9.4.1 The Spectral Theorem (Real Symmetric Case)
We now study operators with special relationships to their adjoints. The most important case is when .
Definition 9.4.1 (Self-adjoint operator). An operator on an inner product space is self-adjoint (or Hermitian) if , i.e., for all .
Note. In an orthonormal basis, is self-adjoint iff its matrix is symmetric (real case) or Hermitian (complex case, ). Self-adjoint operators need not be invertible: for example, the zero operator is self-adjoint.
Proposition 9.4.1 (Invariance of orthogonal complements). Let be self-adjoint and a -invariant subspace (i.e., ). Then is also -invariant.
Proof. Let . For any , we have (since is -invariant), so: since . Thus , i.e., . ∎
Note. This is the key property that makes self-adjoint operators diagonalizable: once we find an eigenvector, its orthogonal complement is invariant, and we can proceed by induction.
To prove the spectral theorem, we need to establish that self-adjoint operators have real eigenvalues and that eigenvectors exist over .
Lemma 9.4.1 (Positivity lemma). Let be self-adjoint on a real inner product space. For any , the operator is invertible.
Proof. We show . For : So for , i.e., is injective, hence invertible. ∎
Corollary 9.4.1 (Quadratics without real roots). If has no real roots (i.e., ), and is self-adjoint, then is invertible.
Proof. Complete the square: .
Since , we have . The operator is self-adjoint (check: ). By the lemma, is invertible. ∎
Theorem 9.4.1 (Spectral theorem (real self-adjoint)). Let be a self-adjoint operator on a finite-dimensional real inner product space. Then:
- All eigenvalues of are real.
- is diagonalizable.
- has an orthonormal basis consisting of eigenvectors of .
Proof. We first show that has an eigenvector. Since is finite-dimensional, choose any nonzero . The vectors (where ) are linearly dependent, so there exists a nonzero polynomial with .
Factor over : since has unique factorization, we can write where the linear factors have real roots and the quadratic factors are irreducible (no real roots, i.e., ).
By the corollary, each is invertible. So Since the quadratic factors are invertible, we can remove them: for some nonzero .
At least one factor must have nontrivial kernel (otherwise their product would be invertible). Thus has an eigenvalue .
(i) Eigenvalues are real: This is automatic since we found the eigenvalue in .
(iii) Orthonormal eigenbasis: We proceed by induction on .
Base case: . Any nonzero vector is an eigenvector; normalize it.
Inductive step: Let be an eigenvector with , normalized so . Let . Since , the orthogonal complement is -invariant.
The restriction is self-adjoint (the inner product on is the restriction of the inner product on ). By induction, has an orthonormal eigenbasis .
Then is an orthonormal eigenbasis of : the are orthonormal, and each is an eigenvector of .
(ii) Diagonalizable: In an eigenbasis, has a diagonal matrix. ∎
Note. The spectral theorem says that self-adjoint operators on real inner product spaces are “as nice as possible”: they are diagonalizable, with real eigenvalues, and the eigenspaces are mutually orthogonal. The matrix in an orthonormal eigenbasis is .
Corollary 9.4.2 (Eigenvectors for distinct eigenvalues are orthogonal). If is self-adjoint and , with , then .
Proof. . So , and implies . ∎
Skew-Adjoint Operators
The “opposite” of self-adjoint is skew-adjoint.
Definition 9.4.2 (Skew-adjoint operator). An operator on an inner product space is skew-adjoint (or anti-self-adjoint) if , i.e., for all .
Note. In an orthonormal basis, is skew-adjoint iff its matrix satisfies (real case) or (complex case). Such matrices are called skew-symmetric or skew-Hermitian.
Proposition 9.4.2 (Properties of skew-adjoint operators). Let be skew-adjoint on a real inner product space.
- for all .
- If is an eigenvalue of (in ), then is purely imaginary.
- is invertible.
Proof. (i) . Over , this gives .
(ii) If with , then by (i), so . More generally, working in the complexification, if is an eigenvalue, the skew-adjoint condition forces .
(iii) Suppose , i.e., . Then by (i), so . Thus is injective, hence invertible. ∎
Proposition 9.4.3 (Cayley transform). Let be skew-adjoint on a real inner product space. Then is orthogonal (i.e., ).
Proof. Since , we have and . Also, and commute (both are polynomials in ).
Then: And: Since and commute, . Thus . ∎
9.4.2 Normal Operators on Complex Spaces
Over , the spectral theorem extends to a larger class of operators: those that commute with their adjoint.
Definition 9.4.3 (Normal operator). An operator on an inner product space is normal if
Example 9.4.1 (Classes of normal operators).
- Self-adjoint operators () are normal: .
- Skew-adjoint operators () are normal: .
- Unitary operators () are normal: .
Lemma 9.4.2 (Characterization of normality). An operator is normal if and only if for all .
Proof. and .
So for all iff for all .
For self-adjoint operators (and is self-adjoint), this implies . Indeed, if is self-adjoint with for all , then for any : and similarly with gives the imaginary part, so for all , hence . ∎
Proposition 9.4.4 (Normal operators and eigenspaces). Let be normal.
- If , then .
- Eigenvectors for distinct eigenvalues are orthogonal.
- .
Proof. (i) Note that is normal (check: , and these commute since and commute).
By the lemma, . If , the left side is zero, so .
(ii) If and with : So , giving .
(iii) Follows from (i): . ∎
Theorem 9.4.2 (Complex spectral theorem). Let be a normal operator on a finite-dimensional complex inner product space (with Hermitian inner product). Then has an orthonormal basis consisting of eigenvectors of .
Proof. Since is algebraically closed, has an eigenvalue . Let be a unit eigenvector with .
Let . We show is -invariant. By part (i) of the proposition, , so is also -invariant.
For and any : Since and , we get . Thus , i.e., .
The restriction is normal (the inner product restricts, and ). By induction on dimension, has an orthonormal eigenbasis . Then is an orthonormal eigenbasis of . ∎
Note. The complex spectral theorem is stronger than the real one: any normal operator (not just self-adjoint) is diagonalizable. The price is working over with Hermitian inner products.
For self-adjoint , the eigenvalues are real. For skew-adjoint , they are purely imaginary. For unitary , they lie on the unit circle .