We have seen the importance of diagonalizable operators in Chapter 5. For an operator on a vector space V to be diagonalizble, it is necessary and sufficient for V to contain a basis of eigenvectors for this operator. As V is an inner product space in this chapter, it is reasonable to seek conditions that guarantee that V has an orthonormal basis of eigenvectors. A very important result that helps achieve our goal is Schur’s theorem (Theorem 6.14). The formulation that follows is in terms of linear operators. The next section contains the more familiar matrix form. We begin with a lemma.
Lemma. Let T be a linear operator on a finite-dimensional inner product space V. If T has an eigenvector, then so does T*.
Suppose that v is an eigenvector of T with corresponding eigenvalue λ. Then for any x∈V
and hence v is orthogonal to the range of T*−ˉλI
Recall (see the exercises of Section 2.1 and see Section 5.4) that a subspace W of V is said to be T-invariant if T(W) is contained in W. If W is T-invariant, we may define the restriction TW: W→W
Let T be a linear operator on a finite-dimensional inner product space V. Suppose that the characteristic polynomial of T splits. Then there exists an orthonormal basis γ
By Exercise 12(a) of Section 5.2, there exists an ordered basis β={w1, w2, …, wn}
As in the proof of Theorem 6.4, span(Sk)=span(S′k)
We now return to our original goal of finding an orthonormal basis of eigenvectors of a linear operator T on a finite-dimensional inner product space V. Note that if such an orthonormal basis β
Let V be an inner product space, and let T be a linear operator on V. We say that T is normal if TT*=T*T
It follows immediately from Theorem 6.10 (p. 356) that T is normal if and only if [T]β
Let T: R2→R2
Note that AA*=I=A*A
Suppose that A is a real skew-symmetric matrix; that is, At=−A
Clearly, the operator T in Example 1 does not even possess one eigenvector. So in the case of a real inner product space, we see that normality is not sufficient to guarantee an orthonormal basis of eigenvectors. All is not lost, however. We show that normality suffices if V is a complex inner product space.
Before we prove the promised result for normal operators, we need some general properties of normal operators.
Let V be an inner product space, and let T be a normal operator on V. Then the following statements are true.
(a) ||T(x)||=||T*(x)||
(b) T−cI
(c) If x is an eigenvector of T corresponding to eigenvalue λ
(d) If λ1
Proof. (a) For any x∈V
The proof of (b) is left as an exercise.
(c) Suppose that T(x)=λx
Hence T*(x)=ˉλx
(d) Let λ1
Since λ1≠λ2
Let T be a linear operator on a finite-dimensional complex inner product space V. Then T is normal if and only if there exists an orthonormal basis for V consisting of eigenvectors of T.
Suppose that T is normal. By the fundamental theorem of algebra (Theorem D.4), the characteristic polynomial of T splits. So we may apply Schur’s theorem to obtain an orthonormal basis β={v1, v2, …, vn}
Furthermore, by the corollary to Theorem 6.5 (p. 345),
It follows that T(vk)=Akkvk
The converse was already proved on page 367.
Interestingly, as the next example shows, Theorem 6.16 does not extend to infinite-dimensional complex inner product spaces.
Consider the inner product space H with the orthonormal set S from Example 9 in Section 6.1. Let V=span(S)
for all integers n. Thus
It follows that U=T*
We show that T has no eigenvectors. Suppose that f is an eigenvector of T, say, T(f)=λf
Hence
Since am≠0
Example 1 illustrates that normality is not sufficient to guarantee the existence of an orthonormal basis of eigenvectors for real inner product spaces. For real inner product spaces, we must replace normality by the stronger condition that T=T*
Let T be a linear operator on an inner product space V. We say that T is self-adjoint (or Hermitian) if T=T*
It follows immediately that if β
Before we state our main result for self-adjoint operators, we need some preliminary work.
By definition, a linear operator on a real inner product space has only real eigenvalues. The lemma that follows shows that the same can be said for self-adjoint operators on a complex inner product space. Similarly, the characteristic polynomial of every linear operator on a complex inner product space splits, and the same is true for self-adjoint operators on a real inner product space.
Lemma. Let T be a self-adjoint operator on a finite-dimensional inner product space V. Then
(a) Every eigenvalue of T is real.
(b) Suppose that V is a real inner product space. Then the characteristic polynomial of T splits.
(a) Suppose that T(x)=λx
So λ=ˉλ
(b) Let n=dim(V), β
We are now able to establish one of the major results of this chapter.
Let T be a linear operator on a finite-dimensional real inner product space V. Then T is self-adjoint if and only if there exists an orthonormal basis β
Suppose that T is self-adjoint. By the lemma, we may apply Schur’s theorem to obtain an orthonormal basis β
So A and A* are both upper triangular, and therefore A is a diagonal matrix. Thus β
The converse is left as an exercise.
We restate this theorem in matrix form in the next section (as Theorem 6.20 on p. 381).
As we noted earlier, real symmetric matrices are self-adjoint, and self-adjoint matrices are normal. The following matrix A is complex and symmetric:
But A is not normal, because (AA*)12=1+i
Label the following statements as true or false. Assume that the underlying inner product spaces are finite-dimensional.
(a) Every self-adjoint operator is normal.
(b) Operators and their adjoints have the same eigenvectors.
(c) If T is an operator on an inner product space V, then T is normal if and only if [T]β
(d) A real or complex matrix A is normal if and only if LA
(e) The eigenvalues of a self-adjoint operator must all be real.
