Isometries

Andrew Thangaraj

Aug-Nov 2020

Recap

  • Vector space \(V\) over a scalar field \(F= \mathbb{R}\) or \(\mathbb{C}\)
  • \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
    • dim null \(T+\) dim range \(T=\) dim \(V\)
    • Solution to \(Ax=b\) (if it exists): \(u+\) null\((A)\)
  • Four fundamental subspaces of a matrix
    • Column space, row space, null space, left null space
  • Eigenvalue \(\lambda\) and Eigenvector \(v\): \(Tv=\lambda v\)
    • There is a basis w.r.t. which a linear map is upper-triangular
    • If there is a basis of eigenvectors, linear map is diagonal w.r.t. it
  • Inner products, norms, orthogonality and orthonormal basis
    • There is an orthonormal basis w.r.t. which a linear map is upper-triangular
    • Orthogonal projection: distance from a subspace
  • Adjoint of a linear map: \(\langle Tv,w\rangle=\langle v,T^*w\rangle\)
    • null \(T=\) \((\)range \(T^*)^{\perp}\)
  • Self-adjoint: \(T=T^*\), Normal: \(TT^*=T^*T\)
    • Eigenvectors corresponding to different eigenvalues are orthogonal
  • Complex/real spectral theorem: \(T\) is normal/self-adjoint \(\leftrightarrow\) orthonormal basis of eigenvectors
  • Positive operators: self-adjoint with non-negative eigenvalues

Rotation and reflection in two dimensions

Rotation by angle \(\theta\)

\(R=\begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}\)

Eigenvalues: \(e^{i\theta}\), \(e^{-i\theta}\)

No real eigenspace (no axis of rotation)

Reflection about \(x\)-axis

\(S=\begin{bmatrix} 1&0\\ 0&-1\end{bmatrix}\)

Eigenvalues: 1 and -1

Eigenspaces: \(x\)-axis is fixed, \(y\)-axis is flipped

Norms before and after transform
\(\lVert Rx\rVert=\lVert x\rVert\)

\(\lVert Sx\rVert=\lVert x\rVert\)

Definition of isometries

\(V\): inner product space over \(F=\mathbb{R}\) or \(\mathbb{C}\)

An operator \(T:V\to V\) is called an isometry if \(\lVert Tv\rVert=\lVert v\rVert\) for all \(v\)

Examples

  1. \(A=\begin{bmatrix} \lambda_1&&0\\ &\ddots&\\ 0&&\lambda_n \end{bmatrix}\) with \(\lvert\lambda_i\rvert=1\)

  2. \(A=\begin{bmatrix} 1&0&0\\ 0&\cos\theta&-\sin\theta\\ 0&\sin\theta&cos\theta \end{bmatrix}\)

Also known as

Real spaces: orthogonal operators

Complex spaces: unitary operators

When is a normal operator an isometry?

\(F=\mathbb{C}\), \(T\): normal operator

\(T=\lambda_1 e_1\overline{e^T_1}+\cdots+\lambda_n e_n\overline{e^T_n}\)

\(\{e_1,\ldots,e_n\}\): orthonormal

Let \(x=x_1e_1+\cdots+x_ne_n\), \(\lVert x\rVert^2=\lvert x_1\rvert^2+\cdots+\lvert x_n\rvert^2\)

\(\lVert Tx\rVert^2=\lvert\lambda_1\rvert^2\lvert x_1\rvert^2+\cdots+\lvert\lambda_n\rvert^2\lvert x_n\rvert^2\)

A normal operator is an isometry if every eigenvalue has absolute value 1.

Interesing result: There are no other isometries

Characterising isometries

\(V\): inner product space over \(F=\mathbb{R}\) or \(\mathbb{C}\)

Suppose \(S:V\to V\) is an operator. The following are equivalent:

  1. \(S\) is an isometry
  2. \(\langle Su,Sv\rangle=\langle u,v\rangle\)
  3. If \(\{e_1,\ldots,e_n\}\) is orthonormal, so is \(\{Se_1,\ldots,Se_n\}\)
  4. There exists one orthonormal basis \(\{e_1,\ldots,e_n\}\) s.t. \(\{Se_1,\ldots,Se_n\}\) is orthonormal
  5. \(S^*S=SS^*=I\) (\(S\) is invertible and \(S^{-1}=S^*\))
  6. \(S^*\) is an isometry

Proof of (1) implies (2)

Real: \(\langle u,v\rangle=\dfrac{\lVert u+v\rVert^2-\lVert u-v\rVert^2}{4}\)

Complex: \(\langle u,v\rangle=\dfrac{\lVert u+v\rVert^2-\lVert u-v\rVert^2+i\lVert u+iv\rVert^2-i\lVert u-iv\rVert^2}{4}\)

Proof (continued)

Proof of (2) implies (3), (4)

\(\langle Se_i,Se_j\rangle=\langle e_i,e_j\rangle\)

Proof of (4) implies (5)

\(\langle S^*Se_i,e_j\rangle=\langle Se_i,Se_j\rangle=\langle e_i,e_j\rangle\) for all \(i,j\)

For \(u=u_1e_1+\cdots+u_ne_n\), \(v=v_1e_1+\cdots+v_ne_n\),

\(\begin{gather} \langle S^*Su,v\rangle\\[5pt] =u_1\overline{v_1}\langle S^*Se_1,e_1\rangle+\cdots+u_i\overline{v_j}\langle S^*Se_i,e_j\rangle+\cdots+u_n\overline{v_n}\langle S^*Se_n,e_n\rangle\\[5pt] =u_1\overline{v_1}\langle e_1,e_1\rangle+\cdots+u_i\overline{v_j}\langle e_i,e_j\rangle+\cdots+u_n\overline{v_n}\langle e_n,e_n\rangle\\[5pt] =\langle u,v\rangle\end{gather}\)

So, \(S^*S=I\), which implies \(SS^*=I\), \(S\) is invertible and \(S^{-1}=S^*\)

Completing the proof and corollaries

Proof of (5) implies (6), (1)

\(\lVert S^*v\rVert^2=\langle S^*v,S^*v\rangle=\langle SS^*v,v\rangle=\langle v,v\rangle=\lVert v\rVert^2\)

\(\lVert Sv\rVert^2=\langle Sv,Sv\rangle=\langle S^*Sv,v\rangle=\langle v,v\rangle=\lVert v\rVert^2\)

Corollary: Every isometry is normal

Proof: \(SS^*=S^*S=I\)

Corollary: A matrix represents an isometry if and only if (1) rows and columns have unit norm, (2) any two rows or columns are orthogonal.

Isometries in complex spaces

\(V\): complex inner product space, \(S:V\to V\)

\(S\) is an isometry if and only if there exists an orthonormal basis of eigenvectors of \(S\) with every eigenvalue having an absolute value 1.

Proof

Converse

Normal with \(\lvert\lambda_i\rvert=1\) implies isometry (example)

Forward result

If \(S\): isometry, then \(S\) is normal and has an orthonormal eigenvector basis \(\{e_1,\ldots,e_n\}\)

Let \(Se_i=\lambda_i e_i\)

Taking norms,

\(\lvert\lambda_i\rvert=\lVert\lambda_i e_i\rVert=\lVert Se_i\rVert=\lVert e_i\rVert = 1\)