Normal Operators

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 \(T^*)^{\perp}\)
  • Self-adjoint: \(T=T^*\)

Commuting operators

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

In general, \(ST\ne TS\)

Example

\(\begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}\begin{bmatrix} 1&0\\ 0&2 \end{bmatrix}\ne\begin{bmatrix} 1&0\\ 0&2 \end{bmatrix}\begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}\)

Operators \(S,T\) are said to commute if \(ST=TS\).

Example: \(I\) commutes with all \(T\)

Exercise: If \(S\) commutes with all \(T\), then \(S=\lambda I\).

Commuting operators have some very important characterisations.

Definition of normal operators

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

An operator \(T:V\to V\) is said to be normal if \(TT^*=T^*T\).

In other words, a normal operator commutes with its adjoint.

  • If \(T\) is self-adjoint, it is normal

  • There are normal operators that are not self-adjoint

Example

\(\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}\begin{bmatrix} 2&3\\ -3&2 \end{bmatrix} = \begin{bmatrix} 2&3\\ -3&2 \end{bmatrix}\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}\)

Norm, Range and Null

For any \(T\), null \(T=\) \((\)range \(T^*)^{\perp}\)
dim range \(T=\) dim range \(T^*\)
dim null \(T=\) dim null \(T^*\)

\(T\) is normal if and only if \(\lVert Tv\rVert=\lVert T^*v\rVert\).

Proof

\(T\) is normal
iff \(TT^*-T^*T=0\)
iff \(\langle (TT^*-T^*T)v,v\rangle=0\) for all \(v\)
iff \(\langle TT^*v,v\rangle=\langle T^*Tv,v\rangle\) for all \(v\)
iff \(\langle T^*v,T^*v\rangle=\langle Tv,Tv\rangle\) for all \(v\)

If \(T\) is normal, null \(T=\) null \(T^*\) and range \(T=\) range \(T^*\)

Eigenvalues and eigenvectors

For any \(T\)

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

Suppose \(T\) is normal. If \(v\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), then \(v\) is an eigenvector of \(T^*\) with eigenvalue \(\bar{\lambda}\).

Proof

\(T\) is normal

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

So, \(T-\lambda I\) is normal

\(\lVert (T^*-\bar{\lambda} I)v\rVert=\lVert (T-\lambda I)v\rVert=0\)

So, \(T^*v=\bar{\lambda}v\)

Orthogonality of eigenvectors

For any \(T\)

\(u\): eigenvector with eigenvalue \(\alpha\)
\(v\): eigenvector with eigenvalue \(\beta\)

If \(\alpha\ne\beta\), \(u\) and \(v\) are linearly independent

If \(T\) is normal and \(\alpha\ne\beta\), \(u\) and \(v\) are orthogonal

Proof

\(Tu=\alpha u\) and \(Tv=\beta v\)

Since \(T\) is normal, \(T^*v=\bar{\beta}v\)

\(\begin{align} (\alpha-\beta)\langle u,v\rangle&=\langle \alpha u,v\rangle - \langle u,\bar{\beta}v\rangle\\ &=\langle Tu,v\rangle - \langle u,T^*v\rangle\\ &=0 \end{align}\)

Note: If \(T\) is self-adjoint, \(T\) is normal. So, above result holds for self-adjoint operators as well.