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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 >



