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

 Messages: [ Previous | Next ]
 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: norm
Posted: Mar 9, 2013 4:58 PM

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...

>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