Self-adjoint Operator

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 \(v\): \(Tv=\lambda v\)
    • 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 \(T^*)^{\perp}\)

Definition of self-adjoint operators

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

An operator \(T:V\to V\) is said to be self-adjoint if \(T=T^*\).

In other words, \(\langle Tv,w\rangle=\langle v,Tw\rangle\) for \(v,w\in V\).

In terms of matrices

\(M(T)\): matrix of \(T\) w.r.t. an orthonormal basis

\(T\) is self-adjoint if \(M(T)\) is equal to its conjugate-transpose.

  • also called Hermitian in complex spaces and symmetric in real spaces

Two properties

If \(T\) is self-adjoint, null \(T=\) \((\)range \(T)^{\perp}\)

Proof

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

If \(T\) is self-adjoint, \(\langle Tv,v\rangle\in\mathbb{R}\)

Proof

\(\langle Tv,v\rangle=\langle v,T^*v\rangle=\langle v,Tv\rangle\)

and

\(\langle Tv,v\rangle=\overline{\langle v,Tv\rangle}\)

When are eigenvalues real?

\(F=\mathbb{R}\)

\(v\): eigenvector of \(T\) with eigenvalue \(\lambda\)

Since \(v\) is real and \(Tv\) is real, \(\lambda\) has to be real.

\(F=\mathbb{C}\)

Eigenvalues can be complex.

Example

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

Eigenvalues: \(i\), \(-i\)

Eigenvectors: \((1,i)\) for \(i\), and \((-1,i)\) for \(-i\)

Eigenvalues of self-adjoint operators

Every eigenvalue of a self-adjoint operator is real.

Proof

\(T=T^*\) and \(\lambda\) is an eigenvalue with eigenvector \(v\ne0\)

\(\lambda\lVert v\rVert^2=\langle \lambda v,v\rangle=\langle Tv,v\rangle=\langle v,Tv\rangle=\langle v,\lambda v\rangle=\bar{\lambda}\lVert v\rVert^2\)

\(\lambda=\bar{\lambda}\)

When is \(Tv\) always orthogonal to \(v\)?

Example

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

\(\langle Ax,x\rangle=0\) for all \(x\in\mathbb{R}^2\)

In real spaces, there are non-trivial \(T\) s.t. \(\langle Tv,v\rangle=0\) for all \(v\).

What about complex inner product spaces?

Example: \(\langle Ax,x\rangle=\langle (i,-1),(1,i)\rangle=2i\) for \(x=(1,i)\)

There exists \(x\in\mathbb{C}^2\) for which \(\langle Ax,x\rangle\ne0\).

Complex inner product spaces and \(\langle Tv,v\rangle\)

\(V\): inner product space over \(\mathbb{C}\).

If \(\langle Tv,v\rangle=0\) for all \(v\), then \(T=0\).

Proof

\(\langle Tv,v\rangle=0\) for all \(v\)

\(\begin{align} \langle Tu,w\rangle &= \dfrac{1}{4}\langle T(u+w),u+w\rangle-\dfrac{1}{4}\langle T(u-w),u-w\rangle\\ &+\dfrac{i}{4}\langle T(u+iw),u+iw\rangle-\dfrac{i}{4}\langle T(u-iw),u-iw\rangle \end{align}\)

So, \(\langle Tu,w\rangle=0\) for all \(u,w\)

Set \(w=Tu\) to get \(\lVert Tu\rVert=0\) for all \(u\)

So, \(T=0\)

Self-adjoint operators and \(\langle Tv,v\rangle\): Result I

\(V\): inner product space over \(\mathbb{C}\).

\(T\) is self-adjoint iff \(\langle Tv,v\rangle\in\mathbb{R}\) for all \(v\).

Proof

\(\langle Tv,v\rangle-\overline{\langle Tv,v\rangle}=\langle Tv,v\rangle-\langle v,Tv\rangle=\langle (T-T^*)v,v\rangle\)

\(\langle Tv,v\rangle\in\mathbb{R}\) implies LHS = 0, which implies \(T=T^*\)

\(T=T^*\) implies RHS = 0, which implies \(\langle Tv,v\rangle\in\mathbb{R}\)

Self-adjoint operators and \(\langle Tv,v\rangle\): Result II

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

If \(T\) is self-adjoint and \(\langle Tv,v\rangle=0\) for all \(v\), then \(T=0\).

Proof

Over \(\mathbb{C}\), result is true even if \(T\) is not self-adjoint.

Over \(\mathbb{R}\), when \(T\): self-adjoint, we have

\(\langle Tu,w\rangle = \dfrac{1}{4}\langle T(u+w),u+w\rangle-\dfrac{1}{4}\langle T(u-w),u-w\rangle\)

So, \(\langle Tu,w\rangle=0\) for all \(u,w\)

Set \(w=Tu\) to get \(\lVert Tu\rVert=0\) for all \(u\)

So, \(T=0\)