S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 4: Categories, Bilinear Forms and Inner Products

16 Adjoints, the Spectral Theorem, and Orthogonal Operators

· Denis Auroux

Original Professor Auroux’s handwritten notes (PDF), behind a password for now:

Recall. A (finite-dimensional!) inner product space is an -vector space together with a symmetric positive definite bilinear form :

(and then we set , etc.).

We’ve seen that there exists an orthonormal basis:

so :

Operators on inner product spaces

Let be an inner product space.

Definition 16.1. Say is an orthogonal operator if it respects the inner product, i.e.

(In other terms, “preserves lengths and angles”.)

Remark (Remarks).

  1. Orthogonal operators map orthonormal bases to orthonormal bases! In particular, orthogonal operators are always invertible!

  2. If is orthogonal then is orthogonal: If are orthogonal then so is (check!). Hence: orthogonal operators form a subgroup of .

  3. If is the matrix representing in an orthonormal basis, then . Indeed, the entries of are the dot products of the columns of !

Adjoint operator

Definition 16.2. Let be a linear operator on an inner product space . There exists a unique linear operator , called the adjoint of , such that

Indeed: given , the linear functional is, using nondegeneracy of , given by the inner product of with some element of , which we call ; then check this has linear dependence on .

Equivalently: defines an isomorphism . Then is the composition

Proposition 16.3.

  1. ;
  2. .

Proof. (1) . Similarly for (2). ∎

Definition 16.4. is self-adjoint if (i.e. for all ).

In an orthonormal basis of , (the transpose gives a row vector, is a column vector). So if the matrix of is and that of is ,

Hence: in an orthonormal basis; is self-adjoint is symmetric.

Note that self-adjoint operators ( symmetric matrices) need not be invertible. For example is a self-adjoint operator…

Proposition 16.5. If is self-adjoint and is an invariant subspace (), then is also an invariant subspace ().

Proof. Let . Then for all , , so Since for all , we get . ∎

Theorem 16.6 (The spectral theorem for real self-adjoint operators). If is self-adjoint then is diagonalizable, with real eigenvalues. Even more, can be diagonalized in an orthonormal basis of !

The proof uses the following key observation:

Lemma 16.7. If is self-adjoint then for all , is invertible.

Proof. For all , So . Hence . ∎

Corollary 16.8. If is a quadratic without real roots and , then is invertible.

Proof. It is enough to show is invertible whenever . Write so by the lemma (applied to ) this is invertible. ∎

Proof of the spectral theorem.

  • First we show the existence of an eigenvector. Pick , . Since are linearly dependent (), there exists a (nonconstant) polynomial such that . This doesn’t factor into degree 1 factors over like it would over , but it factors into linear and quadratic factors: where the are the real roots, and the are irreducible (no real roots), coming from pairs of complex conjugate roots. At least one of these operators must have a nontrivial kernel (else their product is invertible, but !). By the previous corollary, each is invertible, so in fact some must have a nontrivial kernel, hence an eigenvector!

  • Now, diagonalization: we know there’s an eigenvector with eigenvalue ; scaling if needed we may assume . Then is an invariant subspace, hence (by the proposition above) so is . By induction, using the inner product on induced by restricting and observing is still self-adjoint, there is a basis of , (orthonormal if we wish), such that each is an eigenvector of . Then is a basis of in which diagonalizes, and we can assume it is orthonormal.

∎

So:

Remark. This also implies: eigenvectors of for distinct eigenvalues are orthogonal! But we already knew this, because so .

Structure of orthogonal transformations

Next: is there an analogous structure result for orthogonal transformations??

  • In dimension 1: is multiplication by a scalar, so orthogonal .
  • In dimension 2: orthogonal is a rotation or a reflection. (Given an orthonormal basis , is any unit vector in the unit circle ; is also a unit vector and , so there are 2 possibilities:) Rotations (other than by or ) have no eigenvectors. Reflections have eigenvalues , with two orthogonal eigenspaces.

Notation. For , we have subgroups where = invertible linear operators ; = orthogonal ones; and = the subgroup of orientation-preserving orthogonal transformations: those with . (In dimension 1: ; in dimension 2: rotations.)

Since by choosing an orthonormal basis, we usually write , .

has index 2 in : (rotations angles).

Recall: linear operator there exists an invariant subspace of dimension 1 or 2. And if is orthogonal for , then it maps to .

Theorem 16.9. If is an orthogonal operator on a finite-dimensional inner product space, then decomposes into a direct sum of orthogonal invariant subspaces of , and

  • if then ;
  • if then is either a rotation or a reflection.

(In the latter case, we can further decompose into eigenspaces, so we can replace reflections by 1-dimensional blocks.)

This gives a very nice way to think about an individual transformation as built from reflections and rotations on individual subspaces, but it’s pretty useless for understanding the composition of two orthogonal transformations (whose invariant subspaces are unrelated).

Notes not reached in this lecture, covered at the start of {{LINK:L17}}

Hermitian inner products

Now on to the analogue of all this for complex vector spaces: Hermitian inner products.

As previously noted, a bilinear form on a complex vector space can’t be positive definite, since . Solution: abandon -linearity in one of the two variables, and only require “conjugate linear”.

Definition 16.10. A Hermitian form on a complex vector space is such that is sesquilinear:

  • , ;
  • , however (conjugate: );

and is conjugate-symmetric: .

Note. Caution: this is as in Artin. Axler has for the second input.

Conjugate symmetry .

Definition 16.11. A Hermitian inner product is a positive definite (conjugate-symmetric) Hermitian form, i.e. for all , and .

Remark. , , is now a complex antilinear map ! (.)

Still, various things carry over from the real case:

  • positive definite nondegenerate (i.e. ).
  • Given a subspace , its orthogonal is also a subspace, and . (-antilinearity doesn’t affect being a -subspace; positive definite implies .)

Definition 16.12. An orthonormal basis of with a Hermitian inner product is a basis such that

Theorem 16.13. admits an orthonormal basis.

Same proof as in the real case (by induction on : first pick with , then take an orthonormal basis of ) (or Gram–Schmidt…).

Corollary 16.14. Every finite-dimensional Hermitian inner product space is isomorphic to with the standard Hermitian inner product,

In matrix form: , where is the conjugate transpose.