Commutator and its Properties
Let \(\,A,\,B\ \) be elements of a non-abelian algebra,
e.g. complex or real square matrices of size \(\,n\ \)
or linear operators defined on a unitary or Euclidean space.
Definition.
An expression \(\ \ [\,A,B\,]\ :\,=\ AB-BA\ \ \) is called the \(\,\) commutator \(\,\)
of the elements \(\,A\ \) and \(\ B\,.\)
If \(\ [\,A,B\,]\,=\,0\,,\ \) that is \(\ AB=BA\,,\ \ \)
we say that the elements \(\,A\ \) and \(\ B\ \) commute.
Properties of commutators:
\[\begin{split}\begin{array}{cl}
\left[\,A,A\,\right]\ =\ 0\,, & \\ \\
\left[\,B,A\,\right]\ =\ -\ \left[\,A,B\,\right]\,, & \\ \\
\left[\,A_1+A_2\,,\,B\,\right]\ =\
\left[\,A_1\,,B\,\right]\ +\ \left[\,A_2\,,B\,\right]\,, & \\ \\
\left[\,A,\,B_1+B_2\,\right]\ =\ \left[\,A,B_1\,\right]\ +\ \left[\,A,B_2\,\right]\,, & \\ \\
\left[\,\lambda,A\,\right]\ =\ \left[\,A,\lambda\,\right]\ =\ 0\,, &
\lambda\equiv\lambda\,I,\ \ I\ \ \text{-}\ \ \text{the identity element,} \\ \\
\left[\,\lambda\,A,\,B\,\right]\ =\ \left[\,A,\,\lambda\,B\,\right]\ =\
\lambda\ \left[\,A,B\,\right]\,, & \lambda\ \ \text{-}\ \ \text{scalar factor.}
\end{array}\end{split}\]
The commutator \(\ [\,A,B\,]\ \) is thus linear with respect to both variables
\(\,A\ \) and \(\ B\,.\)
One can use mathematical induction to show that:
\[\begin{split}\begin{array}{l}
\left[\,A,\,B_1 B_2\ldots B_{n-1}B_n\,\right]\ \ = \\
\left[A,B_1\right]\,B_2\ldots B_{n-1}B_n\ +\
B_1\left[A,B_2\right]\ldots B_{n-1}B_n\ +\ \ldots\ +\
B_1B_2\ldots B_{n-1}\left[A,B_n\right]\,;
\\ \\
\left[\,A_1A_2\ldots A_{n-1}A_n\,,B\right]\ = \\
\left[A_1\,,B\right]\,A_2\ldots A_{n-1}A_n\ +\
A_1\left[A_2\,,B\right]\ldots A_{n-1}A_n\ +\ \ldots\ +\
A_1A_2\ldots A_{n-1}\left[A_n\,,B\right]\,.
\end{array}\end{split}\]
In particular, for \(\,n=2\ \) one obtains often used formulae:
\[\begin{split}\begin{array}{cc}
\left[\,A,B_1B_2\,\right]\ =\
\left[\,A,B_1\,\right]\,B_2\ +\ B_1\,\left[\,A,B_2\,\right]\,. & \\ \\
\left[\,A_1A_2\,,B\,\right]\ =\
A_1\,\left[\,A_2\,,B\,\right]\ +\ \left[\,A_1\,,B\,\right]\,A_2\,, &
\end{array}\end{split}\]
If \(\ [A,B]=\lambda\,I\,,\ \lambda\in R,\,C,\ \ \) then
putting \(\ B_1=\ldots=B_n=B\,,\ \ A_1=\ldots=A_n=A\,,\ \) we obtain:
\[\left[\,A,B^n\right]\ =\ n\,\lambda\,B^{n-1},\qquad
\left[\,A^n,B\,\right]\ =\ n\,\lambda\,A^{n-1},\qquad n\in N.\]
For matrices \(\,A,\,B\,\in M_n(K)\,,\ \ K=R,\,C,\ \ \) we can list futher properties
\(\\\)
(the last identity also makes sense for linear operators on a unitary or Euclidean space):
\[[\,A,B\,]^{\,T}\ \,=\ \ [\,B^T,A^T\,]\,,\qquad
[\,A,B\,]^{\,*}\ \,=\ \ [\,A^*,B^*\,]\,,\qquad
[\,A,B\,]^{\,+}\ \,=\ \ [\,B^+,A^+\,]\,.\]
Normal Matrices and Normal Operators
Definition.
