We have seen that the theory of the passage from one linear basis of a vector space to another is best studied by means of an associated linear transformation A (Sections 46, 47); the question arises as to what special properties A has when we pass from one orthonormal basis of an inner product space to another. The answer is easy.
Theorem 1. If \mathcal{X}=\{x_{1}, \ldots, x_{n}\} is an orthonormal basis of an n -dimensional inner product space \mathcal{V} , and if U is an isometry on \mathcal{V} , then U\mathcal{X}=\{U x_{1}, \ldots, U x_{n}\} is also an orthonormal basis of \mathcal{V} . Conversely, if U is a linear transformation and \mathcal{X} is an orthonormal basis with the property that U\mathcal{X} is also an orthonormal basis, then U is an isometry.
Proof. Since (U x_{i}, U x_{j})=(x_{i}, x_{j})=\delta_{i j} , it follows that U \mathcal{X} is an orthonormal set along with \mathcal{X} ; it is complete if \mathcal{X} is, since (x, U x_{i})=0 for i=1, \ldots, n implies that (U^{*} x, x_{i})=0 and hence that U^{*} x=x=0 . If, conversely, U\mathcal{X} is a complete orthonormal set along with \mathcal{X} , then we have (U x, U y)=(x, y) whenever x and y are in \mathcal{X} , and it is clear that by linearity we obtain (U x, U y)=(x, y) for all x and y . ◻
We observe that the matrix (u_{i j}) of an isometric transformation, with respect to an arbitrary orthonormal basis, satisfies the conditions \sum_{k} \bar{u}_{k i} u_{k j}=\delta_{i j}, and that, conversely, any such matrix, together with an orthonormal basis, defines an isometry. (Proof: U^{*} U=1 . In the real case the bars may be omitted.) For brevity we shall say that a matrix satisfying these conditions is an isometric matrix .
An interesting and easy consequence of our considerations concerning isometries is the following corollary of Section: Triangular form , Theorem 1.
Theorem 2. If A is a linear transformation on a complex n -dimensional inner product space \mathcal{V} , then there exists an orthonormal basis \mathcal{X} in \mathcal{V} such that the matrix [A; \mathcal{X}] is triangular, or equivalently, if [A] is a matrix, then there exists an isometric matrix [U] such that [U]^{-1}[A][U] is triangular.
Proof. In Section: Triangular form , in the derivation of Theorem 2 from Theorem 1, we constructed a (linear) basis \mathcal{X}=\{x_{1}, \ldots, x_{n}\} with the property that x_{1},\ldots, x_{j} lie in \mathcal{M}_{j} and span \mathcal{M}_{j} for j=1, \ldots, n , and we showed that with respect to this basis the matrix of A is triangular. If we knew that this basis is also an orthonormal basis, we could apply Theorem 1 of the present section to obtain the desired result. If \mathcal{X} is not an orthonormal basis, it is easy to make it into one; this is precisely what the Gram-Schmidt orthogonalization process ( Section: Complete orthonormal sets ) can do. Here we use a special property of the Gram-Schmidt process, namely, that the j -th element of the orthonormal basis it constructs is a linear combination of x_{1}, \ldots, x_{j} and lies therefore in \mathcal{M}_{j} . ◻
EXERCISES
Exercise 1. If (A x)(t)=x(-t) on \mathcal{P} (with the inner product given by (x, y)=\int_{0}^{1} x(t) \overline{y(t)} \,d t ) is the linear transformation A isometric? Is it self-adjoint?
Exercise 2. For which values of \alpha are the following matrices isometric?
2
- \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix} .
- \begin{bmatrix} \alpha & \frac{1}{2}\\ -\frac{1}{2} & \alpha \end{bmatrix} .
Exercise 3. Find a 3 -by- 3 isometric matrix whose first row is a multiple of (1,1,1) .
Exercise 4. If a linear transformation has any two of the properties of being self-adjoint, isometric, or involutory, then it has the third. (Recall that an involution is a linear transformation A such that A^{2}=1 .)
Exercise 5. If an isometric matrix is triangular, then it is diagonal.
Exercise 6. If (x_{1}, \ldots, x_{k}) and (y_{1}, \ldots, y_{k}) are two sequences of vectors in the same inner product space, then a necessary and sufficient condition that there exist an isometry U such that U x_{i}=y_{i} , i=1, \ldots, k , is that (x_{1}, \ldots, x_{k}) and (y_{1}, \ldots, y_{k}) have the same Gramian.
Exercise 7. The mapping \xi \to \frac{\xi+1}{\xi-1} maps the imaginary axis in the complex plane once around the unit circle, missing the point 1 ; the inverse mapping (from the circle minus a point to the imaginary axis) is given by the same formula. The transformation analogues of these geometric facts are as follows.
- If A is skew, then A-1 is invertible.
- If U=(A+1)(A-1)^{-1} , then U is isometric. (Hint: \|(A+1) y\|^{2}=\|(A-1) y\|^{2} for every y .)
- U-1 is invertible.
- If U is isometric and U-1 is invertible, and if A=(U+1)(U-1)^{-1} , then A is skew.
Each of A and U is known as the Cayley transform of the other.
Exercise 8. Suppose that U is a transformation (not assumed to be linear) that maps an inner product space \mathcal{V} onto itself (that is, if x is in \mathcal{V} , then U x is in \mathcal{V} , and if y is in \mathcal{V} , then y=U x for some x in \mathcal{V} ), in such a way that (U x, U y)=(x, y) for all x and y .
- Prove that U is one-to-one and that if the inverse transformation is denoted by U^{-1} , then (U^{-1} x, U^{-1} y)=(x, y) and (U x, y)=(x, U^{-1} y) for all x and y .
- Prove that U is linear. (Hint: (x, U^{-1} y) depends linearly on x .)
Exercise 9. A conjugation is a transformation J (not assumed to be linear) that maps a unitary space onto itself and is such that J^{2}=1 and (J x, J y)=(y, x) for all x and y .
- Give an example of a conjugation.
- Prove that (J x, y)=(J y, x) .
- Prove that J(x+y)=J x+J y .
- Prove that J(\alpha x)=\bar{\alpha} \cdot J x .
Exercise 10. A linear transformation A is said to be real with respect to a conjugation J if A J=J A .
- Give an example of a Hermitian transformation that is not real, and give an example of a real transformation that is not Hermitian.
- If A is real, then the spectrum of A is symmetric about the real axis.
- If A is real, then so is A^{*} .
Exercise 11. Section: Change of orthonormal basis , Theorem 2 shows that the triangular form can be achieved by an orthonormal basis; is the same thing true for the Jordan form?
Exercise 12. If \operatorname{tr} A=0 , then there exists an isometric matrix U such that all the diagonal entries of [U]^{-1}[A][U] are zero. (Hint: see Section: Triangular form , Ex. 6.)
