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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 >>