Let \(\ V\ \) be a unitary space,
\(\ \boldsymbol{A}\in M_n(C)\,,\ F\in\text{End}(V)\,.\)
We say that a matrix \(\ \boldsymbol{A}\ \) is normal \(\,\)
if it commutes with its Hermitian conjugate:
\[[\,\boldsymbol{A},\boldsymbol{A}^+\,]\ =\ 0\qquad\text{that is}\qquad
\boldsymbol{A}\,\boldsymbol{A}^+\ =\ \boldsymbol{A}^+\boldsymbol{A}\,.\]
An operator \(\,F\ \) is normal \(\,\)
if it commutes with its Hermitian conjugation:
\[[\,F,F^+\,]\ =\ 0\qquad\text{that is}\qquad F\,F^+\ =\ F^+F\,.\]
Among complex matrices, normal are Hermitian and unitary matrices; and among real matrices:
symmetric, antisymmetric and orthogonal.
Similarly, Hermitian and unitary operators are normal.
A relation between normal matrices and normal operators describes
Theorem 11.
A linear operator \(\,F\,\) defined on a unitary space \(\,V\,\)
is normal if and only if
in every orthonormal basis \(\,\mathcal{B}\ \) its matrix is normal:
\[F\,F^+\;=\ F^+F\qquad\Leftrightarrow\qquad
\boldsymbol{A}\,\boldsymbol{A}^+\;=\ \boldsymbol{A}^+\boldsymbol{A}\,,\]
where \(\ \ \boldsymbol{A}\,=\,M_{\mathcal{B}}(F)\,.\)
Proof.
This proof is similar to a proof of Theorem 10.: we will use bijectivity and multiplicativity
of the mapping \(\ M_{\mathcal{B}}\ \) and the fact that in an orthonormal basis a matrix
of Hermitian conjugation of an operator is equal to Hermitian conjugate of its matrix.
The following conditions are equivalent:
\begin{eqnarray*}
F\,F^+ & = & F^+F\,, \\
M_{\mathcal{B}}(FF^+) & = & M_{\mathcal{B}}(F^+F)\,, \\
M_{\mathcal{B}}(F)\ M_{\mathcal{B}}(F^+) & = & M_{\mathcal{B}}(F^+)\ M_{\mathcal{B}}(F)\,, \\
M_{\mathcal{B}}(F)\ [\,M_{\mathcal{B}}(F)\,]^+ & = &
[\,M_{\mathcal{B}}(F)\,]^+\ M_{\mathcal{B}}(F)\,, \\
\boldsymbol{A}\,\boldsymbol{A}^+ & = & \boldsymbol{A}^+\boldsymbol{A}\,.
\end{eqnarray*}
It turns out that orthogonality of eigenvectors asssociated with different eigenvalues
is not only a property of Hermitian and unitary operators (as we proved in the previous section),
but is a feature of a wider class of normal operators. This is the statement of
Theorem 12.
Eigenvectors of a normal operator
asssociated with different eigenvalues \(\\\)
are orthogonal.
