hw4_2011_04_22_02_solutions

hw4_2011_04_22_02_solutions - EE263 S Lall 2011.04.22.02...

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Unformatted text preview: EE263 S. Lall 2011.04.22.02 Homework 4 Solutions Due Thursday 4/28. 1. Some basic properties of eigenvalues. 50045 Show that (a) the eigenvalues of A and A T are the same (b) A is invertible if and only if A does not have a zero eigenvalue (c) if the eigenvalues of A are λ 1 ,...,λ n and A is invertible, then the eigenvalues of A- 1 are 1 /λ 1 ,..., 1 /λ n , (d) the eigenvalues of A and T- 1 AT are the same. Hint: you’ll need to use the facts that det A = det( A T ), det( AB ) = det A det B , and, if A is invertible, det A- 1 = 1 / det A . Solution. (a) The eigenvalues of a matrix A are given by the roots of the polynomial det( sI − A ). From determinant properties we know that det( sI − A ) = det( sI − A ) T = det( sI − A T ). We conclude that the eigenvalues of A and A T are the same. (b) First we recall that A is invertible if and only if det( A ) negationslash = 0. But det( A ) negationslash = 0 ⇐⇒ det( − A ) negationslash = 0. i. If 0 is an eigenvalue of A , then det( sI − A ) = 0 when s = 0. It follows that det( − A ) = 0 and thus det( A ) = 0, and A is not invertible. From this fact we conclude that if A is invertible, then 0 is not an eigenvalue of A . ii. If A is not invertible, then det( A ) = det( − A ) = 0. This means that, for s = 0, det( sI − A ) = 0, and we conclude that in this case 0 must be an eigenvalue of A . From this fact it follows that if 0 is not an eigenvalue of A , then A is invertible. (c) From the results of the last item we see that 0 is not an eigenvalue of A . Now consider the eigenvalue/eigenvector pair ( λ i ,x i ) of A . This pair satisfies Ax i = λ i x i . Now, since A is invertible, λ i is invertible. Multiplying both sides by A- 1 and λ- 1 i we have λ- 1 i x i = A- 1 x i , and from this we conclude that the eigenvalues of the inverse are the inverse of the eigenvalues. (d) First we note that det( sI − A ) = det( I ( sI − A )) = det( T- 1 T ( sI − A )). Now, from determinant properties, we have det( T- 1 T ( sI − A )) = det( T- 1 ( sI − A ) T ). But this is also equal to det( sI − T- 1 AT ), and the conclusion is that the eigenvalues of A and T- 1 AT are the same. 2. Norm expressions for quadratic forms. 55020 Let f ( x ) = x T Ax (with A = A T ∈ R n × n ) be a quadratic form. (a) Show that f is positive semidefinite ( i.e. , A ≥ 0) if and only if it can be expressed as f ( x ) = bardbl Fx bardbl 2 for some matrix F ∈ R k × n . Explain how to find such an F (when A ≥ 0). What is the size of the smallest such F ( i.e. , how small can k be)? (b) Show that f can be expressed as a difference of squared norms, in the form f ( x ) = bardbl Fx bardbl 2 − bardbl Gx bardbl 2 , for some appropriate matrices F and G . How small can the sizes of F and G be?...
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hw4_2011_04_22_02_solutions - EE263 S Lall 2011.04.22.02...

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