MIT18_06S10_L22

# MIT18_06S10_L22 - 18.06 Linear Algebra Spring 2010...

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18.06 Linear Algebra, Spring 2010 Transcript – Lecture 22 OK. Shall we start? This is the second lecture on eigenvalues. So the first lecture was -- reached the key equation, A x equal lambda x. x is the eigenvector and lambda's the eigenvalue. Now to use that. And the, the good way to, after we've found -- so, so job one is to find the eigenvalues and find the eigenvectors. Now after we've found them, what do we do with them? Well, the good way to see that is diagonalize the matrix. So the matrix is A. And I want to show -- first of all, this is like the basic fact. This, this formula. That's, that's the key to today's lecture. This matrix A, I put its eigenvectors in the columns of a matrix S. So S will be the eigenvector matrix. And I want to look at this magic combination S inverse A S. So can I show you how that -- what happens there? And notice, there's an S inverse. We have to be able to invert this eigenvector matrix S. So for that, we need n independent eigenvectors. So that's the, that's the case. OK. So suppose we have n linearly independent eigenvectors of A. Put them in the columns of this matrix S. So I'm naturally going to call that the eigenvector matrix, because it's got the eigenvectors in its columns. And all I want to do is show you what happens when you multiply A times S. So A times S. So this is A times the matrix with the first eigenvector in its first column, the second eigenvector in its second column, the n-th eigenvector in its n-th column. And how I going to do this matrix multiplication? Well, certainly I'll do it a column at a time. And what do I get. A times the first column gives me the first column of the answer, but what is it? That's an eigenvector. A times x1 is equal to the lambda times the x1. And that lambda's we're -- we'll call lambda one, of course. So that's the first column. Ax1 is the same as lambda one x1. A x2 is lambda two x2. So on, along to in the n-th column we now how lambda n xn.

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Looking good, but the next step is even better. So for the next step, I want to separate out those eigenvalues, those, those multiplying numbers, from the x-s. So then I'll have just what I want. OK. So how, how I going to separate out? So that, that number lambda one is multiplying the first column. So if I want to factor it out of the first column, I better put -- here is going to be x1, and that's going to multiply this matrix lambda one in the first entry and all zeros. Do you see that that, that's going to come out right for the first column? Because w- we remember how -- how we're going back to that original punchline. That if I want a number to multiply x1 then I can do it by putting x1 in that column, in the first column, and putting that number there. Th- u- what I going to have here? I'm going to have lambda -- I'm going to have x1, x2, .
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## This note was uploaded on 08/22/2011 for the course MATH 1806 taught by Professor Strang during the Fall '10 term at MIT.

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MIT18_06S10_L22 - 18.06 Linear Algebra Spring 2010...

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