Hi, let's talk about a special instance of least squares that winds up being rather important. So consider the instance where x'x = I, and x is full rank. And even maybe what appears to be a little bit stranger, let's consider the case where x is n by n. So this may seem to be an odd thing to do. Recall if here's our, if our y is three-dimensional, y1, y2, y3, and our y is some point in three-dimensional space, and our x is three-dimensional, then Y hat would just be exactly equal to y, because the dimension of the space we're trying to project on is three-dimensional, our original data is three-dimensional, so this extends up to any dimensions. So it seems like if we have a full rank n x n matrix x, what's the point of doing least squares at all? So we'll get to that in a minute. But notice here if I want to solve my least squares equation y minus x beta, and minimize that with respect to beta, then my estimate is of course, we've discussed this, x transpose x inverse x transpose y. But in this case, x transpose x is just pi. And so it works out to just be x transpose y. And looking at this, if my x is equal to n columns, then this means that it's the interproduct of x1, the first column of x with y is the first coefficient. The interproduct with x2 with y and so on. All the way up to x n with y. And my fitted values, now and my fitted values, my y hat, which is now going to be x x transpose y. Well, this works out to be nothing other than the summation i = 1 to n xi <xi, y="">. Now, what's interesting is imagine, imagine the case where instead of having X, I have just a few columns of X. Let's say, X subset. I'll just put a little x sub s. Well, what x sub s transpose X of s, let's suppose it only contains P, a random collection, not a random, but any subset of the columns of x. Well, it also has inter product I just because of this statement up here. Well, then this whole argument just goes through. And then it says that y hat, my y hat fitting the subset of x, rather than the full basis x just fitting the subset of x, that y that is just the sum over the collection of i's in this collection of indices, the subset indices, xi times the inner product of xi and yi. So what this means is, if I have my full design matrix and it's a full ranked basis for RN, my coeffecience just work and it's an ortho-normal basis. My coeffecience work out to be just the inter products between the basis elements and my outcome. And if I want to fit instead of a n dimensional version of this basis, if I just want a p dimensional version of this basis, I can just grab the relevant subset of these coefficients, right?