Andrew Thangaraj
Aug-Nov 2020
\(S,T:V\to V\), 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.
\(V\): 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
\(\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}\begin{bmatrix} 2&3\\ -3&2 \end{bmatrix}\ne\begin{bmatrix} 2&3\\ -3&2 \end{bmatrix}\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}\)
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 \(T^*\)
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, \(T^*v=\bar{\lambda}v\)
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.