Commutativity

The spectral theorem for self-adjoint and for normal operators and the functional calculus may also be used to solve certain problems concerning commutativity. This is a deep and extensive subject; more to illustrate some methods than for the actual results we discuss two theorems from it.

Theorem 1. Two self-adjoint transformations $A$ and $B$ on a finite-dimensional inner product space are commutative if and only if there exists a self-adjoint transformation $C$ and there exist two real-valued functions $f$ and $g$ of a real variable so that $A = f (C)$ and $B = g (C)$ . If such a $C$ exists, then we may even choose $C$ in the form $C = h (A, B)$ , where $h$ is a suitable real-valued function of two real variables.

Proof. The sufficiency of the condition is clear; we prove only the necessity.

Let $A = \sum_{i} α_{i} E_{i}$ and $B = \sum_{j} β_{j} F_{j}$ be the spectral forms of $A$ and $B$ ; since $A$ and $B$ commute, it follows from Section: Spectral theorem , Theorem 3, that $E_{i}$ and $F_{j}$ commute. Let $h$ be any function of two real variables such that the numbers $h (α_{i}, β_{j}) = γ_{i j}$ are all distinct, and write $C = h (A, B) = \sum_{i} \sum_{j} h (α_{i}, β_{j}) E_{i} F_{j} .$ (It is clear that $h$ may even be chosen as a polynomial, and the same is true of the functions $f$ and $g$ we are about to describe.) Let $f$ and $g$ be such that $f (γ_{i j}) = α_{i}$ and $g (γ_{i j}) = β_{j}$ for all $i$ and $j$ . It follows that $f (C) = A$ and $g (C) = B$ , and everything is proved. ◻

Theorem 2. If $A$ is a normal transformation on a finite-dimensional unitary space and if $B$ is an arbitrary transformation that commutes with $A$ , then $B$ commutes with $A^{*}$ .

Proof. Let $A = \sum_{i} α_{i} E_{i}$ be the spectral form of $A$ ; then $A^{*} = \sum_{i} {\bar{α}}_{i} E_{i}$ . Let $f$ be such a function (polynomial) of a complex variable that $f (α_{i}) = {\bar{α}}_{i}$ for all $i$ . Since $A^{*} = f (A)$ , the conclusion follows. ◻

EXERCISES

Exercise 1.

Prove the following generalization of Theorem 2: if $A_{1}$ and $A_{2}$ are normal transformations (on a finite-dimensional unitary space) and if $A_{1} B = B A_{2}$ , then $A_{1}^{*} B = B A_{2}^{*}$ .
Theorem 2 asserts that the relation of commutativity is sometimes transitive: if $A^{*}$ commutes with $A$ and if $A$ commutes with $B$ , then $A^{*}$ commutes with $B$ . Does this formulation remain true if $A^{*}$ is replaced by an arbitrary transformation $C$ ?

Exercise 2.

If $A$ commutes with $A^{*} A$ , does it follow that $A$ is normal?
If $A^{*} A$ commutes with $A A^{*}$ , does it follow that $A$ is normal?

Exercise 3.

A linear transformation $A$ is normal if and only if there exists a polynomial $p$ such that $A^{*} = p (A)$ .
If $A$ is normal and commutes with $B$ , then $A$ commutes with $B^{*}$ .
If $A$ and $B$ are normal and commutative, then $A B$ is normal.

Exercise 4. If $A$ and $B$ are normal and similar, then they are unitarily equivalent.

Exercise 5.

If $A$ is Hermitian, if every proper value of $A$ has multiplicity $1$ , and if $A B = B A$ , then there exists a polynomial $p$ such that $B = p (A)$ .
If $A$ is Hermitian, then a necessary and sufficient condition that there exist a polynomial $p$ such that $B = p (A)$ is that $B$ commute with every linear transformation that commutes with $A$ .

Exercise 6. Show that a commutative set of normal transformations on a finite-dimensional unitary space can be simultaneously diagonalized.

EXERCISES

Quick Links

Social Media