hw1s - CME302 Homework 1 Solutions 2 x y W = W(x y = W x W...

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CME302 Homework 1 Solutions September 30, 2009 T&B 2.1 Suppose A C m × m is unitary and upper triangular (u.t.). The lower triangular (l.t.) case follows by symmetry of the problem. We show A is in fact diagonal: 1. By definition, A - 1 = A * . 2. The inverse of a nonsingular u.t. matrix U is also u.t. Consider solving Uv i = e i , where e i is the i th column of the identity matrix. For a particular i , only the first i components of v i are nonzero (as evident in the back substitution algorithm). By construction, V = ( v 1 | v 2 | · · · | v m ) = U - 1 . 3. The Hermitian transpose of a u.t. matrix is l.t.. 4. A matrix that is both u.t. (by 2) and l.t. (by 3) is diagonal. T&B 2.3a We show that the eigenvalues of a Hermitian matrix are real. First, ( x * Ax ) * = x * A * x = x * Ax as A * = A . If α C is such that ¯ α = α , then the imaginary part is zero (written Im( α ) = 0) and so in fact α R . Consequently, both x * Ax and x * x are real scalars. If x is an eigenvector of A , then by definition Ax = λx , and so x * Ax = λx * x . As x * Ax, x * x R , so too is λ . T&B 2.3b We show that the eigenvectors of a Hermitian matrix are orthogonal. Let Ax = λx and Ay = μy ; λ negationslash = μ by assumption. First, ( Ay ) * = y * A * = y * A = μy * as A is Hermitian. Next, premultiplying by y * , y * Ax = λy * x . The l.h.s. (short for “left hand side”) is ( y * A ) x = μy * x , and so ( μ λ ) y * x = 0. As μ negationslash = λ by assumption, y * x = 0: y is orthogonal to x .
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