Lemma. \(\,\) For a normal operator \(\ F\in\text{End}(V):\)
(1)\[Fx=\lambda\,x\quad\Leftrightarrow\quad F^+\,x=\lambda^*\,x\,,\qquad x\in V,\quad\lambda\in C.\]
Proof of the lemma. \(\,\)
Note first that if \(\,F\ \) is a normal operator, then
for every \(\,x\in V:\)
\[\|\,Fx\,\|^2\ =\ \langle Fx,Fx\rangle\ =\ \langle F^+F\,x,x\rangle\ =\
\langle FF^+x,x\rangle\ =\ \langle F^+x,F^+x\rangle\ =\ \|\,F^+x\,\|^2\,,\]
which gives equality of the norms:
(2)\[\|\,Fx\,\|\ =\ \|\,F^+x\,\|\,,\quad x\in V\,.\]
Further, if an operator \(\ F\ \) is normal,
then normal is also the operator \(\ F-\lambda\,I\,,\ \) where \(\ \,\lambda\in C,\ \ I\ \) the idenity operator:
\[\begin{split}\begin{array}{cl}
\quad\left[\,(F-\lambda\,I),\,(F-\lambda\,I)^+\,\right]\ = &
\\ \\
=\ \left[\,F-\lambda\,I,\,F^+-\lambda^*\,I\,\right]\ = &
\\ \\
=\ \left[\,F,F^+\,\right]-\left[\,F,\,\lambda^*\,I\,\right]-
\left[\,\lambda\,I,F^+\,\right]+\left[\,\lambda\,I,\,\lambda^*\,I\,\right]\ = &
\\ \\
=\ \left[\,F,F^+\,\right]-\lambda^*\left[\,F,I\,\right]-
\lambda\,\left[\,I,F^+\,\right]+\lambda\,\lambda^*\,\left[\,I,I\,\right]\ = &
\left[\,F,F^+\,\right]\ =\ 0\,.
\end{array}\end{split}\]
Substitution \(\ F\rightarrow F-\lambda\,I\ \) in the equation (2) leads to
\[\begin{split}\begin{array}{ccc}
& \|\,(F-\lambda\,I)\,x\,\|\ =\ \|\,(F-\lambda\,I)^+\,x\,\| &
\\ \\
\text{that is}
& \|\,F x-\lambda\,x\,\|\ =\ \|\,F^+x-\lambda^*\,x\,\|\,, & \lambda\in C\,,\ \ x\in V\,.
\end{array}\end{split}\]
Now the following equalities finish the proof:
\[\begin{split}\begin{array}{ccc}
Fx\ =\ \lambda\,x & & \\ \\
Fx-\lambda\,x\,=\,\theta & & \\ \\
\|\,Fx-\lambda\,x\,\|\,=\,0 & \quad\Leftrightarrow & \quad\|\,F^+x-\lambda^*\,x\,\|\,=\,0 \\ \\
& & \quad F^+x-\lambda^*\,x\,=\,\theta \\ \\
& & \quad F^+x\ =\ \lambda^*\,x\,.
\end{array}\end{split}\]
Proof of theorem 12. \(\,\) Assume that \(\,F\ \) is a normal operator, and let
\(\quad Fx_1\,=\ \lambda_1\,x_1\,,\quad Fx_2\,=\ \lambda_2\,x_2\,,\quad
x_1,\,x_2\,\in\,V\!\smallsetminus\!\{\theta\}\,,\ \ \lambda_1\neq\lambda_2\,.\ \,\) Then
\[\begin{split}\begin{array}{l}
\langle\,x_1,Fx_2\rangle\ =\
\langle\,x_1,\lambda_2\,x_2\rangle\ =\
\lambda_2\ \langle\,x_1,x_2\rangle\,,
\\ \\
\langle\,x_1,Fx_2\rangle\ =\
\langle\,F^+x_1,x_2\rangle\ =\
\langle\,\lambda_1^*\,x_1,x_2\rangle\ =\
\lambda_1\ \langle\,x_1,x_2\rangle\,.
\end{array}\end{split}\]
Hence,
\(\ \ (\lambda_2-\lambda_1)\,\langle\,
x_1,x_2\rangle = 0\,,\ \)
and thus \(\ \langle\,x_1,x_2\rangle=0\ \) as required.
