S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 9: Inner Product Spaces

9.5 Orthogonal and Unitary Operators

9.5.1 Definitions and Characterizations

Orthogonal (real) and unitary (complex) operators are the isometries of inner product spaces: they preserve lengths and angles.

Definition 9.5.1 (Orthogonal operator). Let be a real inner product space. An operator is orthogonal if for all .

Definition 9.5.2 (Unitary operator). Let be a complex inner product space (with Hermitian inner product ). An operator is unitary if for all .

Note. We use “orthogonal” over and “unitary” over , though the definitions are parallel. Some authors use “orthogonal” for both.

Theorem 9.5.1 (Characterizations of orthogonal/unitary operators). For an operator on a finite-dimensional inner product space, the following are equivalent:

  1. is orthogonal (resp. unitary).
  2. (equivalently, ).
  3. .
  4. for all .
  5. maps some orthonormal basis to an orthonormal basis.
  6. maps every orthonormal basis to an orthonormal basis.

Proof. (i) (ii): for all . By nondegeneracy, .

(ii) (iii): implies is invertible with . Then .

(ii) (iv): .

(iv) (i): Use the polarization identity. Over : If preserves norms: .

(i) (vi): If is orthonormal, then .

(vi) (v): Trivial.

(v) (ii): Let be orthonormal with orthonormal. For any : Since is orthonormal, , so (the dual basis property in an orthonormal basis). Thus . ∎

Corollary 9.5.1 (Orthogonal/unitary operators are invertible). If is orthogonal or unitary, then , which is also orthogonal/unitary.

Corollary 9.5.2 (Orthogonal/unitary operators form a group). The set of orthogonal (resp. unitary) operators on forms a subgroup of .

Proof. Identity: , so . Closure: if and , then . Inverses: , so . ∎

Matrix Characterization

Proposition 9.5.1 (Orthogonal and unitary matrices). Let be an orthonormal basis of , and let be the matrix of in this basis.

  1. (Real case) is orthogonal iff , i.e., is an orthogonal matrix.
  2. (Complex case) is unitary iff , i.e., is a unitary matrix.

Equivalently: the columns of form an orthonormal set (with respect to the standard inner product on ).

Proof. In an orthonormal basis, . So iff .

The columns of are (standard basis vectors). The -entry of is . So iff columns are orthonormal. ∎

Corollary 9.5.3 (Rows of orthogonal/unitary matrices). If is orthogonal (resp. unitary), then the rows of also form an orthonormal set.

Proof. implies is invertible with . Thus , which says the columns of (= rows of ) are orthonormal. ∎

Proposition 9.5.2 (Determinant of orthogonal/unitary operators). If is orthogonal or unitary, then .

For orthogonal (real case), .

Proof. . But .

Over , , so , giving . ∎

9.5.2 Reflections

A particularly simple class of orthogonal operators are reflections across hyperplanes.

Definition 9.5.3 (Reflection). Let be a real inner product space. A reflection is an orthogonal operator such that and .

Note. The condition says is an involution. Since is orthogonal, it is also self-adjoint (). The eigenvalues of are (since ), and means the -eigenspace is 1-dimensional.

Proposition 9.5.3 (Formula for reflections). Every reflection can be written as where with . Geometrically, reflects across the hyperplane .

Proof. Define . We verify is a reflection.

Linear: Clear from the formula.

Orthogonal:

Involution: .

Eigenspaces: , so spans the -eigenspace. If , then , so is the -eigenspace.

Conversely, any reflection has one-dimensional. Let span this kernel with . Then , and on , acts as (since is diagonalizable with eigenvalues , and the -eigenspace is spanned by ). For with the second term in : ∎

Proposition 9.5.4 (Reflections map equal-length vectors). Given any two vectors with , there exists a reflection with .

Proof. Case 1: . If , any reflection fixes the origin, so any reflection works. If (and ), choose any unit vector orthogonal to and set . Then .

Case 2: . Let and define .

Compute:

Since , we have , so

Substituting: ∎

Theorem 9.5.2 (Orthogonal operators as products of reflections). Every orthogonal operator on is a product of at most reflections.

Proof. Induction on . For , the identity is the empty product.

For : Let be orthogonal. Pick a unit vector . Then .

By the previous proposition, there exists a reflection with . Let . Then is orthogonal and .

Since , the subspace is -invariant (if , then ).

The restriction is orthogonal on the -dimensional space . By induction, for reflections on .

Extend each to by setting and . Then is a reflection on (the -eigenspace of in becomes the -eigenspace of in ).

We have , so , giving (since ). This is a product of at most reflections. ∎

9.5.3 The Classical Matrix Groups

The orthogonal and unitary operators form important matrix groups.

