Isometries in Vector Spaces

Def: Let $V$ be a dimensional inner product space over a field $\mathbb{F}$, and $T\in \mathcal{L}(V)$. If $\parallel Tv\parallel =\parallel v\parallel$ for all $v\in V$, then $T$ is an isometry. We call $T$ a unitary operator if $\mathbb{F}\mathbb{=}\mathbb{C}$ and an orthogonal operator if $\mathbb{F}\mathbb{=}\mathbb{R}$. In infinite dimensional vector spaces, if the operator that preserves norms, and is surjective, then it is an isometry.

Th: Let $T$ be linear operator on a finite dimensional inner product space $V$. Then following statements are equivalent:

• $T^{\ast }T=I$
• $T{T}^{\ast }=I$
• $⟨Tu,Tv⟩=⟨u,v⟩$ for all $u,v\in V$
• If $\beta$ is an orthonormal basis for $V$, then $T[\beta ]$ is an orthonormal basis for $V$.
• There exists on orthonormal basis $\beta$ for $V$ such that $T[\beta ]$ is an orthonormal basis for $V$.
• $\parallel Tv\parallel =\parallel v\parallel$ for all $v\in V$.
• $T^{\ast }$ is an isometry
• $T$ is invertible and $T^{-1}=T^{\ast }$

Cor: Let $T$ be linear operator on a finite dimensional real inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors of $T$ with corresponding eigenvalues of absolute value $1$ iff $T$ is orthogonal and self-adjoint.

Cor: Let $T$ be linear operator on a finite dimensional complex inner product space $V$. Then $V$ has an orthonormal basis of eigenvectors of $T$ with corresponding eigenvalues of absolute value $1$ iff $T$ is unitary.

Partial Isometries

Def: Let $V$ be a finite dimensional inner product space. A linear operator $U$ on $V$ is called a partial isometry if there exists a subspace $W$ of $V$ such that $\parallel U(w)\parallel =\parallel w\parallel$ for all $w\in W$, and $U(v)=0$ for $v\in W^{\mathrm{\perp }}$. This doesn’t imply that $W$ is $U-$invariant.

• $⟨Ux,Uy⟩=⟨x,y⟩$ , for all $x,y\in W$
• If $\beta$ is an orthonormal basis for $W$, then $U[\beta ]$ is also an orthonormal basis for $R(U)$.
• There exists a basis $\gamma$ for $V$ such that the first $\dim W$ columns of $[U{]}_{\gamma }$ form an orthonormal set and the remaining columns are zero.
• Let $\gamma$ be an orthonormal basis for $R(U{)}^{\mathrm{\perp }}$, and $\beta$ be an orthonormal basis of $W$, then $R[\beta ]\cup \gamma$ is an orthonormal basis for $V$.
• Suppose $\beta$ be an orthonormal basis of $W$, and $\gamma$ be an orthonormal basis for $R(U{)}^{\mathrm{\perp }}$. Let $T\in \mathcal{L}(V)$, such that if $v\in \beta$ then $TUv=v$ and if $v\in \gamma$, then $Tv=0$. Then $T=U^{\ast }$.
• $U^{\ast }$ is a a partial isometry.
• $U^{\ast }U$ is orthogonal projection on $W$.
• $U{U}^{\ast }U=U$

Def: A square matrix $A$ is called on orthogonal matrix if $A{A}^{\mathrm{\top }}=A^{\mathrm{\top }}A=I$ and unitary if $A{A}^{\ast }=A^{\ast }A=I$.

The set of all orthogonal transformations/matrices is called the Orthogonal Group, and the set of all unitary transformations/matrices is called the Unitary Group

Def: $A$ and $B$ are unitarily/orthogonally equivalent iff there exists a unitary/orthogonal matrix $P$ such that $A=P^{\ast }BP$. This forms an equivalence relation on ${\mathcal{M}}_n(\mathbb{C})/{\mathcal{M}}_n(\mathbb{R})$

Th: Let $A$ be a complex $n\times n$. Then $A$ is normal iff $A$ is unitarily equivalent to a diagonal matrix.

Th: Let $A$ be a real $n\times n$. Then $A$ is symmetric iff $A$ is orthogonally equivalent to a diagonal matrix.

