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? >
answered by quasi

