S. D. V. Stephens

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

15 Inner Product Spaces, Orthogonal and Self-Adjoint Operators

· Denis Auroux

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

Recall: bilinear forms

  • A bilinear form corresponds to , This gives an isomorphism .
  • is nondegenerate if is an isomorphism.
  • In a basis , represent by a matrix with entries : is symmetric iff is symmetric (), and nondegenerate iff is invertible.

Definition 15.1. If is a subspace of a vector space equipped with a bilinear form , we define its orthogonal complement This is a vector subspace.

Warning: this is most intuitive if is symmetric or skew. Otherwise we have to worry about “left-orthogonal” vs. “right-orthogonal” to .

Lemma 15.2. If is nondegenerate then (else ).

Proof. where this map is the composition of and the restriction , . By the rank theorem, . If is nondegenerate then is an isomorphism and is surjective, so ; in general . ∎

Example 15.3.

  • with the standard dot product : then , the “usual” orthogonal complement, because (see below) and .
  • But for , if is the 1-dimensional subspace spanned by any nonzero vector , then !! (Because .)

Inner product spaces

Definition 15.4. An inner product space is a vector space over together with a symmetric, definite positive bilinear form .

Symmetric: . Definite positive: for all , and iff .

This definition only makes sense over an ordered field, so that “” makes sense. In practice this means . We can’t do this over (we’ll see a workaround: Hermitian forms).

  • Let , , be the linear map corresponding to . Then definite positive is injective (since for all , ! ) (assuming ) is an iso , i.e. is nondegenerate. (The converse is false: nondegenerate positive.)

Proposition 15.5. a finite-dimensional inner product space, a subspace .

Proof.

  • We’ve seen: is nondegenerate, so .
  • Since is positive definite, . So . Since the dimensions add up to , this implies .

∎

Definition 15.6.

  • The norm of a vector is .
  • are orthogonal if .

Observe: .

  • If and are orthogonal then (Pythagorean theorem).
  • In general, by analogy with the law of triangles, we define the angle between 2 vectors, This only makes sense if ?

Theorem 15.7 (Cauchy–Schwarz inequality). For all , .

Proof. The inequality is unaffected by scaling, so we can assume . Decompose along where . Explicitly, , with and orthogonal to . Then , so This is the desired inequality for . ∎

Orthonormal bases

Definition 15.8. Let be finite-dimensional over with inner product . A basis of is said to be orthonormal if

In such a basis, .

Theorem 15.9. Every finite-dimensional inner product space (over ) has an orthonormal basis.

Proof 1. By induction on : choose , and let , so . Then let , so . Let be an orthonormal basis for (the restriction of to is an inner product!). Then is an orthonormal basis for (check!). ∎

Proof 2. Start with any basis of and use the Gram–Schmidt process. First set . Then take , which is (and nonzero by independence of the ), and set and so on: set Then is an orthonormal basis. ∎

So: every finite-dimensional inner product space over is isomorphic (as an inner product space, not just as a vector space) to the standard , .

Notes not reached in this lecture, covered at the start of Lecture 16

Operators on inner product spaces

Let be an inner product space. There are two special classes of linear operators on of interest to us.

Definition 15.10. Say is an orthogonal operator if it respects the inner product, i.e. for all .

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

Remarks.

  1. Orthogonal operators map orthonormal bases to orthonormal bases! In particular, orthogonal operators are always invertible!
  2. If is orthogonal then is orthogonal: where the first equality is because is orthogonal. If are orthogonal then so is (check!).

Proposition 15.11. Hence: orthogonal operators form a subgroup of .

  1. 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 15.12. 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

Definition 15.13. 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 ,

Proposition 15.14. 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 15.15. If is self-adjoint and is an invariant subspace (), then is also an invariant subspace ().

Proof. Let . Then for all , , so where the first equality is because and the second because , . Since for all , we get . ∎

Theorem 15.16 (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 (to be seen next time) uses the following key observation:

Lemma 15.17. If is self-adjoint then for all (i.e. ), is invertible.

Proof. For all , So . Hence . ∎

Corollary 15.18. If is a quadratic without real roots and , then is invertible.

Proof. It is enough to show is invertible whenever . Write where is self-adjoint; so by the lemma (applied to ) this is invertible. ∎