Question

Suppose the columns of A are not independent. How could you find a matrix B so that P = B(BtransB)inverse Btrans does give the projection onto the column space of A?

Answer 1

(question Strang 4.3.28)

Answer 2

I don't have your text. So, I may be assuming something you're not allowed to assume.

If A is linearly independent, then P = AA+, where A+ is the pseudo-inverse. In terms of your equation, A would just equal B. However, A is not independent. So, you want to find a matrix B (related to A) such that B is linearly independent. Then, this equation holds.

I'm not sure how rigorous/formal you need to be, so I'll outline the explanation and you can formalize it yourself:

Claim: If A is mxn with k linearly independent columns (k < n), then B is mxk where the linearly independent columns of A make up the columns of B.

Why?

Well, the column space of A is made up of the linearly independent columns of A. P is your matrix which makes that happen. So, you would expect the only part of A that matters is the linearly independent columns. But, does the size match?

If A is mxn, then P = (mxn)(nxm)(mxn)(nxm) = (mxm). So, P is mxm. We expect P to be mxm by our definition of B, which is mxk (k<n):

P = (mxk)(kxm)(mxk)(kxm) = (mxm). So, the dimensions match.

So, PB would be mxk, which is the column space of A (if A has k linearly independent columns.)

The last thing would be: can you guarantee you'll always be able to find a B in that way? Yes. Again, if A is mxn with k linearly independent columns, then post multiply with a nxk matrix with identity vectors to get the columns you want.

That might be a bit of a longer explanation than you need, but it's a pretty cool problem.

Answer 3

Wow, thanks so much for your response. Unfortunately, I've already turned my work in, but I'm definitely looking over your explanation.