Effect of orthonormal projection on rank

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Homework Statement


Given rank(R) and a QR factorization A = QR, what is the rank(A)


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The Attempt at a Solution


I want to know if multiplication by a full rank orthonormal matrix Q and an upper trapezoidal matrix R yields rank(R)=rank(Q*R)=rank(A)

This is mostly guesswork by me but I'd like to use it for a question I need to answer.
 
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Well, the rank of a matrix is the dimension of the image, right? If the image of R is a subspace S of dimension rank(R), then what's the dimension of Q(S) if Q is full rank?
 
They are equal? As the only way Q(S) would be dissimilar would be if rank(Q)<rank(R).

But does not the reason for this have anything to do with Q being orthnormal? Otherwise couldn't Q act on R and cause some of the image to overlap effectively reducing the rank?
 
Q is full rank, so it's one to one. So yes, rank(QR)=dim(Q(S))=dim(S)=rank(R). So rank(QR)=rank(R).
 
Thank you very much =)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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