Th: A matrix that is both unitary and upper triangular then it is diagonal.

Schur’s Theorem Again

Let $A\in {\mathcal{M}}_n(\mathbb{F})$ be a matrix whose characteristic polynomial splits over $\mathbb{F}$.

• If $\mathbb{F}\mathbb{=}\mathbb{C}$, then $A$ is unitarily equivalent to a complex upper triangular matrix
• If $\mathbb{F}\mathbb{=}\mathbb{R}$, then $A$ is orthogonally equivalent to a real upper triangular matrix.

Cor: Suppose $V$ is a complex inner product space and $T\in \mathcal{L}(V)$. The the following are equivalent:

• $T$ is an isometry
• There is an orthonormal basis of $V$ consisting of eigenvectors of $T$ whose corresponding eigenvalues all have absolute value $1$.

QR Factorization

Th: Let $A\in {\mathcal{M}}_{n\times m}(\mathbb{C})$ with linearly independent columns. Then $A$ can be factorized as $A=QR$, where $Q\in {\mathcal{M}}_{n\times m}(\mathbb{F})$ with orthogonal columns and $R\in {\mathcal{M}}_m(\mathbb{F})$ upper triangleular matrix with positive diagonal entries.

We can consider the $i$th column of $A$ as $a_i$, then we can apply the Gram-Schmidt orthonormalization process to $\{a_i{\}}_{i=1}^m$, the following vectors $\{q_i{\}}_{i=1}^m$ will form the matrix $Q$.

Similarly given the set of orthonormal vectors we can calculate the matrix $R$, as follows $R_{ij}=⟨q_i,a_j⟩$ for $1\le i\le j\le m$, otherwise $R_{ij}=0$.

If $A\in {\mathcal{M}}_n(\mathbb{C})$, then $Q$ will be a unitary/orthogonal matrix.

Given a linear system $Ax=b$ when is invertible. We can decompose $A=QR$, and solve the system $QRx=b$, or $Rx=Q^{\ast }b$, this an be solved easily since $R$ is upper triangleular.

Rigid Motions

Def: Let $V$be a real inner product space. A function $f:V\to V$ is called a rigid motion if for all $x,y\in V$

Th: Let $f:V\to V$ be a rigid motion on a finite dimensional real inner product space $V$. The there exists a unique orthogonal operator $T$ on $V$ and a unique translation $g$ on $V$ such that $f=g\circ T$.

Th: Let $f:V\to V$ be a rigid motion then $f$ is invertible and $f^{-1}$ is also a rigid motion.

Cor: Th: Let $f:V\to V$ be a rigid motion on a finite dimensional real inner product space $V$. The there exists a unique orthogonal operator $T$ on $V$ and a unique translation $g$ on $V$ such that $f=T\circ g$.

Orthogonal Operators on ${\mathbb{R}}^2$

Th: Let $T$ be an orthogonal operator in ${\mathbb{R}}^2$. Then exactly one of the following conditions is satisfied:

• $T$ is a rotation, and $detT=1$
• $T$ is a reflection about a line through the origin, and $detT=-1$

Cor: Any rigid motion on ${\mathbb{R}}^2$ is either a rotation followed by a translation or a reflection about a line through the origin followed by a translation.

Householder Operators

Def: Let $V$ be a finite dimensional complex/real inner product space, and let $u$ be a unit vector in $V$. Define the Householder operator $H_u:V\to V$ by $H_u(x)=x-2⟨x,u⟩u$ for all $x\in V$.

• $H_u$ is linear
• $H_u(x)=x$ iff $x$ is orthogonal to $u$
• $H_u(u)=-u$
• $H_u^{\ast }=H_u$ and $H_u^2=I$, and hence $H_u$ is a unitary/orthogonal operator. If $V$ is real inner product space, then $H_u$ is a reflection.
• $det{H}_u=-1$.
• $H_u$ has eigenvalues of $\pm 1$. We know that $E(1,H_u)=\mathrm{span}(u{)}^{\mathrm{\perp }}$, and $E(-1,H_u)=\mathrm{span}(u)$.

This type of operator can be used to calculate the $QR$ factorization of $n\times m$ matrix, when $n\le m$.