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MIT18_06S10_L23

# MIT18_06S10_L23 - 18.06 Linear Algebra Spring 2010...

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18.06 Linear Algebra, Spring 2010 Transcript – Lecture 23 -- and lift-off on differential equations. So, this section is about how to solve a system of first order, first derivative, constant coefficient linear equations. And if we do it right, it turns directly into linear algebra. The key idea is the solutions to constant coefficient linear equations are exponentials. So if you look for an exponential, then all you have to find is what's up there in the exponent and what multiplies the exponential and that's the linear algebra. So -- and the result -- one thing we will fine -- it's completely parallel to powers of a matrix. So the last lecture was about how would you compute A to the K or A to the 100? How do you compute high powers of a matrix? Now it's not powers anymore, but it's exponentials. That's the natural thing for differential equation. Okay. But can I begin with an example? And I'll just go through the mechanics. How would I solve the differential -- two differential equations? So I'm going to make it -- I'll have a two by two matrix and the coefficients are minus one two, one minus two and I'd better give you some initial condition. So suppose it starts u at times zero -- this is u1, u2 -- let it -- let it -- suppose everything is in u1 at times zero. So -- at -- at the start, it's all in u1. But what happens as time goes on, du2/dt will -- will be positive, because of that u1 term, so flow will move into the u2 component and it will go out of the u1 component. So we'll just follow that movement as time goes forward by looking at the eigenvalues and eigenvectors of that matrix. That's a first job. Before you do anything else, find the -- find the matrix and its eigenvalues and eigenvectors. So let me do that. Okay. So here's our matrix. Maybe you can tell me right away what -- what are the eigenvalues and -- eigenvalues anyway. And then we can check. But can you spot any of the eigenvalues of that matrix? We're looking for two eigenvalues. Do you see -- I mean, if I just wrote that matrix down, what -- what do you notice about it? It's singular, right. That -- that's a singular matrix. That tells me right away that one of the eigenvalues is lambda equals zero. I can -- that's a singular matrix, the second column is minus two times the first column, the determinant is zero, it's -- it's singular, so zero is an eigenvalue and the other

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eigenvalue will be -- from the trace. I look at the trace, the sum down the diagonal is minus three. That has to agree with the sum of the eigenvalue, so that second eigenvalue better be minus three. I could, of course -- I could compute -- why don't I over here -- compute the determinant of A minus lambda I, the determinant of this minus one minus lambda two one minus two minus lambda matrix. But we know what's coming. When I do that multiplication, I get a lambda squared.
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MIT18_06S10_L23 - 18.06 Linear Algebra Spring 2010...

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