Properties of Adjoint of a Linear Map

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 \(v\): \(Tv=\lambda v\)
    • Some linear maps are diagonalizable
  • Inner products, norms, orthogonality and orthonormal basis
    • Upper triangular matrix for a linear map over an orthonormal basis
    • Orthogonal projection gives closest vector in the subspace
    • Least squares solution to a linear equation is orthogonal projection
  • Adjoint of a linear map: \(\langle Tv,w\rangle=\langle v,T^*w\rangle\)

Properties

\(V,W,U\): finite-dimensional inner product spaces over \(F=\mathbb{R}\) or \(\mathbb{C}\)

  1. \((S+T)^* = S^* + T^*\), where \(S,T:V\to W\) are linear maps.

  2. \((\lambda T)^* = \bar{\lambda}T^*\), where \(\lambda\in F\).

  3. \((T^*)^*=T\).

  4. \(I^*=I\), where \(I\) is the identity operator.

  5. \((ST)^* = T^* S^*\), where \(S:W\to U\) and \(T:V\to W\) are linear maps.

Sample Proofs

  1. \(\langle v,(S+T)^* w\rangle=\langle (S+T)v,w\rangle=\langle Sv,w\rangle+\langle Tv,w\rangle=\langle v,S^* w\rangle+\langle v,T^* w\rangle=\langle v,(S^* +T^*)w\rangle\)

  2. \(\langle v,(\lambda T)^* w\rangle=\langle \lambda T v, w\rangle=\lambda\langle T v, w\rangle=\lambda\langle v, T^* w\rangle=\langle v, \bar{\lambda} T^* w\rangle\)

Null and range of adjoint

\(T:V\to W\), a linear map and \(T^*:W\to V\), adjoint of \(T\)

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

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

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

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

  5. dim range \(T=\) dim range \(T^*\)

Proof

\(w\in\) null \(T^*\)

iff \(T^* w=0\), iff \(\langle v,T^* w\rangle=0\) for all \(v\in V\), iff \(\langle Tv,w\rangle=0\) for all \(v\in V\)

iff \(w\in\) \((\)range \(T)^{\perp}\)

(4) is complement of (1), (3) is (1) with \(T\) set as \(T^*\), (2) is complement of (3)

For (5), use fundamental theorem of linear maps on (3)

Conjugate transpose: Matrix of adjoint

\(V,W\): finite-dimensional inner product spaces over \(F=\mathbb{R}\) or \(\mathbb{C}\)

\(T:V\to W\), a linear map

Orthonormal basis for \(V\): \(B_V=\{e_1,\ldots,e_n\}\)

Orthonormal basis for \(W\): \(B_W=\{f_1,\ldots,f_m\}\)

Matrix of \(T\) w.r.t. \(B_V\) and \(B_W\) is \(M(T,B_V,B_W)\)

\[M(T^*,B_W,B_V)=\text{conjugate-transpose}(M(T,B_V,B_W))\]

Conjugate-transpose of a matrix: transpose and conjugate each element

Proof

\((i,j)\)-th entry of \(M(T,B_V,B_W)\): \(\langle Te_j,f_i\rangle=\langle e_j,T^* f_i\rangle=\overline{\langle T^* f_i,e_j\rangle}\)

\((j,i)\)-th entry of \(M(T^* ,B_W,B_V)\): \(\langle T^* f_i,e_j\rangle\)

Adjoint and four fundamental spaces of a matrix

\(A\): \(m\times n\) matrix over \(\mathbb{R}\)

Represents \(T:\mathbb{R}^n\to \mathbb{R}^m\) w.r.t. standard basis

\(T^*\): represented by the conjugate-transpose\((A)\) or \(\overline{A^T}\)

Since entries of \(A\) are real: \(\overline{A^T}=A^T\)

  1. range \(T^* =\) colspace \(A^T =\) rowspace \(A =\) \((\)null \(A)^{\perp}\)

  2. null \(T^* =\) null \(A^T =\) left-null \(A =\) \((\)colspace \(A)^{\perp}\)

  3. range \(T =\) colspace \(A =\) rowspace \(A^T =\) \((\)left-null \(A)^{\perp}\)

  4. null \(T =\) null \(A =\) left-null \(A^T =\) \((\)rowspace \(A)^{\perp}\)

Composition of operators and their null spaces

\(T:V\to W\) and \(S:W\to U\) are linear maps

\(ST:V\to U\) is a linear map

Suppose \(N_W=\) range \(T\cap\) null \(S\) (a subspace of \(W\))

dim null \(ST=\) dim null \(T+\) dim \(N_W\)

Proof

\(v\in\) null \(ST\) iff \(S(Tv)=0\) iff \(Tv\in\) null \(S\) iff \(Tv\in N_W\)

null \(ST=\{v:Tv\in N_W\}\)

Basis for null \(T\): \(\{v_1,\ldots,v_k\}\)

Basis for \(N_W\): \(\{w_1,\ldots,w_r\}\), \(r=\) dim \(N_W\)

Let \(v_{k+i}\in V\) be s.t. \(Tv_{k+i}=w_i\), \(i=1,\ldots,r\)

Basis for null \(ST\): \(\{v_1,\ldots,v_k,v_{k+1},\ldots,v_{k+r}\}\)

Composition of linear maps: special case

\(T:V\to W\) and \(S:W\to U\) are linear maps

\(ST:V\to U\) is a linear map

if range \(T\) and null \(S\) intersect only at 0, null \(ST=\) null \(T\)

  • \(S\) retains all “details” of \(T\)

    • \(S\) restricted to range \(T\) is one-to-one

  • By fundamental theorem of linear maps, dim range \(ST=\) dim range \(T\)

The operators \(TT^*\) and \(T^*T\)

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

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

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

So, null \(T^*\) and range \(T\) intersect only at \(0\)

Therefore, null \(T^* T=\) null \(T\) and dim range \(T=\) dim range \(T^* T=\) dim range \(T^*\)

\(TT^*: W\to W\)

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

So, null \(T\) and range \(T^*\) intersect only at \(0\)

Therefore, null \(TT^* =\) null \(T^*\) and dim range \(T=\) dim range \(TT^*=\) dim range \(T^*\)