hw4_2009_10_18_13_solutions

hw4_2009_10_18_13_solutions - EE263 S Lall 2009.10.18.13...

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Unformatted text preview: EE263 S. Lall 2009.10.18.13 edited Homework 4 Solutions Due Thursday 10/22. 1. A Pythagorean inequality for the matrix norm. 55035 Suppose that A ∈ R m × n and B ∈ R p × n . Show that vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble ≤ radicalbig bardbl A bardbl 2 + bardbl B bardbl 2 . Under what conditions do we have equality? Solution. Suppose that v ∈ R n with bardbl v bardbl = 1 is the maximum gain direction of bracketleftbigg A B bracketrightbigg . Hence vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble = vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbigg v vextenddouble vextenddouble vextenddouble vextenddouble . But vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbigg v vextenddouble vextenddouble vextenddouble vextenddouble 2 = vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg Av Bv bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble 2 = bardbl Av bardbl 2 + bardbl Bv bardbl 2 ≤ bardbl A bardbl 2 + bardbl B bardbl 2 (1) since in general the maximum gain directions of A and B are different from the maximum direction of bracketleftbigg A B bracketrightbigg (equality holds when maximum gain input directions of A and B are the same.) Therefore we have shown that vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble ≤ radicalbig bardbl A bardbl 2 + bardbl B bardbl 2 . Equality holds when the maximum gain direction of A and B are the same. If v 1 ∈ R n with bardbl v 1 bardbl = 1 is the maximum gain direction of both A and B then vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble 2 ≥ vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbigg v 1 vextenddouble vextenddouble vextenddouble vextenddouble 2 = vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg Av 1 Bv 1 bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble 2 = bardbl Av 1 bardbl 2 + bardbl Bv 1 bardbl 2 = bardbl A bardbl 2 + bardbl B bardbl 2 , and since also vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble 2 ≤ bardbl A bardbl 2 + bardbl B bardbl 2 we should have vextenddouble vextenddouble vextenddouble vextenddouble bracketleftbigg A B bracketrightbiggvextenddouble vextenddouble vextenddouble vextenddouble 2 = bardbl A bardbl 2 + bardbl B bardbl 2 ....
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This note was uploaded on 08/23/2010 for the course EE 263 taught by Professor Boyd,s during the Fall '08 term at Stanford.

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hw4_2009_10_18_13_solutions - EE263 S Lall 2009.10.18.13...

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