show that if W is a nonsingular matrix then ||Wx|| is a vector norm.
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Anonymous2009-12-21 6:39
Satisfies all the definitions. ez game
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Anonymous2009-12-21 17:22
yes obviously, but how do you show that ||Wx|| >= 0, remember W is a general nonsingular matrix so it obviously works if W is diagonal, but how do you do it in general?
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Anonymous2009-12-21 18:17
||W*x|| is >= 0 for ANY W, even if it's singular... || || is a norm and anything you plug into it will give a nonnegative answer.
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Anonymous2009-12-21 18:26
>>4
So what you really want to show is that:
1.) For vector x and any constant c, ||W(cx)|| = |c| ||Wx||
2.) For any vectors x and y, ||W(x+y)|| <= ||Wx|| + ||Wy||
3.) ||Wx|| = 0 if and only if x = 0
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Anonymous2009-12-21 22:04
Seeing posts on /sci/ regarding norms is the norm, nowadays. :\
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Anonymous2009-12-21 22:09
>>4
so why does the question specify that W must be nonsingular
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Anonymous2009-12-21 22:33
>>7
Because if W is singular, you can find a nonzero x such that Wx = 0. That's the definition of singular. That would mean that ||Wx|| = 0 which violates property 3