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216 Section 6.1 Chapter 6 Hermitian, Orthogonal, And 216 SECTION 6.1 CHAPTER 6 HERMITIAN, ORTHOGONAL, AND UNITARY OPERATORS In Chapter 4, we saw advantages in using bases consisting of eigenvectors of linear opera- tors in a number of applications. Chapter 5 illustrated the benefit of orthonormal bases. Unfortunately, eigenvectors of linear operators are not usually orthogonal, and vectors in an orthonormal basis are not likely to be eigenvectors of any pertinent linear operator. There are operators, however, for which eigenvectors are orthogonal, and hence it is possible to have a basis that is simultaneously orthonormal and consists of eigenvectors. This chapter introduces some of these operators. 6.1 Hermitian Operators § When the basis for an n-dimensional real, inner product space is orthonormal, the inner product of two vectors u and v can be calculated with formula 5.48. If v not only represents a vector, but also denotes its representation as a column matrix, we can write the inner product as the product of two matrices, one a row matrix and the other a column matrix, (u, v) = uT v. If A is an n n real matrix, the inner product of u and the vector Av is × (u, Av) = uT (Av) = (uT A)v = (AT u)T v = (AT u, v). (6.1) This result, (u, Av) = (AT u, v), (6.2) allows us to move the matrix A from the second term to the first term in the inner product, but it must be replaced by its transpose AT . A similar result can be derived for complex, inner product spaces. When A is a complex matrix, we can use equation 5.50 to write T T T (u, Av) = uT (Av) = (uT A)v = (AT u)T v = A u v = (A u, v). (6.3) This is the complex counterpart of equation 6.2, T (u, Av) = (A u, v), (6.4) but this time, we must also take complex conjugates of entries in AT . We identify this matrix in the following definition. Definition 6.1 The Hermitian conjugate of a complex matrix A is the transpose of its complex conjugate T A . For example, the Hermitian conjugate of 1+2i 3 3 i 1 2i 2+3i 2 − − 2 3i 2i 1+6i is 3 2i 4+2i . −2 4 2i 6+7i 3 + i 1− 6i 6 7i − − − − − A real matrix is symmetric if it is equal to its transpose. The complex counterpart of a symmetric matrix is contained in the next definition. Definition 6.2 A square complex matrix A is said to be Hermitian if it is equal to its Hermitian conjugate, T A = A . (6.5) In a Hermitian matrix, entries on opposite sides of the diagonal are complex conjugates, and diagonal entries are real (see Exercise 7). A square real matrix is Hermitian if it is symmetric. SECTION 6.1 217 If L is a linear operator on a complex, inner product space V , we can associate a complex matrix A with L so that mapping a vector v can be written as matrix multiplication, L(v) = Av. We use Hermitian matrices to define Hermitian operators. Definition 6.3 A linear operator H on a complex inner product space is said to be Hermitian if its matrix is Hermitian. We use the letter H to represent a Hermitian operator to distinguish that it is Hermitian. The matrix A of a Hermitian operator satisfies equation 6.5, and for such an operator, equation 6.4 implies that u,H(v) = H(u), v . (6.6) Hermitian operators are the first of our operators for which eigenvectors are orthogonal. This is proved in the following theorem. Theorem 6.1 Eigenvalues of a Hermitian operator on an inner product space are real and eigenvectors corresponding to distinct eigenvalues are orthogonal. Proof Let λ and v be an eigenpair for a Hermitian operator H. Not out of necessity, but to simplify calculations, suppose that v has length one. Then λ = λ(v, v) (the length of v is one) = (v, λv) (property 5.19b) = v,H(v) (λ is an eigenvalue for H with eigenvector v) = H(v), v (equation 6.6) = (λv, v) (λ is an eigenvalue for H with eigenvector v) = λ(v, v) (property 5.19a) = λ (the length of v is one). Hence, λ is real. Now suppose that v1 and v2 are eigenvectors corresponding to (real) distinct eigenvalues λ1 and λ2. Then λ1(v1, v2) = (λ1v1, v2) (λ1 is real and property 5.19a) = H(v1), v2 (λ1 is an eigenvalue for H with eigenvector v1) = v1,H(v2) (equation 6.6) = (v1, λ2v2) (λ2 is an eigenvalue for H with eigenvector v2) = λ2(v1, v2). (property 5.19b) Since λ1 = λ2, this implies that (v1, v2) = 0, and the eigenvectors are orthogonal. 