hw7_solutions

hw7_solutions - EE263 Summer 2010-11 Laurent Lessard EE263...

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Unformatted text preview: EE263 Summer 2010-11 Laurent Lessard EE263 homework 7 1. Some true/false questions. Determine if the following statements are true or false. For each statement, either provide a proof that it is always true, or a counterexample demonstrating that it may fail. You cant assume anything about the dimensions of the matrices (unless its explicitly stated), but you can assume that the dimensions are such that all expressions make sense. For example, the statement A + B = B + A is true, because no matter what the dimensions of A and B are (they must, however, be the same), and no matter what values A and B have, the statement holds. As another example, the statement A 2 = A is false, because it fails for the matrix bracketleftbig 2 bracketrightbig . There are also matrices for which it does hold, e.g. , an identity matrix. But that doesnt make the statement true. (a) If A R 3 3 satisfies A + A T = 0, then A is singular. Solution. True . A general 3 3 skew symmetric matrix ( i.e. , one that satisfies A T = A ) has the form A = a b a c b c . Evidently we have Ax = 0, with x = ( c, b,a ). Alternatively, one can compute the determinant explicitly and show that it is identically zero. (b) If A k = 0 for some integer k 1, then I A is nonsingular. Solution. True . We first observe that all eigenvalues of A must be zero. The eigenvalues of I A are each one minus an eigevalue of A , i.e. , they are all equal to one. In particular, 0 is not an eigenvalue of I A , so it is nonsingular. (c) If A,B R n n are both diagonalizable, then AB is diagonalizable. Solution. False . Consider A = bracketleftbigg 1 1 1 1 bracketrightbigg and B = bracketleftbigg 1 / 2 1 / 2 1 bracketrightbigg . Clearly both A and B are diagonalizable, but AB = bracketleftbigg 1 1 1 bracketrightbigg is not diagonalizable. (d) If A,B R n n , then every eigenvalue of AB is an eigenvalue of BA . Solution. True . Take any eigenvalue of AB , and let v be an eigenvector, i.e. , ABv = v . Suppose negationslash = 0, then BA ( Bv ) = B ( ABv ) = ( Bv ) . Since Bv negationslash = 0 (otherwise = 0), Bv is an eigenvector of BA associated with the eigenvalue . Now suppose = 0 and we need to show that BA is also singular. Suppose BA is nonsingular, then both A and B is full rank. But this will imply that Bv = 0 and thus v = 0, a contradiction. (e) If A,B R n n , then every eigenvector of AB is an eigenvector of BA . 1 Solution. False. Consider A = bracketleftbigg 1 1 1 1 bracketrightbigg and B = bracketleftbigg 1 1 1 bracketrightbigg . Then AB = bracketleftbigg 1 2 1 2 bracketrightbigg has (1 , 1) as an eigenvector, which is clearly not an eigenvector of BA = bracketleftbigg 2 2 1 1 bracketrightbigg ....
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hw7_solutions - EE263 Summer 2010-11 Laurent Lessard EE263...

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