6.5 Unitary and Orthogonal Operators and their Matrices In this section we focus on length-preserving transformations of an inner product space. Throughout, as usual, we assume V is an inner product space. Definition 6.36. A linear isometry of an inner product space V over F is a linear map T satisfying x V, T(x) = x 8 2 jj jj jj jj It should be clear that every eigenvalue of an isometry must have modulus 1: if T(w) = lw, then w 2 = T(w) 2 = lw 2 = l 2 w 2 jj jj jj jj jj jj j j jj jj Example 6.37. Let T = L (R2), where A = 1 4 3 . Then A 2 L 5 3− 4 2 2 2 x 1 4x 3y 1 x T = − = (4x 3y)2 + (3x + 4y)2 = x2 + y2 = y 5 3x + 4y 25 − y The matrix in this example is very special in that its inverse is its transpose: 1 1 4 3 1 4 3 T A− = = = A 16 + 9 3 4 5 3 4 25 25 − − We call such matrices orthogonal. Definition 6.38. A unitary operator T on an inner product space V is an invertible linear map satis- fying T∗T = I = TT∗.A unitary matrix is a matrix satisfying A∗ A = I. • If V is real, we usually call these orthogonal operators/matrices: this isn’t necessary, since unitary encompasses both real and complex spaces. Note that an orthogonal matrix satisfies AT A = I. • If b is an orthonormal basis of a finite-dimensional V, then T (V) is unitary if and only if 2 L the matrix [T]b is unitary. • We need only assume T∗T = I (or TT∗ = I) if V is finite-dimensional: if b is an orthonormal basis, then T∗T = I [T∗] [T] = I [T] [T∗] = I TT∗ = I () b b () b b () If V is infinite-dimensional, we need T∗ to be both the left- and right-inverse of T. This isn’t an empty requirement: see Exercise 6.5.12.. 1 i 2+2i Example 6.39. The matrix A = 3 2 2i i is easily seen to be unitary: − 1 i 2 + 2i i 2 2i 1 i2 + 4 + 8i + 4i2 2 2i A∗ A = − − = − − 9 2 2i i 2 + 2i i 9 2 + 2i i − − − 1 Theorem 6.40. Let T be a linear operator on V. 1. If T is a unitary/orthogonal operator, then it is a linear isometry. 2. If T is a linear isometry and V is finite-dimensional, then T is unitary/orthogonal. Proof. 1. If T is unitary, then x, y V, x, y = T∗T(x), y = T(x),T(y) (†) 8 2 h i h i h i In particular taking x = y shows that T is an isometry. 2. (I T T) = I (T T) = I T T is self-adjoint. By the spectral theorem, there exists an − ∗ ∗ ∗ − ∗ ∗ − ∗ orthonormal basis of V of eigenvectors of I T T. For any such x with eigenvalue l, − ∗ 2 2 2 0 = x T(x) = x, x T(x),T(x) = x, (I T∗T)x = l x jj jj − jj jj h i − h i h − i jj jj = l = 0 ) Since I T T = 0 on a basis, T T = I. Since V is finite-dimensional, we also have TT = I − ∗ ∗ ∗ whence T is unitary. The finite-dimensional restriction is important in part 2: we use the existence of adjoints, the spectral theorem, and that a left-inverse is also a right-inverse. Again, see Exercise 6.5.12. for an example of a non-unitary isometry in infinite dimensions. The proof shows a little more: Corollary 6.41. On a finite dimensional space, being unitary is equivalent to each of the following: (a) Preservation of the inner producta (†). In particular, in a real inner product space isomteries x,y also preserve the angle q between vectors since cos q = xh yi . jj jjjj jj (b) The existence of an orthonormal basis b = w ,..., w such that T(b) = T(w ),...,T(w ) f 1 ng f 1 n g is also orthonormal. (c) That every orthonormal basis b of V is mapped to an orthonormal basis T(b). a(†) is in fact equivalent to being an isometry in infinite dimensions: recall the polarization identity. While (a) is simply (†), claims (b) and (c) are also worth proving explicitly: see Exercise 6.5.8. If b is n the standard orthonormal basis of F and T = LA, then the columns of A form the orthonormal set T(b). This makes identifying unitary/orthogonal matrices easy: Corollary 6.42. A matrix A M (R) is orthogonal if and only if its columns form an orthonormal 2 n basis of Rn with respect to the standard (dot) inner product. A matrix A M (C) is unitary if and only if its columns form an orthonormal basis of Cn with 2 n respect to the standard (hermitian) inner product. 