Adjoint of an Operator and Operator-Adjoint Product

Andrew Thangaraj

Aug-Nov 2020

Recap

  • Vector space \(V\) over a scalar field \(F\)
    • \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course
  • \(m\times n\) 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 \(v\): \(Tv=\lambda v\)
    • 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 \(T^*)^{\perp}\)

Eigenvalues of \(ST\) and \(TS\)

\(S:V\to W\), \(T:W\to V\)

\(ST:W\to W\) and \(TS: V\to V\) are operators

If \(\lambda\ne0\) is an eigenvalue of \(ST\), then it is an eigenvalue of \(TS\) as well.

Proof

\(v\): \(STv=\lambda v\), \(\lambda\ne0\)

Then, \(Tv\ne0\),

and \(\lambda Tv=T(\lambda v)=T(STv)=(TS)Tv\)

So, \(Tv\): eigenvector of \(TS\) with eigenvalue \(\lambda\)

More on \(AB\) and \(BA\)

\(A\): \(m\times n\) and \(B\): \(n\times m\)

\(A\), \(B\): represent suitable linear maps

\(AB\): \(m\times m\), \(BA\): \(n\times n\)

\(t^n\text{det}(tI_m-AB)=t^m\text{det}(tI_n-BA)\)

Proof

\(C=\begin{bmatrix} tI_m&A\\ B&I_n \end{bmatrix}\), \(D=\begin{bmatrix} I_m&0\\ -B&tI_n \end{bmatrix}\)

det\((CD)=\) det\((DC)\)

det\(\begin{bmatrix} tI_m-AB&A\\ 0&tI_n \end{bmatrix}=\) det\(\begin{bmatrix} tI_m&A\\ 0&tI_n-BA \end{bmatrix}\)

\(n>m\)

\(\{\)Eigenvalues of \(BA\}=\{\)Eigenvalues of \(AB\}\cup(n-m)\) zeros

Operator and its adjoint

\(V\): finite-dimensional inner product space

\(T:V\to V\), an operator

\(T^*:V\to V\), adjoint of \(T\)

Properties

  • \(\langle Tu,v\rangle=\langle u,T^*v\rangle\)

  • null \(T^*=\) \((\)range \(T)^{\perp}\)

  • dim range \(T=\) dim range \(T^*\)

  • dim null \(T=\) dim null \(T^*\)

Eigenvalues

\(T:V\to V\), an operator

if \(\lambda\) is an eigenvalue of \(T\), \(\bar{\lambda}\) is an eigenvalue of \(T^*\)

Proof

dim range \((T-\lambda I)<\) dim \(V\)

\((T-\lambda I)^* = T^*-\bar{\lambda}I\)

Since dim range \((T-\lambda I)=\) dim range \((T^*-\bar{\lambda}I)\),

dim range \((T^*-\bar{\lambda}I)<\) dim \(V\)

Eigenvalues of \(TT^*\) and \(T^*T\)

\(T:V\to W\), \(T^*:W\to V\)

\(TT^*: W\to W\), \(T^*T: V\to V\)

Nonzero eigenvalues of \(TT^*\) and \(T^*T\) are the same

Differences only in additional zero eigenvalues for one of the two

Let dim \(V\ge\) dim \(W\)

Singular values of a linear map \(T\): eigenvalues of \(TT^*\)

Least squares, adjoint and pseudo-inverse

\(T:V\to W\), linear map and \(w\in W\)

Least squares: Solve \(\min\limits_v\,\lVert Tv-w\rVert\)

Solution: \(Tv-w\) is orthogonal to range \(T\), or

\(Tv-w\in\) \((\)range \(T)^{\perp}\)

Since null \(T^*=\) \((\)range \(T)^{\perp}\), we require

\(T^*(Tv-w)=0\)

Normal equation: \(T^*Tv=T^*w\)

Suppose \(T^*T\) is invertible

Pseudo-inverse of \(T\): \((T^* T)^{-1}T^*\)

Why? \(((T^* T)^{-1}T^*)T=I\)