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Math. 115A H. B. Enderton
The Spectral Theorem
For the final topic in this course, we combine our work in Chapter 5 (on diagonalization) with our work in Chapter 6 (on inner product spaces).
We have seen that a linear operator T on a finite-dimensional vector space V is diagonalizable if and only if V has a basis of eigenvectors of T. Now suppose that V is actually an inner product space. Then we might hope to obtain an orthonormal basis of eigenvectors, so that T , relative to this orthonormal basis, is represented by a diagonal matrix or even a real diagonal matrix.
For example, in Problem Set VIII, a certain symmetric matrix A has the property that LA is rep- resented by a diagonal matrix, relative to an orthonormal basis for R^3. And a certain Hermitian matrix B, over C, has the property that LB is represented by a real diagonal matrix, relative to an orthonormal basis for C^3. We want to investigate this phenomenon in general.
Certainly orthonormal bases are the “nicest” bases. Suppose that β is an orthonormal ordered basis {b 1 , b 2 ,... , bn} for an n-dimensional inner product space V. Then coordinate vectors can be found by using the Fourier coefficients:
[v]β =
〈v, b 1 〉 .. . 〈v, bn〉
so that v = 〈v, b 1 〉b 1 + · · · + 〈v, bn〉bn.
Relative to β, a linear operator T is represented by the matrix A = [T ]ββ whose entries are given by the equation: (A)ij = 〈T (bj ), bi〉
(As a special case, the standard ordered basis {e 1 ,... , en} for Cn^ or Rn^ is orthonormal, and (A)ij = 〈Aej , ei〉 = etiAej .) And the coordinate map [ ]β relative to an orthonormal basis is an isometry, i.e., it preserves the inner product: 〈u, v〉 = 〈[u]β , [v]β 〉
where the inner product on the left is in V , and the inner product on the right is in Cn^ or Rn.
The following fact gives us cause for optimism regarding symmetric matrices.
Lemma. Consider the space Rn^ with the usual inner product, and assume that A is an n × n (real) symmetric matrix. Then any two eigenvectors for different eigenvalues are orthogonal.
Proof. We make use of the fact that for any real matrix B, symmetric or not, 〈Bx, y〉 = 〈x, Bty〉.
So for a symmetric matrix A, 〈Ax, y〉 = 〈x, Ay〉.
Now suppose that x and y are eigenvectors for the distinct eigenvalues λ and μ, respectively. Then λ〈x, y〉 = 〈λx, y〉 = 〈Ax, y〉 = 〈x, Ay〉 = 〈x, μy〉 = μ〈x, y〉. Since λ 6 = μ, we must have 〈x, y〉 = 0 and hence x ⊥ y. a
Thus for a real symmetric matrix, its various eigenspaces will all be orthogonal to each other. For each individual eigenspace Eλ, we can make an orthonormal basis, using the Gram–Schmidt process. When we combine all these bases for all the different Eλ’s, the resulting set will still be orthonormal (by the lemma).
On the other hand, if A is a real n × n matrix that is not symmetric, then as we will see, there is no orthonormal basis for Rn^ that diagonalizes A. So it is natural for us to focus our attention on symmetric matrices.
As mentioned above, for a symmetric matrix, 〈Ax, y〉 = 〈x, Ay〉
for x and y in Rn. In fact this property characterizes exactly the symmetric matrices: If the displayed equation is always true of A, then (A)ij = 〈Aej , ei〉 = 〈ej , Aei〉 = 〈Aei, ej 〉 = (A)ji. This is the clue as to which linear operators have the best chance of being represented in the way we seek: those operators T with the property that 〈T (u), v〉 = 〈u, T (v)〉
for all vectors u and v. Call such operators self-adjoint.
(Digression: Section 6.3 defines the adjoint T ∗^ of T to be the unique operator such that 〈T (u), v〉 = 〈u, T ∗(v)〉
always holds. Then the self-adjoint operators can be defined by requiring that T = T ∗. This definition is equivalent to the definition given above.)
In the complex inner product space Cn, the usual inner product is defined by the equation 〈x, y〉 = y∗x. A complex n × n matrix A is called Hermitian if A = A∗. In particular, a real matrix is Hermitian if and only if it is symmetric. Note that in a Hermitian matrix, all the entries on the main diagonal must be real. Recall that for any matrix A, we have 〈Ax, y〉 = 〈x, A∗y〉. Thus for a Hermitian matrix A,
〈Ax, y〉 = 〈x, Ay〉
for x and y in Cn.
