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

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 fom Posts: 1,968 Registered: 12/4/12
Re: norm
Posted: Mar 10, 2013 7:30 AM

On 3/10/2013 6:26 AM, fom 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?
>

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