18.06 Linear Algebra, Spring 2010
Transcript – Lecture 23
 and liftoff 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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 Fall '10
 Strang
 Differential Equations, Linear Algebra, Algebra, Exponential Function, Linear Equations, Equations, Derivative, Orthogonal matrix

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