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Unitary-And-Hermitian-Matrices.Pdf 11-24-2014 Unitary Matrices and Hermitian Matrices Recall that the conjugate of a complex number a + bi is a bi. The conjugate of a + bi is denoted a + bi or (a + bi)∗. − In this section, I’ll use ( ) for complex conjugation of numbers of matrices. I want to use ( )∗ to denote an operation on matrices, the conjugate transpose. Thus, 3 + 4i = 3 4i, 5 6i =5+6i, 7i = 7i, 10 = 10. − − − Complex conjugation satisfies the following properties: (a) If z C, then z = z if and only if z is a real number. ∈ (b) If z1, z2 C, then ∈ z1 + z2 = z1 + z2. (c) If z1, z2 C, then ∈ z1 z2 = z1 z2. · · The proofs are easy; just write out the complex numbers (e.g. z1 = a+bi and z2 = c+di) and compute. The conjugate of a matrix A is the matrix A obtained by conjugating each element: That is, (A)ij = Aij. You can check that if A and B are matrices and k C, then ∈ kA + B = k A + B and AB = A B. · · You can prove these results by looking at individual elements of the matrices and using the properties of conjugation of numbers given above. Definition. If A is a complex matrix, A∗ is the conjugate transpose of A: ∗ A = AT . Note that the conjugation and transposition can be done in either order: That is, AT = (A)T . To see this, consider the (i, j)th element of the matrices: T T T [(A )]ij = (A )ij = Aji =(A)ji = [(A) ]ij. Example. If 1 2i 4 1 + 2i 2 i 3i ∗ − A = − , then A = 2 + i 2 7i . 4 2 + 7i 6 + 6i − − − 3i 6 6i − − Since the complex conjugate of a real number is the real number, if B is a real matrix, then B∗ = BT . Remark. Most people call A∗ the adjoint of A — though, unfortunately, the word “adjoint” has already been used for the transpose of the matrix of cofactors in the determinant formula for A−1. (Sometimes 1 people try to get around this by using the term “classical adjoint” to refer to the transpose of the matrix of cofactors.) In modern mathematics, the word “adjoint” refers to a property of A∗ that I’ll prove below. This property generalizes to other things which you might see in more advanced courses. The ( )∗ operation is sometimes called the Hermitian — but this has always sounded ugly to me, so I won’t use this terminology. Since this is an introduction to linear algebra, I’ll usually refer to A∗ as the conjugate transpose, which at least has the virtue of saying what the thing is. Proposition. Let U and V be complex matrices, and let k C. ∈ (a) (U ∗)∗ = U. (b) (kU + V )∗ = kU ∗ + V ∗. (c) (UV )∗ = V ∗U ∗. (d) If u, v Cn, their dot product is given by ∈ ∗ u v = v u. · Proof. I’ll prove (a), (c), and (d). For (a), I use the fact noted above that ( )and( )T can be done in either order, along with the facts that A = A and (AT )T = A. I have ∗ ∗ (U ) = [(U T )]T = [(U T )T ] = U = U. This proves (a). For (c), I have ∗ ∗ ∗ (UV ) = (UV )T = V T U T = V T U T = V U . · · For (d), recall that the dot product of complex vectors u =(u1,u2,...,un) and v =(v1, v2,...,vn) is u v = u1v1 + u2v2 + + unvn. · ··· Notice that you take the complex conjugates of the components of v before multiplying! This can be expressed as the matrix multiplication u1 u2 ∗ u v = [ v1 v2 vn ] . = v u. · ··· . . un Example. In this example, use the complex dot product. (a) Compute (1 + 3i, 2 + i) (4 5i, 2 + 3i). · − (b) Find (2 + i, 3 5i) . k − k (c) Find a nonzero vector (a, b) which is orthogonal to (1 + 8i, 2 3i). − (a) 1 + 3i (1+3i, 2 + i) (4 5i, 2 + 3i)=[4+5i 2 3i ] =(4+5i)(1+3i)+(2 3i)(2 + i) = 4 + 13i. · − − 2 + i − − 2 It’s a common notational abuse to write the number “ 4 + 13i” instead of writing it as a 1 1 matrix “[ 4 + 13i]”. − × − (b) 2 (2 + i, 3 5i) = (2+ i, 3 5i) (2 + i, 3 5i)=(2 i)(2 + i)+(3+5i)(3 5i)=4+1+9+25=39. k − k − · − − − Hence, (2 + i, 3 5i) = √39. k − k The following formula is evident from this example: (a + bi, c + di) = a2 + b2 + c2 + d2. k k p This extends in the obvious way to vectors in Cn. (c) I need (a, b) (1+8i, 2 3i) = 0. · − In matrix form, this is a [ 1 8i 2 + 3i ] = 0. − b Note that the vector (1 + 8i, 2 3i) was conjugated and transposed. Doing the matrix multiplication,− (1 8i)a +(2+3i)b = 0. − I can get a solution (a, b) by switching the numbers 1 8i and 2 + 3i and negating one of them: (a, b)=(2+3i, 1 + 8i). − − There are two points about the equation u v = v∗u which might be confusing. First, why is it necessary to conjugate and transpose v? The reason for· the conjugation goes back to the need for inner products to be positive definite (so u u is a nonnegative real number). The reason for the transpose· is that I’m using the convention that vectors are column vectors. So if u and v are n-dimensional column vectors and I want the product to be a number — i.e. a 1 1 matrix — I have to multiply an n-dimensional row vector (1 n) and an n-dimensional column vector× (n 1). To get the row vector, I have to transpose the column vector.× × Finally, why do u and v switch places in going from the left side to the right side? The reason you write v∗u instead of u∗v is because inner products are defined to be linear in the first variable. If you use u∗v you get a product which is linear in the second variable. Of course, none of this makes any difference if you’re dealing with real numbers. So if x and y are vectors in Rn, you can write x y = xT y or x y = yT x. · · Definition. A complex matrix U is unitary if UU ∗ = I. Notice that if U happens to be a real matrix, U ∗ = U T , and the equation says UU T = I — that is, U is orthogonal. In other words, unitary is the complex analog of orthogonal. By the same kind of argument I gave for orthogonal matrices, UU ∗ = I implies U ∗U = I — that is, U ∗ is U −1. Proposition. Let U be a unitary matrix. (a) U preserves inner products: x y =(Ux) (Uy). Consequently, it also preserves lengths: Ux = x . · · k k k k (b) An eigenvalue of U must have length 1. 3 (c) The columns of a unitary matrix form an orthonormal set. Proof. (a) ∗ ∗ ∗ ∗ ∗ (Ux) (Uy)=(Uy) (Ux) = y U Ux = y Ix = y x = x y. · · Since U preserves inner products, it also preserves lengths of vectors, and the angles between them. For example, 2 2 x = x x =(Ux) (Ux) = Ux , so x = Ux . k k · · k k k k k k (b) Suppose x is an eigenvector corresponding to the eigenvalue λ of U. Then Ux = λx, so Ux = λx = λ x . k k k k | |k k But U preserves lengths, so Ux = x , and hence λ = 1. k k | k | | (c) Suppose ↑ ↑ ↑ U = c1 c2 cn . · ↓ ↓ ↓ Then U ∗U = I means T 1 0 0 0 c1 ··· ← T → 0 1 0 0 c2 ↑ ↑ ↑ ··· ← → c1 c2 c 0 0 1 0 . n = . · . ··· . T . cn ↓ ↓ ↓ ← → 0 0 0 1 ··· T th Here ck is the complex conjugate of the k column ck, transposed to make it a row vector. If you look at the dot products of the rows of U ∗ and the columns of U, and note that the result is I, you see that the equation above exactly expresses the fact that the columns of U are orthonormal. T For example, take the first row c1 . Its product with the columns c1, c2, and so on give the first row of the identity matrix, so c1 c1 = 1, c1 c2 = 0,...,c1 cn = 0. · · · This says that c1 has length 1 and is perpendicular to the other columns. Similar statements hold for c2, ... , cn. Example. Find c and d so that the following matrix is unitary: 1 (1+2i) c √ 7 . 1 (1 i) d √7 − I want the columns to be orthogonal, so their complex dot product should be 0. First, I’ll find a vector that 1 is orthogonal to the first column. I may ignore the factor of ; I need √7 (a, b) (1+2i, 1 i) = 0 · − a [ 1 2i 1 + i ] = 0 − b This gives (1 2i)a +(1+ i)b = 0. − 4 I may take a = 1 + i and b = 1 + 2i. Then − (1 + i, 1 + 2i) = √7. k − k So I need to divide each of a and b by √7 to get a unit vector. Thus, 1 1 (c, d) = (1 + i), ( 1 + 2i) . √7 √7 − Proposition. (Adjointness) let A M(n, C) and let u, v Cn. Then ∈ ∈ ∗ Au v = u A v. · · Proof. ∗ ∗ ∗ ∗ ∗ ∗ ∗ u A v =(A v) u = v (A ) u = v Au = Au v. · · Remark. If ( , ) is any inner product on a vector space V and T : V V is a linear transformation, the adjoint T ∗ of· T· is the linear transformation which satisfies → ∗ (T (u), v)=(u,T (v)) for all u, v V.
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