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YAMT: Quick Linear Algebra Help

D

DVK916

Banned


Let B be positive definite matrix and d be a given vector. Then for any arbitrary nonzero vector x:

Prove

Max (x'd)^2/(x'Bx) = d'*B^-1*d
x != 0

And determin for which vector x is it maximized.

Note: x' is x transpose, x != 0 means x not equal to zero, and B^-1 is inverse maxtrix B.
 
I think I am so suppose to use the Cauche Squartz inequality, but don't know how to use that to prove the problem.
 
I applied the Cauchy Schwartz inequality, but the last step is to prove x'x*d'd/(x'Bx) = d'B^(-1)d. If you can prove that, then the result follows.

EDIT: ok, so that last step might not be true, but the Cauchy Schwartz inequality is:

(x'd)^2 <= x'x*d'd.

Divide both sides by x'Bx (which is always positive since B is positive definite), and you are making some progress. I need to think about how to complete the problem, but the Duke vs. G'town game calls.
 
haha...don't remember a damn thing from that class. good luck...looks like you have chuckywang helping you out 🙂
 
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