Before continuing with the program of studying the analogies between complex numbers and linear transformations, we take time out to pick up some important auxiliary results about inner product spaces.
Theorem 1. A necessary and sufficient condition that a linear transformation \(A\) on an inner product space be \(0\) is that \((A x, y)=0\) for all \(x\) and \(y\) .
Proof. The necessity of the condition is obvious; sufficiency follows from setting \(y\) equal to \(A x\) . ◻
Theorem 2. A necessary and sufficient condition that a self-adjoint linear transformation \(A\) on an inner product space \(A\) be \(0\) is that \((A x, x)=0\) for all \(x\) .
Proof. Necessity is obvious. The proof of sufficiency begins by verifying the identity \begin{align} (A x, y)+(A y, x)=(A(x+y),(x+y))-(A x, x)-(A y, y). \tag{1} \end{align}
(Expand the first term on the right side.) Since \(A\) is self-adjoint, the left side of this equation is equal to \(2 \operatorname{Re}(A x, y)\) . The assumed condition implies that the right side vanishes, and hence that \(\operatorname{Re}(A x, y)=0\) . At this point it is necessary to split the proof into two cases. If the inner product space is real (that is, \(A\) is symmetric), then \((A x, y)\) is real, and therefore \((A x, y)=0\) . If the inner product space is complex (that is, \(A\) is Hermitian), then we find a complex number \(\theta\) such that \(|\theta|=1\) and \(\theta(A x, y)=|(A x, y)|\) . (Here \(x\) and \(y\) are temporarily fixed.) The result we already have, applied to \(\theta x\) in place of \(x\) , yields \begin{align} 0 &= \operatorname{Re}(A(\theta x), y)\\ &= \operatorname{Re} \theta(A x, y)\\ &= \operatorname{Re}|(A x, y)|\\ &= |(A x, y)|. \end{align} In either case, therefore, \((A x, y)=0\) for all \(x\) and \(y\) , and the desired result follows from Theorem 1. ◻
It is useful to ask how important is the self-adjointness of \(A\) in Theorem 2; the answer is that in the complex case it is not important at all.
Theorem 3. A necessary and sufficient condition that a linear transformation \(A\) on a unitary space be \(0\) is that \((A x, x)=0\) for all \(x\) .
Proof. As before, necessity is obvious. For the proof of sufficiency we use the so-called polarization identity: \begin{align} \alpha \bar{\beta}(A x, y)+\bar{\alpha} \beta(A y, x) &= \big(A(\alpha x+\beta y),(\alpha x+\beta y)\big)\\ & \quad -|\alpha|^{2}(A x, x)\\ & \quad -|\beta|^{2}(A y, y). \tag{2} \end{align}
(Just as for (1), the proof consists of expanding the first term on the right.) If \((A x, x)\) is identically zero, then we obtain, first choosing \(\alpha=\beta=1\) , and then \(\alpha=i\) ( \(=\sqrt{-1}\) ), \(\beta=1\) \begin{align} (A x, y)+(A y, x) & =0 \\ i(A x, y)-i(A y, x) & =0. \end{align} Dividing the second of these two equations by \(i\) and then forming their arithmetic mean, we see that \((A x, y)=0\) for all \(x\) and \(y\) , so that, by Theorem 1, \(A=0\) . ◻
This process of polarization is often used to get information about the "bilinear form" \((A x, y)\) when only knowledge of the "quadratic form" \((A x, x)\) is assumed.
It is important to observe that, despite its seeming innocence, Theorem 3 makes very essential use of the complex number system; it and many of its consequences fail to be true for real inner product spaces. The proof, of course, breaks down at our choice of \(\alpha=\sqrt{-1}\) . For an example consider a \(90^{\circ}\) rotation of the plane; it clearly has the property that it sends every vector \(x\) into a vector orthogonal to \(x\) .
We have seen that Hermitian transformations play the same role as real numbers; the following theorem indicates that they are tied up with the concept of reality in deeper ways than through the formal analogy that suggested their definition.
Theorem 4. A necessary and sufficient condition that a linear transformation \(A\) on a unitary space be Hermitian is that \((A x, x)\) be real for all \(x\) .
Proof. If \(A=A^{*}\) , then \[(A x, x)=(x, A^{*} x)=(x, A x)=\overline{(A x, x)},\] so that \((A x, x)\) is equal to its own conjugate and is therefore real. If, conversely, \((A x, x)\) is always real, then \[(A x, x)=\overline{(A x, x)}=(x, A^{*} x)=(A^{*} x, x),\] so that \(\big([A-A^{*}] x, x\big)=0\) for all \(x\) , and, by Theorem 3, \(A=A^{*}\) . ◻
Theorem 4 is false for real inner product spaces. This is to be expected, for, in the first place, its proof depends on a theorem that is true for unitary spaces only, and, in the second place, in a real space the reality of \((A x, x)\) is automatic, whereas the identity \((A x, y)=(x, A y)\) is not necessarily satisfied.