quasi
Posts:
10,969
Registered:
7/15/05


Re: norm
Posted:
Mar 10, 2013 7:00 AM


David C. Ullrich wrote: >quasi 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...
Me neither.
My answer was a hurried one  I was on my way out.
A simple counterexample:
Let A be the 3 x 2 matrix
( 0 0 ) A = ( 1 0 ) ( 0 1 )
Then A is columnwise orthonormal.
Let x be the 3 x 1 column vector
( 1 ) x = ( 0 ) ( 0 )
Then x = 1 but A'x = 0.
In fact, if A is any p x q columnwise orthonormal matrix, then rank(A) = q, hence p >= q. Suppose p > q. Then the columns of A' are linearly dependent, hence there is some nonzero vector x in R^p such that A'x = 0, and thus, for that x, the norm of x is not the same as norm of A'x.
On the other hand, if p = q, then A columnwise orthonormal implies A is also rowwise orthonormal, consequently, assuming p = q, we do have
A'x = x
for all x in R^p.
If p > q, we can at least say
A'x <= x
for all x in R^p.
quasi

