 Now, I want to do an example with complex vectors here. Because one has to be careful, when you do your foyer coefficients, you have like the vi.xj over vi.vi. When you're working with complex inner products, that is the Hermitian product, the order does matter. If you switch these two around on the top, you'll actually get the conjugate, and then you won't actually have an orthogonal basis that you need. So do make sure you get the order correct. And so let's go through the details of that right here. So if we take our first vector u, we're just going to leave it alone. We don't really need to modify whatsoever. So x1, the first vector we're going to construct right here, will just be u1, i, and 0. So to find the second one, we're going to call this one, not y, we'll call it x2. I guess that makes more sense. x2 is we're going to take the previous vector v, and we're going to subtract from it u.v over u.u right here. And remember, as we're taking the Hermitian product, we take the conjugate of the first factor. So I'm going to be a little bit more detailed in the inner product here. So if you take u.v, you're going to get 1 bar times 1 plus i bar times 0 plus 0 bar times negative i. And this sits all above 1 bar times 1 plus i bar times i plus 0 bar times 0. Of course, the bar is just the complex conjugate. Times this by the 1, i, and 0. Like so. Now there's a lot of zeros here. So this does simplify simple enough. 1, 0, negative i. And so the conjugate of a real number, of course, is just itself. So you can get 1 plus 0 plus 0 for the numerator. For the denominator, you get 1 plus 1. i bar times i is just going to be negative i times i, which is 1. And then 0 there, and 0. And so that fraction, make sure that looks like an i. So the fraction we get negative 1 half times the vector 1, i, and 0. And so even though these are complex numbers, this will just be a linear combination like anything else. We're going to get 1 minus a half, which is a half. We get 0 minus i halves, which is to be minus i halves. And then we're going to get negative i minus 0. So we get negative i right there. And so this should be orthogonal to the original vector. Again, if you don't like the fractions, take out the 1 half. That gives you 1 negative i and negative 2 i. Like so. And we can check, and we're going to take our x2, because we don't want the 1 half scale. We're good without it. And so we're going to take x2 to be this vector 1 negative i and negative 2 i. And so let's check to see, in fact, if we got an orthogonal set or not. So as a reminder, the first vector x1, which we didn't change, that was the vector 1 i 0. And x2 is what we just got a moment ago. 1 negative i, negative 2 i. And so if we take the inner product of these things. Remember, we're taking the Hermitian product here, x1 dot x2. We're going to take 1 bar times 1. We're going to get i bar times negative i. And then we're going to get 0 bar times negative 2 i. Well, 0 times anything is easy enough. That's 0 even for complex numbers. You get 1 times 1, which is a 1. And then you're going to get a negative i times a negative i. That's a double negative. So you get 1 minus i squared. I'm sorry, i1 plus i squared. Negative i squared itself is negative once you get 1 minus 1, which is 0. So this in fact shows us that x1 is orthogonal to x2, like we wanted to. And this span didn't change. We still have the same spanning. The same span that we did before. So if we take the subspace span by those two vectors, u and v, x1, x2 will span the exact same thing. So we have to be a little bit more careful with complex vectors. But other than that, it works out pretty nicely.