(I_{n}R)=R^{\top}R=I_{n},$$

so $QR\in\mathrm{O}(n)$.

Clause isometry. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §adjoint with $m=n$, $A=Q$ and $w=Qy$, and by the rules $(AB)z=A(Bz)$ and $I_{n}x=x$ of Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector,

$$(Qx)\cdot(Qy)=x\cdot\bigl(Q^{\top}(Qy)\bigr)=x\cdot\bigl((Q^{\top}Q)y\bigr)=x\cdot(I_{n}y)=x\cdot y.$$

With $x$ in place of $y$ this gives $(Qx)\cdot(Qx)=x\cdot x$. By Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §square, $\lVert Qx\rVert\ge0$ and $\lVert Qx\rVert^{2}=(Qx)\cdot(Qx)=x\cdot x$, while $\lVert x\rVert$ is the only nonnegative real number whose square is $x\cdot x$; hence $\lVert Qx\rVert=\lVert x\rVert$.

Clause columns. Let $P\in\mathcal{M}_{n}(\mathbb{R})$ and $i,j\in[n]$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §column, $(Pe_{i})_{k}=P_{ki}$ and $(Pe_{j})_{k}=P_{kj}$ for every $k\in[n]$; so by The Dot Product on $\mathbb{R}^n$ §dot-product, Transpose of a Real Matrix §transpose and Product of Real Matrices §product,

$$(Pe_{i})\cdot(Pe_{j})=\sum_{k=1}^{n}P_{ki}P_{kj}=\sum_{k=1}^{n}(P^{\top})_{ik}P_{kj}=(P^{\top}P)_{ij}.\tag{1}$$

By Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality and Identity Matrix and Standard Basis Vectors §identity, $P^{\top}P=I_{n}$ holds if and only if $(P^{\top}P)_{ij}=\delta_{ij}$ for all $i,j\in[n]$, that is, by (1), if and only if $(Pe_{i})\cdot(Pe_{j})=\delta_{ij}$ for all $i,j\in[n]$. This proves clause columns.

Clause from-columns. Let $u$ be as in clause from-columns. The map $(i,j)\mapsto(u_{j})_{i}$ from $[n]\times[n]$ to $\mathbb{R}$ is a matrix $P\in\mathcal{M}_{n}(\mathbb{R})$ with $P_{ij}=(u_{j})_{i}$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §matrices and Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §entries, and it is the only matrix with these entries by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality. For $i,j\in[n]$, $(Pe_{j})_{i}=P_{ij}=(u_{j})_{i}$ by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §column, so $Pe_{j}=u_{j}$ by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §equality. Hence $(Pe_{i})\cdot(Pe_{j})=u_{i}\cdot u_{j}=\delta_{ij}$ for all $i,j\in[n]$, and $P\in\mathrm{O}(n)$ by clause columns.

Clause determinant. By Rules for the Determinant: Rows, Scalar Multiples and the Transpose §identity, by The Determinant is Multiplicative §multiplicative with $Q^{\top}$ and $Q$ in place of $A$ and $B$, and by Rules for the Determinant: Rows, Scalar Multiples and the Transpose §transpose,

$$1=\det I_{n}=\det(Q^{\top}Q)=\det(Q^{\top})\cdot\det Q=(\det Q)^{2}.$$

Hence $(\det Q-1)(\det Q+1)=0$, so $\det Q=1$ or $\det Q=-1$.

Clause extension. Let $k$ and $u$ be as stated.

Let first $k=0$. Then $[k]$ is empty, and the condition on $v$ reduces to $\lVert v\rVert=1$. Since $n\ge1$, $1\in[n]$; let $v=e_{1}$. By Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §square and Identity Matrix and Standard Basis Vectors §basis, $\lVert e_{1}\rVert\ge0$ and $\lVert e_{1}\rVert^{2}=\sum_{i=1}^{n}\delta_{i1}^{2}=1$; as $1$ is the only nonnegative real number whose square is $1$, $\lVert e_{1}\rVert=1$.

Let now $k\ge1$, so $k\in\mathbb{N}$ and $k<n$. The map $(i,j)\mapsto(u_{i})_{j}$ from $[k]\times[n]$ to $\mathbb{R}$ is a matrix $U\in\mathcal{M}_{k\times n}(\mathbb{R})$ with entries $U_{ij}=(u_{i})_{j}$, by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §matrices and Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §entries. By A Homogeneous Linear System with More Unknowns than Equations Has a Nonzero Solution §nontrivial with $k$ in place of $m$ and $U$ in place of $A$, there is $w\in\mathbb{R}^{n}$ with $w\neq0$ and $Uw=0$, the origin of $\mathbb{R}^{k}$, all of whose components are $0$ by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §origin. For $i\in[k]$, by Matrix-Vector Product §product and The Dot Product on $\mathbb{R}^n$ §dot-product,

