Let us now study the algebraic structure of the class of all linear transformations on an inner product space \(\mathcal{V}\) . In many fundamental respects this class resembles the class of all complex numbers. In both systems, notions of addition, multiplication, \(0\) , and \(1\) are defined and have similar properties, and in both systems there is an involutory anti-automorphism of the system onto itself (namely, \(A \to A^{*}\) and \(\zeta \to \bar{\zeta}\) ). We shall use this analogy as a heuristic principle, and we shall attempt to carry over to linear transformations some well-known concepts from the complex domain. We shall be hindered in this work by two difficulties in the theory of linear transformations, of which, possibly surprisingly, the second is much more serious; they are the impossibility of unrestricted division and the non-commutativity of general linear transformations.
The three most important subsets of the complex number plane are the set of real numbers, the set of positive real numbers, and the set of numbers of absolute value one. We shall now proceed systematically to use our heuristic analogy of transformations with complex numbers, and to try to discover the analogues among transformations of these well-known numerical concepts.
When is a complex number real? Clearly a necessary and sufficient condition for the reality of \(\zeta\) is the validity of the equation \(\zeta=\bar{\zeta}\) . We might accordingly (remembering that the analogue of the complex conjugate for linear transformations is the adjoint) define a linear transformation \(A\) to be real if \(A=A^{*}\) . More commonly linear transformations \(A\) for which \(A=A^{*}\) are called self-adjoint ; in real inner product spaces the usual word is symmetric , and, in complex inner product spaces, Hermitian . We shall see that self-adjoint transformations do indeed play the same role as real numbers.
It is quite easy to characterize the matrix of a self-adjoint transformation with respect to an orthonormal basis \(\mathcal{X}=\{x_{1}, \ldots, x_{n}\}\) . If the matrix of \(A\) is \((\alpha_{i j})\) , then we know that the matrix of \(A^{*}\) with respect to the dual basis of \(\mathcal{X}\) is \((\alpha_{i j}^{*})\) , where \(\alpha_{i j}^{*}=\overline{\alpha_{j i}}\) ; since an orthonormal basis is self-dual and since \(A=A^{*}\) , we have \[\alpha_{i j}=\overline{\alpha_{j i}}.\] We leave it to the reader to verify the converse: if we define a linear transformation \(A\) by means of a matrix \((\alpha_{i j})\) and an arbitrary orthonormal coordinate system \(\mathcal{X}=\{x_{1}, \ldots, x_{n}\}\) , via the usual equations \begin{align} A\Big(\sum_{j} \xi_{j} x_{j}\Big) &= \sum_{i} \eta_{i} x_{i},\\ \eta_{i} &= \sum_{j} \alpha_{i j} \xi_{j} \end{align}
and if the matrix \((\alpha_{i j})\) is such that \(\alpha_{i j}=\overline{\alpha_{j i}}\) , then \(A\) is self-adjoint.
The algebraic rules for the manipulation of self-adjoint transformations are easy to remember if we think of such transformations as the analogues of real numbers. Thus, if \(A\) and \(B\) are self-adjoint, so is \(A+B\) ; if \(A\) is self-adjoint and different from \(0\) , and if \(\alpha\) is a non-zero scalar, then a necessary and sufficient condition that \(\alpha A\) be self-adjoint is that \(\alpha\) be real; and if \(A\) is invertible, then both or neither of \(A\) and \(A^{-1}\) are self-adjoint. The place where something always goes wrong is in multiplication; the product of two self-adjoint transformations need not be self-adjoint. The positive facts about products are given by the following two theorems.
Theorem 1. If \(A\) and \(B\) are self-adjoint, then a necessary and sufficient condition that \(A B\) (or \(B A\) ) be self-adjoint is that \(A B=B A\) (that is that \(A\) and \(B\) commute).
Proof. If \(A B=B A\) , then \[(A B)^{*}=B^{*} A^{*}=B A=A B.\] If \((A B)^{*}=A B\) , then \[A B=(A B)^{*}=B^{*} A^{*}=B A.\] ◻
Theorem 2. If \(A\) is self-adjoint, then \(B^{*} A B\) is self-adjoint for all \(B\) ; if \(B\) is invertible and \(B^{*} A B\) is self-adjoint, then \(A\) is self-adjoint.