6 It is important to note that different eigenvectors corresponding to the same eigenvalue are not usually orthogonal. Indeed, if v is an eigenvector of an operator, then so also is 3 v, and these vectors are certainly not orthogonal. When an eigenvalue has two, or more, linearly independent eigenvectors, it is possible to have orthogonal eigenvectors corresponding to this eigenvalue. Here are some examples to illustrate the theorem and this remark. Example 6.1 The symmetric matrix corresponding to a Hermitian operator on a real inner product space is 2 1 1 − A = 1 2 1 . 11 2 − 218 SECTION 6.1 Show that eigenvalues are real, and eigenvectors corresponding to different eigenvalues are orthogonal. Solution Eigenvalues are given by 2 λ 1 1 − − 2 0 = det(A λI) = det 1 2 λ 1 = λ(λ 3) . − 1− 1 2 λ − − − − Thus, λ = 0 and λ = 3 are the eigenvalues. Eigenvectors corresponding to λ = 0 are v3(1, 1, 1). Eigenvectors corresponding to λ = 3 are v2(1, 1, 0) + v3( 1, 0, 1). If we take − − the inner product of any eigenvector v1 = v(1, 1, 1) corresponding to λ = 0, and any − eigenvector v2 = u(1, 1, 0) + w( 1, 0, 1) corresponding to λ = 3, we obtain − (v1, v2) = v(1, 1, 1),u(1, 1, 0) + w( 1, 0, 1) = vu(1 1+0)+ vw( 1+0+1)=0. − − − − They are orthogonal. The inner product of the eigenvectors (1, 1, 0) and ( 1, 0, 1) corre- sponding to λ =3 is (1, 1, 0), ( 1, 0, 1) = 1. They are not orthogonal. Notice,− however, − − the vector (1, 1, 0)+ 2( 1, 0, 1)= ( 1, 1, 2) is an eigenvector corresponding to λ =3 and it is orthogonal to the eigenvector− (1,−1, 0). • Example 6.2 The Hermitian matrix corresponding to a Hermitian operator on a complex inner product space is 2 0 i A = 0 10 . i 0 2 − Find its eigenvalues and eigenvectors. Demonstrate orthogonality of eigenvectors corre- sponding to different eigenvalues. Solution Eigenvalues are given by 2 λ 0 i − 2 0 = det(A λI) = det 0 1 λ 0 = (λ 1) (λ 3). − i −0 2 λ − − − − − Thus, λ = 1 and λ = 3 are eigenvalues. Eigenvectors corresponding to λ = 1 are v2(0, 1, 0)+ v3( i, 0, 1). Eigenvectors corresponding to λ = 3 are v3(i, 0, 1). If we take the inner product − of any eigenvector v = v2(0, 1, 0) + v3( i, 0, 1) corresponding to λ = 1, and any eigenvector − w = w3(i, 0, 1) corresponding to λ = 3, we obtain (v, w) = v2(0, 1, 0) + v3( i, 0, 1),w3(i, 0, 1) = v2w3(0) + v3w3( 1+1)=0. − − They are orthogonal. • Eigenvectors corresponding to eigenvalues with algebraic multiplicity larger than one need not be orthogonal, but from nonorthogonal eigenvectors, the Gram-Schmidt process can be used to construct orthogonal eigenvectors corresponding to the same eigenvalue. Here is an example. Example 6.3 The symmetric (Hermitian) matrix of an operator on a 3-dimensional space is 5 2 4 − A = 2 8 2 . −4 2 5 Show that A has two eigenvalues one with (algebraic) multiplicity 2. Find orthonormal eigenvectors corresponding to the eigenvalue with algebraic multiplicity 2. Solution Eigenvalues are given by SECTION 6.1 219 5 λ 2 4 − − 2 0 = det 2 8 λ 2 = λ(λ 9) . −4− 25 λ − − − Eigenvalues are λ = 0 and λ = 9. Eigenvectors corresponding to λ = 9 satisfy 4 2 4 v1 − − 0 = (A 9I)v = 2 1 2 v2 . − − − 4 2 4 v3 − The reduced row echelon form for the augmented matrix of this system is 1 1/2 1 0 − 0 0 0 0 . 0 0 0 0 When we convert to equations, v2 v1 + v3 =0. 2 − Eigenvectors are therefore v2/2 + v3 1 1 − v2 − v = v2 = 2 + v3 0 . 2 v3 0 1 Linearly independent eigenvectors corresponding to λ = 9 are v1 = (1, 2, 0) and v2 = − (1, 0, 1). We can construct an eigenvector corresponding to λ = 9 orthogonal to v1 using Gram-Schmidt, (1, 0, 1), (1, 2, 0) 1 4 2 v3 = (1, 0, 1) − (1, 2, 0)= (1, 0, 1) (1, 2, 0) = , , 1 . − 5 − − 5 − 5 5 Orthonormal eigenvectors corresponding to λ = 9 are (1, 2, 0) (4, 2, 5) vˆ1 = − and vˆ3 = . √5 3√5 • We now know that eigenvectors corresponding to different eigenvalues of a Hermitian operator are orthogonal, and should an eigenvalue with algebraic multiplicity greater than 1 have more than one linearly independent eigenvector, then orthogonal eigenvectors can be constructed from them. The following theorem shows that the geometric multiplicity of every eigenvalue of a Hermitian operator on a finite-dimensional vector space must be equal to its algebraic multiplicity. In other words, the number of linearly independent eigenvectors of a Hermitian operator on a finite-dimensional space must be equal to the dimension of the space.
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