2 Examples 6.43. 1. The matrix A = cos q sin q M (R) is orthogonal for any q. Example 6.37 is q sin q −cos q 2 2 this with q = tan 1 3 . More generally (Exercise 6.5.6.), it can be seen that every real orthogonal − 4 2 2 matrix has the form A or × q cos q sin q B = q sin q cos q − for some angle q. The effect of the linear map LAq is to rotate counter-clockwise by q, while that 1 of LBq is to reflect across the line making angle 2 q with the positive x-axis. p2 p3 1 1 2. A = p2 0 2 M3(R) is orthogonal: check the columns!. p6 − p2 p3 1 ! 2 − − 3. The matrix A = 1 1 i is unitary: indeed it maps the standard basis to the orthonormal basis p2 i 1 1 1 1 i T(b) = , p i p 1 2 2 It is also easy to check that the characteristic polynomial is 1 t i 2 p2 p2 1 1 1 pi/4 p(t) = det − 1 = t + = t = (1 i) = e± i t − p2 2 ) p2 ± p2 p2 − ! whence the eigenvalues of T both have modulus 1. 4. Here is an example of an infinite-dimensional unitary operator. On the space C[ p, p], the − function T( f (x)) = eix f (x) is linear. Moreover p p ix 1 ix 1 ix ix e f (x), g(x) = e f (x)g(x) dx = f (x)e− g(x) dx = f (x), e− g(x) 2p p 2p p D E Z− Z− D E ix 1 whence T∗( f (x)) = e− f (x). Indeed T∗ = T− and so T is a unitary operator. Since C[ p, p] is infinite-dimensional, we don’t expect all parts of the Corollary to hold: − • Being unitary, T preserves the inner product. However, in contrast to a unitary operator on a finite-dimensional complex space, T has no eigenvalues/eigenvectors, since T( f ) = l f x, eix f (x) = l f (x) f (x) 0 () 8 () ≡ • T certainly maps any orthonormal set to an orthonormal set, however it can be seen that C[ p, p] has no orthonormal basis!a − aAn orthonormal set b = f : k Z can be found so that every function f equals an infinite series in the sense that f k 2 g f ∑ a f = 0. However, these are not finite sums and so b is not a basis. Moreover, given that the norm is defined by jj − k kjj an integral, this also isn’t quite the same as saying that f = ∑ ak fk as functions. Indeed there is no guarantee that such an infinite series is itself continuous! For these reasons, when working with Fourier series, one tends to consider a broader class than the continuous functions. 3 Unitary and Orthogonal Equivalence Suppose A M (R) is symmetric (self-adjoint) AT = A. By the spectral theorem, A has an orthonor- 2 n mal eigenbasis b = w ,..., w : Aw = l w . If we write U = (w w ), then the columns of U f 1 ng j j j 1 ··· n are orthonormal and thus U is an orthogonal matrix. We can therefore write l1 0 1 . ···. T A = UDU− = U 0 . .. 1 U 0 l B ··· nC @ A The same approach works if A M (C) is normal: we now have A = UDU where U is unitary. 2 n ∗ 1+i 1+i Example 6.44. The matrix A = 1 i 1+i is normal as can easily be checked. Its characteristic polynomial is − − p(t) = t2 2(1 + i)t + 4i = (t 2i)(t 2) − − − with corresponding orthonormal eigenvectors 1 1 1 1 w = , w = 2 p i 2i p i 2 − 2 We conclude that 1 1 1 1 2 0 1 1 1 − 1 1 1 0 1 i A = = p i i 0 2i p i i i i 0 i 1 i 2 − 2 − − − This is an example of unitary equivalence: Definition 6.45. Square matrices A, B are unitarily equivalent if there exists a unitary matrix U such T that B = U∗ AU. Orthogonal equivalence is similar: B = U AU. The above discussion proves half the following: Theorem 6.46. A M (C) is normal if and only if it is unitarily equivalent to a diagonal matrix 2 n (the matrix of its eigenvalues). Similarly, A M (R) is symmetric if and only if it is orthogonally equivalent to a diagonal matrix.