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Topic: norm
Replies: 8   Last Post: Mar 10, 2013 1:45 PM

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Posts: 12,067
Registered: 7/15/05
Re: norm
Posted: Mar 10, 2013 7:00 AM
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David C. Ullrich wrote:
>quasi wrote:
>>novis wrote:
>>>quasi wrote:
>>>> novis wrote:

>>>> >Suppose A is a p x q columnwise orthonormal matrix and suppose
>>>> >x is any vector in R^p. Then what is the relation between
>>>> >||x|| and ||Ax|| ?

>>>> A is a p x q matrix, so regarded as a function,
>>>>    A maps R^q to R^p.
>>>> Thus,
>>>>    x is in R^q
>>>> not in R^p as you specified, and
>>>>    Ax is in R^p
>>>> Also, since A is columnwise orthonormal, it follows that
>>>> p >= q.
>>>> As far as norm comparison, since A is orthonormal,
>>>>    |Ax| = |x|
>>>> where the norms are the usual Euclidean norms in R^p and R^q,
>>>> respectively.

>>>Well I was talking about A transpose x or ||A'x||. Can you please
>>>show how ||x||=||A'x||?

>>OK, I missed your use of the symbol ' denoting transpose.
>>So A' is a map from R^p to R^q.
>>As before, since A is columnwise orthonormal, rank(A) = q,
>>hence p >= q.
>>For x in R^p, A'x is in R^q, and yes, it's true that
>> |A'x| = |x|.

>I don't think so...

Me neither.

My answer was a hurried one -- I was on my way out.

A simple counterexample:

Let A be the 3 x 2 matrix

( 0 0 )
A = ( 1 0 )
( 0 1 )

Then A is columnwise orthonormal.

Let x be the 3 x 1 column vector

( 1 )
x = ( 0 )
( 0 )

Then |x| = 1 but A'x = 0.

In fact, if A is any p x q columnwise orthonormal matrix, then
rank(A) = q, hence p >= q. Suppose p > q. Then the columns of
A' are linearly dependent, hence there is some nonzero vector
x in R^p such that A'x = 0, and thus, for that x, the norm of
x is not the same as norm of A'x.

On the other hand, if p = q, then A columnwise orthonormal
implies A is also rowwise orthonormal, consequently,
assuming p = q, we do have

|A'x| = |x|

for all x in R^p.

If p > q, we can at least say

|A'x| <= |x|

for all x in R^p.


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