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

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 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: norm
Posted: Mar 10, 2013 1:45 PM

On Sun, 10 Mar 2013 06:26:40 -0500, fom <fomJUNK@nyms.net> wrote:

>On 3/9/2013 3:58 PM, David C. Ullrich wrote:
>> On Sat, 09 Mar 2013 14:47:45 -0500, quasi <quasi@null.set> 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...

>
>Why not?
>
>(||A||)^2 = max((||Ax||)^2/(||x||)^2), x<>0
>
>(||A||)^2 = max((x'A'Ax)/(x'x)), x<>0
>
>For A with orthonormal columns, A'A=I
>
>(||A||)^2 = 1
>
>||A|| = 1

The question was about A'. Yes, ||A'|| = 1.
That does not imply that |Ax| = |x|.

>
>More generally (or less suspiciously), one might look at
>singular value decompositions,
>
>http://en.wikipedia.org/wiki/Singular_value_decomposition#Norms
>
>The norm I constructed above is taken to be the square root of
>the largest eigenvalue for A'A.
>

>>
>
>>> where the norms are the usual Euclidean norms in R^q and R^p
>>> respectively.
>>>
>>> Is this homework?
>>>
>>> In any case, I don't have time to help you on this right now,
>>> maybe tomorrow.
>>>
>>> quasi

>>

Date Subject Author
3/8/13 kaushik.sinha.cs@gmail.com
3/9/13 quasi
3/9/13 kaushik.sinha.cs@gmail.com
3/9/13 quasi
3/9/13 David C. Ullrich
3/10/13 quasi
3/10/13 fom
3/10/13 fom
3/10/13 David C. Ullrich