Normal Operators ---------------- Commutator and its Properties ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Let :math:`\,A,\,B\ ` be elements of a non-abelian algebra, e.g. complex or real square matrices of size :math:`\,n\ ` or linear operators defined on a unitary or Euclidean space. .. admonition:: Definition. An expression :math:`\ \ [\,A,B\,]\ :\,=\ AB-BA\ \ ` is called the :math:`\,` *commutator* :math:`\,` of the elements :math:`\,A\ ` and :math:`\ B\,.` If :math:`\ [\,A,B\,]\,=\,0\,,\ ` that is :math:`\ AB=BA\,,\ \ ` we say that the elements :math:`\,A\ ` and :math:`\ B\ ` *commute*. **Properties of commutators:** .. math:: \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} The commutator :math:`\ [\,A,B\,]\ ` is thus linear with respect to both variables :math:`\,A\ ` and :math:`\ B\,.` One can use mathematical induction to show that: .. math:: \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} In particular, for :math:`\,n=2\ ` one obtains often used formulae: .. math:: \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} If :math:`\ [A,B]=\lambda\,I\,,\ \lambda\in R,\,C,\ \ ` then putting :math:`\ B_1=\ldots=B_n=B\,,\ \ A_1=\ldots=A_n=A\,,\ ` we obtain: .. :math:`\quad\left[\,A,B^n\,\right]\ =\ n\,\lambda\,B^{n-1},\quad \left[\,A^n,B\,\right]\ =\ n\,\lambda\,A^{n-1},\qquad n\in N.` .. math:: \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 :math:`\,A,\,B\,\in M_n(K)\,,\ \ K=R,\,C,\ \ ` we can list futher properties :math:`\\` (the last identity also makes sense for linear operators on a unitary or Euclidean space): .. math:: [\,A,B\,]^{\,T}\ \,=\ \ [\,B^T,A^T\,]\,,\qquad [\,A,B\,]^{\,*}\ \,=\ \ [\,A^*,B^*\,]\,,\qquad [\,A,B\,]^{\,+}\ \,=\ \ [\,B^+,A^+\,]\,. Normal Matrices and Normal Operators ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ .. admonition:: Definition. Let :math:`\ V\ ` be a unitary space, :math:`\ \boldsymbol{A}\in M_n(C)\,,\ F\in\text{End}(V)\,.` We say that a matrix :math:`\ \boldsymbol{A}\ ` is *normal* :math:`\,` if it commutes with its Hermitian conjugate: .. math:: [\,\boldsymbol{A},\boldsymbol{A}^+\,]\ =\ 0\qquad\text{that is}\qquad \boldsymbol{A}\,\boldsymbol{A}^+\ =\ \boldsymbol{A}^+\boldsymbol{A}\,. An operator :math:`\,F\ ` is *normal* :math:`\,` if it commutes with its Hermitian conjugation: .. math:: [\,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 .. admonition:: Theorem 11. A linear operator :math:`\,F\,` defined on a unitary space :math:`\,V\,` is normal if and only if in every orthonormal basis :math:`\,\mathcal{B}\ ` its matrix is normal: .. math:: F\,F^+\;=\ F^+F\qquad\Leftrightarrow\qquad \boldsymbol{A}\,\boldsymbol{A}^+\;=\ \boldsymbol{A}^+\boldsymbol{A}\,, where :math:`\ \ \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 :math:`\ 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: .. math:: :nowrap: \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 .. admonition:: Theorem 12. Eigenvectors of a normal operator asssociated with different eigenvalues :math:`\\` are orthogonal. .. :math:`\;` **Lemma.** :math:`\,` For a normal operator :math:`\ F\in\text{End}(V):` .. math:: :label: Lemma Fx=\lambda\,x\quad\Leftrightarrow\quad F^+\,x=\lambda^*\,x\,,\qquad x\in V,\quad\lambda\in C. **Proof of the lemma.** :math:`\,` Note first that if :math:`\,F\ ` is a normal operator, then for every :math:`\,x\in V:` .. :math:`\ \|\,Fx\,\| = \|\,F^+x\,\|\,,\ x\in V.\ ` Wynika to stąd, że .. math:: \|\,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: .. math:: :label: norm_eq \|\,Fx\,\|\ =\ \|\,F^+x\,\|\,,\quad x\in V\,. Further, if an operator :math:`\ F\ ` is normal, then normal is also the operator :math:`\ F-\lambda\,I\,,\ ` where :math:`\ \,\lambda\in C,\ \ I\ ` the idenity operator: .. to również operator :math:`\ F-\lambda\,I\ ` jest normalny: .. math:: \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} Substitution :math:`\ F\rightarrow F-\lambda\,I\ ` in the equation :eq:`norm_eq` leads to .. math:: \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} Now the following equalities finish the proof: .. math:: \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} **Proof of theorem 12.** :math:`\,` Assume that :math:`\,F\ ` is a normal operator, and let :math:`\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 .. math:: \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} Hence, :math:`\ \ (\lambda_2-\lambda_1)\,\langle\, x_1,x_2\rangle = 0\,,\ ` and thus :math:`\ \langle\,x_1,x_2\rangle=0\ ` as required.