Change of orthonormal basis

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 (Sections 46, 47); the question arises as to what special properties has when we pass from one orthonormal basis of an inner product space to another. The answer is easy.

Theorem 1. If is an orthonormal basis of an -dimensional inner product space , and if is an isometry on , then is also an orthonormal basis of . Conversely, if is a linear transformation and is an orthonormal basis with the property that is also an orthonormal basis, then is an isometry.

Proof. Since , it follows that is an orthonormal set along with ; it is complete if is, since for implies that and hence that . If, conversely, is a complete orthonormal set along with , then we have whenever and are in , and it is clear that by linearity we obtain for all and . ◻

We observe that the matrix of an isometric transformation, with respect to an arbitrary orthonormal basis, satisfies the conditions and that, conversely, any such matrix, together with an orthonormal basis, defines an isometry. (Proof: . 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 is a linear transformation on a complex -dimensional inner product space , then there exists an orthonormal basis in such that the matrix is triangular, or equivalently, if is a matrix, then there exists an isometric matrix such that is triangular.

Proof. In Section: Triangular form , in the derivation of Theorem 2 from Theorem 1, we constructed a (linear) basis with the property that lie in and span for , and we showed that with respect to this basis the matrix of 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 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 -th element of the orthonormal basis it constructs is a linear combination of and lies therefore in . ◻

EXERCISES

Exercise 1. If on (with the inner product given by ) is the linear transformation isometric? Is it self-adjoint?

Exercise 2. For which values of are the following matrices isometric?

Exercise 3. Find a -by- isometric matrix whose first row is a multiple of .

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 such that .)

Exercise 5. If an isometric matrix is triangular, then it is diagonal.

Exercise 6. If and are two sequences of vectors in the same inner product space, then a necessary and sufficient condition that there exist an isometry such that , , is that and have the same Gramian.

Exercise 7. The mapping maps the imaginary axis in the complex plane once around the unit circle, missing the point ; 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 is skew, then is invertible.
If , then is isometric. (Hint: for every .)
is invertible.
If is isometric and is invertible, and if , then is skew.

Each of and is known as the Cayley transform of the other.

Exercise 8. Suppose that is a transformation (not assumed to be linear) that maps an inner product space onto itself (that is, if is in , then is in , and if is in , then for some in ), in such a way that for all and .

Prove that is one-to-one and that if the inverse transformation is denoted by , then and for all and .
Prove that is linear. (Hint: depends linearly on .)

Exercise 9. A conjugation is a transformation (not assumed to be linear) that maps a unitary space onto itself and is such that and for all and .

Give an example of a conjugation.
Prove that .
Prove that .
Prove that .

Exercise 10. A linear transformation is said to be real with respect to a conjugation if .

Give an example of a Hermitian transformation that is not real, and give an example of a real transformation that is not Hermitian.
If is real, then the spectrum of is symmetric about the real axis.
If is real, then so is .

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 , then there exists an isometric matrix such that all the diagonal entries of are zero. (Hint: see Section: Triangular form , Ex. 6.)

EXERCISES

Quick Links

Social Media