(f) The identity and zero operators are self-adjoint.
(g) Every normal operator is diagonalizable.
(h) Every self-adjoint operator is diagonalizable.
For each linear operator T on an inner product space V, determine whether T is normal, self-adjoint, or neither. If possible, produce an orthonormal basis of eigenvectors of T for V and list the corresponding eigenvalues.
(a) V=R2
(b) V=R3
(c) V=C2
(d) V=P2(R)
(e) V=M2×2(R)
(f) V=M2×2(R)
Give an example of a linear operator T on R2
Let T and U be self-adjoint operators on an inner product space V. Prove that TU is self-adjoint if and only if TU=UT
Prove (b) of Theorem 6.15.
Let V be a complex inner product space, and let T be a linear operator on V. Define
(a) Prove that T1
(b) Suppose also that T=U1+iU2
(c) Prove that T is normal if and only if T1T2=T2T1
Let T be a linear operator on an inner product space V, and let W be a T-invariant subspace of V. Prove the following results.
(a) If T is self-adjoint, then TW
(b) W⊥
(c) If W is both T- and T*-invariant, then (TW)*=(T*)W
(d) If W is both T- and T*-invariant and T is normal, then TW
Let T be a normal operator on a finite-dimensional complex inner product space V, and let W be a subspace of V. Prove that if W is T-invariant, then W is also T*-invariant. Hint: Use Exercise 24 of Section 5.4.
Let T be a normal operator on a finite-dimensional inner product space V. Prove that N(T)=N(T*)
Let T be a self-adjoint operator on a finite-dimensional inner product space V. Prove that for all x∈V
Deduce that T−iI
Assume that T is a linear operator on a complex (not necessarily finite-dimensional) inner product space V with an adjoint T*. Prove the following results.
(a) If T is self-adjoint, then 〈T(x), x〉
(b) If T satisfies 〈T(x), x〉=0
(c) If 〈T(x), x〉
Let T be a normal operator on a finite-dimensional real inner product space V whose characteristic polynomial splits. Prove that V has an orthonormal basis of eigenvectors of T. Hence prove that T is self-adjoint.
An n×n
Simultaneous Diagonalization. Let V be a finite-dimensional real inner product space, and let U and T be self-adjoint linear operators on V such that UT=TU
Let A and B be symmetric n×n
Prove the Cayley-Hamilton theorem for a complex n×n
Now if T=LA
The following definitions are used in Exercises 17 through 23.
A linear operator T on a finite-dimensional inner product space is called positive definite [positive semidefinite] if T is self-adjoint and 〈T(x), x〉>0 [〈T(x), x〉≥0]
An n×n
Let T and U be self-adjoint linear operators on an n-dimensional inner product space V, and let A=[T]β
(a) T is positive definite [semidefinite] if and only if all of its eigenvalues are positive [nonnegative].
(b) T is positive definite if and only if
(c) T is positive semidefinite if and only if A=B*B
(d) If T and U are positive semidefinite operators such that T2=U2
(e) If T and U are positive definite operators such that TU=UT
(f) T is positive definite [semidefinite] if and only if A is positive definite [semidefinite].
Because of (f), results analogous to items (a) through (d) hold for matrices as well as operators.
Let T: V→W
(a) T*T and TT* are positive semidefinite. (See Exercise 15 of Section 6.3.)
(b) rank(T*T)=rank(TT*)=rank(T).
Let T and U be positive definite operators on an inner product space V. Prove the following results.
(a) T+U
(b) If c>0
(c) T−1
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Let V be an inner product space with inner product 〈⋅, ⋅〉
Let V be a finite-dimensional inner product space, and let T and U be self-adjoint operators on V such that T is positive definite. Prove that both TU and UT are diagonalizable linear operators that have only real eigenvalues. Hint: Show that UT is self-adjoint with respect to the inner product 〈x, y〉′=〈T(x), y〉. To show that TU is self-adjoint, repeat the argument with T−1 in place of T.
This exercise provides a converse to Exercise 20. Let V be a finite-dimensional inner product space with inner product 〈⋅, ⋅〉, and let 〈⋅, ⋅〉′ be any other inner product on V.
(a) Prove that there exists a unique linear operator T on V such that 〈x, y〉′=〈T(x), y〉 for all x and y in V. Hint: Let β={v1, v2, …, vn} be an orthonormal basis for V with respect to 〈⋅, ⋅〉, and define a matrix A by Aij=〈vj, vi〉′ for all i and j. Let T be the unique linear operator on V such that [T]β=A.
(b) Prove that the operator T of (a) is positive definite with respect to both inner products.
Let U be a diagonalizable linear operator on a finite-dimensional inner product space V such that all of the eigenvalues of U are real. Prove that there exist positive definite linear operators T1 and T′1 and self-adjoint linear operators T2 and T′2 such that U=T2T1=T′1T′2. Hint: Let 〈⋅, ⋅〉 be the inner product associated with V, β a basis of eigenvectors for U, 〈⋅, ⋅〉′ the inner product on V with respect to which β is orthonormal (see Exercise 22(a) of Section 6.1), and T1 the positive definite operator according to Exercise 22. Show that U is self-adjoint with respect to 〈⋅, ⋅〉′ and U=T−11U*T1 (the adjoint is with respect to 〈⋅, ⋅〉). Let T2=T−11U*.