Question

In Recitation video for the "Projection onto Subspaces" lecture (chapter 2), the short cut method for finding the Projection matrix was to write b as sum of its components:

Ib = Pb + Pnb (Pn is the normal projection)

From this we got: I = P + Pn. Is this a valid operation? I thought we could only do this if b was invertible?

Answer 1

I see your point!

If you’re given that XB=B for a specific B, you can only deduce that X=I if B is invertible.

(To make this tie in with what you’re discussing here set X=P+Pn.)

But it’s a different story if you’re told that XB=B for *all* B, or even (for a vector b) that Xb=b for all b; in that case you *can* deduce that X=I; to see this take b as successively (1,0,0,…,0), (0,1,0,…,0), …, (0,0,0,…,1); the first one pulls out the first column of X and shows that it is equal to column 1 of I, the second shows the same thing for the second column, and so on. The same argument shows more generally that if Xb=Yb for all b then we must have X=Y (without requiring invertibility of X, Y or b).

Josh.

Answer 2

Yup, makes sense. Thanks a lot!