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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 7:26 AM

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

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