Linear Algebra Least Squares Question

flyingmuskrat
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Homework Statement



Suppose the columns of A are not independent. How could you find a matrix B so that P=B(BTB)^-1BT does give the projection onto the column space of A? (The usual formula will fail when AT A is not invertible).

T is transpose.

Homework Equations


The Attempt at a Solution



I think this is a thought question or something? Do you rearrange the columns...or just delete the dependent columns? But wouldn't that mess up the answer? Idk I got the rest on my p-set but this one I just have nooo idea. I feel like it's really obvious and I'm just missing it. And it's not gram-schmidt or something because that's the section after.
 
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how about considering a basis for the column space?
 
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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