MIT18_06S10_L15 - 18.06 Linear Algebra, Spring 2010...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
18.06 Linear Algebra, Spring 2010 Transcript – Lecture 15 OK, guys the -- we're almost ready to make this lecture immortal. OK. Are we on? All right. This is an important lecture. It's about projections. Let me start by just projecting a vector b down on a vector a. So just to so you see what the geometry looks like in when I'm in -- in just two dimensions, I'd like to find the point along this line so that line through a is a one- dimensional subspace, so I'm starting with one dimension. I'd like to find the point on that line closest to a. Can I just take that problem first and then I'll explain why I want to do it and why I want to project on other subspaces. So where's the point closest to b that's on that line? It's somewhere there. And let me connect that and -- and what's the whole point of my picture now? What's the -- where does orthogonality come into this picture? The whole point is that this best point, that's the projection, P, of b onto the line, where's orthogonality? It's the fact that that's a right angle. That this -- the error -- this is like how much I'm wrong by -- this is the difference between b and P, the whole point is -- that's perpendicular to a. That's got to give us the equation. That's got to tell us -- that's the one fact we know, that's got to tell us where that projection is. Let me also say, look -- I've drawn a triangle there. So if we were doing trigonometry we would do like we would have angles theta and distances that would involve sine theta and cos theta that leads to lousy formulas compared to linear algebra. The formula that we want comes out nicely and what's the -- what do we know? We know that P, this projection, is some multiple of a, right? It's on that line. So we know it's in that one-dimensional subspace, it's some multiple, let me call that multiple x, of a. So really it's that number x I'd like to find. So this is going to be simple in 1-D, so let's just carry it through, and then see how it goes in high dimensions. OK. The key fact is -- the key to everything is that perpendicular. The fact that a is perpendicular to a is perpendicular to e. Which is (b-ax), xa. I don't care what -- xa. That that equals zero.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Do you see that as the central equation, that's saying that this a is perpendicular to this -- correction, that's going to tell us what x is. Let me just raise the board and simplify that and out will come x. OK. So if I simplify that, let's see, I'll move one to -- one term to one side, the other term will be on the other side, it looks to me like x times a transpose a is equal to a transpose b. Right? I have a transpose b as one f- one term, a transpose a as the other, so right away here's my a transpose a. But it's just a number now. And I divide by it.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/22/2011 for the course MATH 1806 taught by Professor Strang during the Fall '10 term at MIT.

Page1 / 13

MIT18_06S10_L15 - 18.06 Linear Algebra, Spring 2010...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online