Proof. If \(A=A^{*}\) , then \[(B^{*} A B)^{*}=B^{*} A^{*} B^{* *}=B^{*} A B.\] If \(B\) is invertible and \(B^{*} A B=(B^{*} A B)^{*}=B^{*} A^{*} B\) , then (multiply by \(B^{*-1}\) on the left and \(B^{-1}\) on the right) \(A=A^{*}\) . ◻
A complex number \(\zeta\) is purely imaginary if and only if \(\bar{\zeta}=-\zeta\) . The corresponding concept for linear transformations is identified by the word skew ; if a linear transformation \(A\) on an inner product space is such that \(A^{*}=-A\) , then \(A\) is called skew symmetric or skew Hermitian according as the space is real or complex. Here is some evidence for the thoroughgoing nature of our analogy between complex numbers and linear transformations: an arbitrary linear transformation \(A\) may be expressed, in one and only one way, in the form \(A=B+C\) , where \(B\) is self-adjoint and \(C\) is skew. (The representation of \(A\) in this form is sometimes called the Cartesian decomposition of \(A\) .) Indeed, if we write \begin{align} & B=\frac{A+A^{*}}{2}, \tag{1}\\ & C=\frac{A-A^{*}}{2}, \tag{2} \end{align} then we have \(B^{*}=\frac{A^{*}+A}{2}=B\) and \(C^{*}=\frac{A^{*}-A}{2}=-C\) , and, of course, \(A=B+C\) . From this proof of the existence of the Cartesian decomposition, its uniqueness is also clear; if we do have \(A=B+C\) , then \(A^{*}=B-C\) , and, consequently, \(A\) , \(B\) , and \(C\) are again connected by (1) and (2).
In the complex case there is a simple way of getting skew Hermitian transformations from Hermitian ones, and vice versa: just multiply by \(i\) ( \(=\sqrt{-1}\) ). It follows that, in the complex case, every linear transformation \(A\) has a unique representation in the form \(A=B+i C\) , where \(B\) and \(C\) are Hermitian. We shall refer to \(B\) and \(C\) as the real and imaginary parts of \(A\) .
EXERCISES
Exercise 1. Give an example of two self-adjoint transformations whose product is not self-adjoint.
Exercise 2. Consider the space \(\mathcal{P}_{n}\) with the inner product given by \((x, y)=\int_{0}^{1} x(t) \overline{y(t)} \,d t\) .
- Is the multiplication operator \(T\) (defined by \((T x)(t)=t x(t)\) ) self-adjoint?
- Is the differentiation operator \(D\) self-adjoint?
Exercise 3.
- Prove that the equation \((x, y)=\sum_{j=0}^{n} x\big(\frac{j}{n}\big) \overline{y\big(\frac{j}{n}\big)}\) defines an inner product in the space \(\mathcal{P}_{n}\) .
- Is the multiplication operator \(T\) (defined by \((T x)(t)=t x(t)\) ) self-adjoint (with respect to the inner product defined in (a))?
- Is the differentiation operator \(D\) self-adjoint?
Exercise 4. If \(A\) and \(B\) are linear transformations such that \(A\) and \(A B\) are self-adjoint and such that \(\mathcal{N}(A) \subset \mathcal{N}(B)\) , then there exists a self-adjoint transformation \(C\) such that \(C A=B\) .
Exercise 5. If \(A\) and \(B\) are congruent and \(A\) is skew, does it follow that \(B\) is skew?
Exercise 6. If \(A\) is skew, does it follow that so is \(A^{2}\) ? How about \(A^{3}\) ?
Exercise 7. If both \(A\) and \(B\) are self-adjoint, or else if both are skew, then \(A B+B A\) is self-adjoint and \(A B-B A\) is skew. What happens if one of \(A\) and \(B\) is self-adjoint and the other skew?
Exercise 8. If \(A\) is a skew-symmetric transformation on a Euclidean space, then \((A x, x)=0\) for every vector \(x\) . Converse?
Exercise 9. If \(A\) is self-adjoint, or skew, and if \(A^{2} x=0\) , then \(A x=0\) .
Exercise 10.
- If \(A\) is a skew-symmetric transformation on a Euclidean space of odd dimension, then \(\operatorname{det} A=0\) .
- If \(A\) is a skew-symmetric transformation on a finite-dimensional Euclidean space, then \(\rho(A)\) is even.