$$0=(Uw)_{i}=\sum_{j=1}^{n}(u_{i})_{j}w_{j}=u_{i}\cdot w.$$

By Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §vanishing and Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §square, $\lVert w\rVert\neq0$ and $\lVert w\rVert\ge0$, so $\lVert w\rVert>0$. Let $t=1/\lVert w\rVert$, so $t>0$, and let $v=tw$. By Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §homogeneity, $\lVert v\rVert=|t|\,\lVert w\rVert=t\lVert w\rVert=1$. For $i\in[k]$, by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §symmetric and The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §second, $v\cdot u_{i}=u_{i}\cdot(tw)=t\,(u_{i}\cdot w)=0$. This proves the first sentence of the clause.

For the second sentence, suppose moreover that $u_{i}\cdot u_{j}=\delta_{ij}$ for all $i,j\in[k]$, let $v\in\mathbb{R}^{n}$ satisfy $\lVert v\rVert=1$ and $v\cdot u_{i}=0$ for every $i\in[k]$, and let $u'$ be the map of the clause, given by Finite Sets: Removing an Element, Injective Self-Maps, Infinite Sets, Sets of Permutations, Pairs and Extended Tuples §extension with $\mathbb{R}^{n}$ in place of $Y$, $k$ in place of $n$, $u$ in place of $t$ (an element of $(\mathbb{R}^{n})^{k}$ by Tuples in a Set: the Set of n-Tuples and Their Components §membership) and $v$ in place of $y$. Since $k+1\le n$, $[k+1]\subseteq[n]$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §inclusion, so $\delta_{ij}$ is defined for $i,j\in[k+1]$. Let $i,j\in[k+1]$. By Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting §successor, $[k+1]=[k]\cup\{k+1\}$ and $k+1\notin[k]$, so each of $i$ and $j$ either lies in $[k]$ or equals $k+1$. If $i,j\in[k]$, then $u'_{i}\cdot u'_{j}=u_{i}\cdot u_{j}=\delta_{ij}$. If $i\in[k]$ and $j=k+1$, then $i\neq j$, and by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §symmetric and Identity Matrix and Standard Basis Vectors §identity, $u'_{i}\cdot u'_{j}=u_{i}\cdot v=v\cdot u_{i}=0=\delta_{ij}$; if $i=k+1$ and $j\in[k]$, then likewise $u'_{i}\cdot u'_{j}=v\cdot u_{j}=0=\delta_{ij}$. If $i=j=k+1$, then by Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square §square and Identity Matrix and Standard Basis Vectors §identity, $u'_{i}\cdot u'_{j}=v\cdot v=\lVert v\rVert^{2}=1=\delta_{ij}$.

Clause expansion. Let $z=Q^{\top}x\in\mathbb{R}^{n}$. By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector and clause inverse, $Qz=Q(Q^{\top}x)=(QQ^{\top})x=I_{n}x=x$. For $i\in[n]$, by Matrix-Vector Product §product, Transpose of a Real Matrix §transpose, Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §column and The Dot Product on $\mathbb{R}^n$ §dot-product,

$$z_{i}=\sum_{l=1}^{n}(Q^{\top})_{il}x_{l}=\sum_{l=1}^{n}x_{l}Q_{li}=\sum_{l=1}^{n}x_{l}(Qe_{i})_{l}=x\cdot(Qe_{i}).$$

Hence, for $l\in[n]$, by Matrix-Vector Product §product and Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §column in the first three equalities, and in the last by Rules for Pointwise Operations on Real-Valued Maps: a Commutative Ring, Scalar Multiples and Evaluation §evaluation with $[n]$ in place of $I$ and of $A$, the map $i\mapsto\bigl(x\cdot(Qe_{i})\bigr)\,Qe_{i}$ from $[n]$ to $\mathbb{R}^{[n]}$ in place of $F$ and $l$ in place of $p$, which gives $\bigl(\sum_{i=1}^{n}\bigl(x\cdot(Qe_{i})\bigr)\,Qe_{i}\bigr)_{l}=\sum_{i=1}^{n}\bigl(\bigl(x\cdot(Qe_{i})\bigr)\,Qe_{i}\bigr)_{l}$, together with $(tw)_{l}=t\,w_{l}$ for $t\in\mathbb{R}$ and $w\in\mathbb{R}^{n}$ from Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §operations,

$$x_{l}=(Qz)_{l}=\sum_{i=1}^{n}Q_{li}z_{i}=\sum_{i=1}^{n}\bigl(x\cdot(Qe_{i})\bigr)(Qe_{i})_{l}=\Bigl(\sum_{i=1}^{n}\bigl(x\cdot(Qe_{i})\bigr)\,Qe_{i}\Bigr)_{l},$$

and the clause follows by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §equality.