Thus multiplication LA by a Hermitian n × n matrix A is a self-adjoint linear operator on Cn, by the above. Multiplication by a real symmetric matrix is a self-adjoint linear operator on Rn.
Theorem 0. Let T be a linear operator on an n-dimensional real or complex inner product space V , and let β be an orthonormal ordered basis for V. Then T is self-adjoint if and only if its representing matrix A relative to β is a symmetric matrix (in the real case), or a Hermitian matrix (in the complex case).
Proof. Suppose that β is an orthonormal ordered basis 〈b 1 , b 2 ,... , bn〉 and let A = [T ]ββ. On the one hand, if T is self-adjoint then
(A)ij = 〈T (bj ), bi〉 = 〈bj , T (bi)〉 = 〈T (bi), bj 〉 = (A)ji
so that A is a Hermitian matrix (and if real, is symmetric).
On the other hand, if A is a Hermitian matrix then 〈T (u), v〉 = 〈[T (u)]β , [v]β 〉 = 〈A[u]β , [v]β 〉 = 〈[u]β , A[v]β 〉 = 〈[u]β , [T (v)]β 〉 = 〈u, T (v)〉
so that T is a self-adjoint operator. a
In particular, any real diagonal matrix D is automatically symmetric (and Hermitian). So Theorem 0 tells us that if we hope to have T represented by D relative to an orthonormal basis, then T must be self-adjoint. In other words, at most the self-adjoint operators have the property we want. What is left is the other direction: to show that any self-adjoint operator is indeed represented by a real diagonal matrix relative to an orthonormal basis.
In Chapter 5, we saw that there were two potential barriers to diagonalization. One is that the characteristic polynomial might have non-real roots (in a real vector space). The other barrier is that the dimension of some eigenspace Eλ might fall short of the algebraic multiplicity of λ in the characteristic polynomial. Theorem 1 below will remove the first barrier. Finally, Theorem 2 will remove the second.
Our first objective is to prove that all eigenvalues of a self-adjoint linear operator are real numbers.
Theorem 1. (a) Any eigenvalue of a self-adjoint linear operator is real. (b) For a real symmetric matrix, any root of its characteristic polynomial is real. (c) For a Hermitian matrix, any root of its characteristic polynomial is real.
Corollary. (a) For a real n × n matrix A, there is an orthonormal basis for Rn^ relative to which LA is represented by a (real) diagonal matrix if and only if A is symmetric.
(b) For a complex n × n matrix B, there is an orthonormal basis for Cn^ relative to which LB is represented by a real diagonal matrix if and only if B is Hermitian.
Examples. See Problem Set VIII. For a real symmetric matrix A, we can make a (real) change- of-basis matrix Q with orthonormal columns (such matrices are called orthogonal) such that Q−^1 AQ is a real diagonal matrix. For a Hermitian matrix B, we can make a (complex) change-of-basis matrix U with orthonormal columns (such matrices are called unitary) such that U −^1 BU is a real diagonal matrix.
The spectral theorem is Theorem 6.17 on page 374. See also page 401. The spectrum of T is its list λ 1 ,... , λk of eigenvalues. The idea is that properly viewed (i.e., relative to just the right orthonormal basis), T can be taken apart into a linear combination of projections. For example, in terms of the representing matrix,
λ 0 0 0 0 λ 0 0 0 0 μ 0 0 0 0 ν
=^ λ
+^ μ
+^ ν
In terms of linear operators, in this example,
T = λPλ + μPμ + νPν
where Pλ the operation of orthogonal projection onto the eigenspace Eλ, and similarly for μ and ν. The sum Pλ + Pμ + Pν equals the identity operator (one says we have a “resolution of the identity”). The equation T = λPλ + μPμ + νPν
is referred to as the “spectral decomposition” of T.
(Digression: We have focused here on real diagonal matrices. If you are willing to accept complex diagonal matrices, then the condition of self-adjointness can be weakened to a condition called normality, discussed in the book.)
One question not yet addressed is this: What do we do if T is not diagonalizable? Give up? If T has some eigenvectors, then we can form a basis with as many eigenvectors as we can. The resulting matrix may not be diagonal, but it may be better than whatever we started with. Beyond that, there is something called Jordan normal form. In 115B, that form will be studied.
