which, in this case since A is self-adjoint, is just A. Consider now the restriction of A to M denoted by AM where A\M₁: M² → M². It is clear that, if A is self-adjoint, then so must be the restriction of A to M¹. Now, since dim M
which, in this case since A is self-adjoint, is just A. Consider now the restriction of A to M denoted by AM where A\M₁: M² → M². It is clear that, if A is self-adjoint, then so must be the restriction of A to M¹. Now, since dim M
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![THEOREM 2.1. Let X be a finite dimensional inner product space and let A be a
self-adjoint linear transformation on X: A: X→ X. Then there exists an orthonor-
mal basis for X consisting of eigenvectors of A.
Proof. First it will be shown that the above statement is not vacuous; i.e., it will
be shown that, since A is self-adjoint, it always possesses eigenvectors. There are
two cases to consider: Either X is a complex space or it is a real space. In the former,
the characteristic equation factors completely, and A must have eigenvalues. In
the latter, however, it is not clear that A will always have eigenvalues and so we
turn our attention toward the complexification of X and to the corresponding
extension of A, A+, it being clear that A will have eigenvalues. Since A is self-
adjoint, we have, by property (d) above,
A+ = (A*)+ = (A+)*,
which shows that A being self-adjoint implies that A is self-adjoint too. As shown
at the end of the first chapter, however, the eigenvalues of a self-adjoint transforma-
tion must be real; hence all the eigenvalues of A* are real. Since they are all real
and by the observation, immediately preceding Theorem 2.1, they must also be
eigenvalues of A. Thus we have shown that the consideration of the eigenvalues of a
self-adjoint transformation is nonvacuous in either case, and we shall now proceed
with the rest of the theorem.
Let λ be an eigenvalue of A and suppose x is a corresponding eigenvector. Thus
we have Ax = 2x. Since x cannot be zero, we can choose x₁ = x/x|| and if, in
addition, dim X = 1, then we are done. If not, we shall proceed by induction;
i.e., we shall assume that the theorem is true for all spaces of dimension less than
X and then show that it follows for X from this assumption. Letting M = [x₂] =
{ax₁ αe F}, the space spanned by x₁, we have the following direct sum decompo-
sition of X:
X = M + M¹,
and we can note immediately that we must have dim M¹ <dim X. As shown in
Chapter 1, M being invariant under A implies that M is invariant under 4*,
which, in this case since A is self-adjoint, is just A. Consider now the restriction of A
to M denoted by AM where AM₁ M² → M². It is clear that, if A is self-adjoint,
then so must be the restriction of A to M¹. Now, since dim M <dim X, we can
apply the induction hypothesis to assert the existence of an orthonormal basis for
M¹ consisting of eigenvectors for A/M₁, {X₂, X3,...,x. Eigenvectors of AM,
however, must also be eigenvectors of A. Hence for the entire space we have
{x₁, x2, ..., x} as orthonormal basis of eigenvectors of A.
{X1,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15b7304-cc73-4505-92c3-23aa2fda4f71%2F104e4f44-6a98-4448-ac1a-2162cb106119%2Fkdq5mz_processed.png&w=3840&q=75)
Transcribed Image Text:THEOREM 2.1. Let X be a finite dimensional inner product space and let A be a
self-adjoint linear transformation on X: A: X→ X. Then there exists an orthonor-
mal basis for X consisting of eigenvectors of A.
Proof. First it will be shown that the above statement is not vacuous; i.e., it will
be shown that, since A is self-adjoint, it always possesses eigenvectors. There are
two cases to consider: Either X is a complex space or it is a real space. In the former,
the characteristic equation factors completely, and A must have eigenvalues. In
the latter, however, it is not clear that A will always have eigenvalues and so we
turn our attention toward the complexification of X and to the corresponding
extension of A, A+, it being clear that A will have eigenvalues. Since A is self-
adjoint, we have, by property (d) above,
A+ = (A*)+ = (A+)*,
which shows that A being self-adjoint implies that A is self-adjoint too. As shown
at the end of the first chapter, however, the eigenvalues of a self-adjoint transforma-
tion must be real; hence all the eigenvalues of A* are real. Since they are all real
and by the observation, immediately preceding Theorem 2.1, they must also be
eigenvalues of A. Thus we have shown that the consideration of the eigenvalues of a
self-adjoint transformation is nonvacuous in either case, and we shall now proceed
with the rest of the theorem.
Let λ be an eigenvalue of A and suppose x is a corresponding eigenvector. Thus
we have Ax = 2x. Since x cannot be zero, we can choose x₁ = x/x|| and if, in
addition, dim X = 1, then we are done. If not, we shall proceed by induction;
i.e., we shall assume that the theorem is true for all spaces of dimension less than
X and then show that it follows for X from this assumption. Letting M = [x₂] =
{ax₁ αe F}, the space spanned by x₁, we have the following direct sum decompo-
sition of X:
X = M + M¹,
and we can note immediately that we must have dim M¹ <dim X. As shown in
Chapter 1, M being invariant under A implies that M is invariant under 4*,
which, in this case since A is self-adjoint, is just A. Consider now the restriction of A
to M denoted by AM where AM₁ M² → M². It is clear that, if A is self-adjoint,
then so must be the restriction of A to M¹. Now, since dim M <dim X, we can
apply the induction hypothesis to assert the existence of an orthonormal basis for
M¹ consisting of eigenvectors for A/M₁, {X₂, X3,...,x. Eigenvectors of AM,
however, must also be eigenvectors of A. Hence for the entire space we have
{x₁, x2, ..., x} as orthonormal basis of eigenvectors of A.
{X1,
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