Definition 9.5.4 (Orthogonal and special orthogonal groups).

  • , the orthogonal group.
  • , the special orthogonal group.

Definition 9.5.5 (Unitary and special unitary groups).

  • , the unitary group.
  • , the special unitary group.

Proposition 9.5.5 (Index of in ). is a normal subgroup of of index 2. We have an exact sequence

Proof. The determinant map is a surjective homomorphism (surjective since and , or use a reflection with ). Its kernel is . ∎

Note. Elements of are called rotations or proper orthogonal transformations. They preserve orientation. Elements of reverse orientation.

Example 9.5.1 (Low-dimensional cases).

  • , .
  • consists of rotation matrices . Thus (the circle group) via .
  • .

Theorem 9.5.3 (Structure of orthogonal operators). Let where is a finite-dimensional real inner product space. Then decomposes as an orthogonal direct sum where is the -eigenspace, is the -eigenspace, and each is a 2-dimensional -invariant subspace on which acts as rotation by angle .

Proof. The characteristic polynomial of has real coefficients. Its roots (possibly complex) come in conjugate pairs. If is an eigenvalue with and , then is also an eigenvalue.

Let , an irreducible quadratic over . By the real Jordan form theory (or direct argument), is an even-dimensional -invariant subspace.

Decompose into generalized eigenspaces for real eigenvalues () and for conjugate pairs of complex eigenvalues. Since is orthogonal and hence normal, it is diagonalizable over , so generalized eigenspaces are actual eigenspaces.

For the conjugate pair , the real subspace is -invariant. Choose a unit eigenvector for . Then and are real vectors spanning a 2-dimensional -invariant subspace on which acts as rotation by .

Repeating this decomposition and using orthogonality of distinct eigenspaces gives the result. ∎

Note. In matrix form, choosing an orthonormal basis adapted to this decomposition, is block-diagonal: where .

This is the “real canonical form” for orthogonal matrices.

Hermitian Forms and Real Forms

When a complex vector space is viewed as a real vector space, a Hermitian form decomposes into real and imaginary parts with specific properties.

Proposition 9.5.6 (Decomposition of Hermitian forms). Let be a complex vector space with Hermitian form , and let denote viewed as a real vector space (of twice the dimension). Define by Let be scalar multiplication by . Then:

  1. is a symmetric bilinear form on .
  2. is a skew-symmetric bilinear form on .
  3. for all .
  4. is nondegenerate iff is nondegenerate iff is nondegenerate.

Proof. (i) is -bilinear since is sesquilinear and is -linear. For symmetry:

(ii) Similarly, .

(iii) .

(iv) Suppose is nondegenerate and for all . Then for all , so for all (since is surjective). Thus for all , giving . The other implications are similar. ∎

Note. A linear operator is -linear (i.e., ) iff it comes from a -linear operator on . The proposition implies: for nondegenerate , any two of the following imply the third:

  1. preserves .
  2. preserves .
  3. is -linear.

This gives the relation .

Exercise 9.5.1 (Inner product spaces exercises).

  1. Let be the space of polynomials of degree at most 2, with inner product . Apply Gram–Schmidt to the basis to find an orthonormal basis.
  2. Let be self-adjoint on a real inner product space, and suppose with maximizes the Rayleigh quotient among unit vectors. Prove that is an eigenvector of . (Do not use the spectral theorem.)
  3. Let be an invertible linear operator on a finite-dimensional real inner product space. Show that maps the unit sphere to an ellipsoid. (Hint: consider .)
  4. For an invertible operator , define the dilatation . Show that equals the ratio of the largest to smallest eigenvalues of .
  5. Let be the matrix of a (symmetric positive-definite) inner product on with respect to the standard basis, so the inner product is . For , prove that the largest entry of occurs on the diagonal.
Show solution

(1) Start with , , .

, so .

. . . So .

. . . . So .

The orthonormal basis is .

(2) Let . For and small, consider . Since and : Since is self-adjoint, (real), so: Thus: At , is a maximum, so .

Differentiating: . So for all .

Thus , meaning . So for some .

(3) We have . Now where is self-adjoint and positive-definite.

By the spectral theorem, there exists an orthonormal eigenbasis with , where .

For : .

So . Setting , this is , an ellipsoid with semi-axes along .

(4) We have . The operator is self-adjoint and positive-definite (since is invertible).

By the spectral theorem, let be an orthonormal eigenbasis with , . For with (i.e., ):

This is maximized when all weight is on the largest eigenvalue: . It is minimized when all weight is on the smallest: .

Thus .

(5) For , let with (positive-definiteness requires positive diagonal) and .

We show . Suppose not: and . Then , contradicting .

So the largest entry in magnitude is on the diagonal. Since and , the largest entry is , on the diagonal.