All right, in this example, we're going to do another eigenvalue computation.
In this example, we're going to work with the matrix A, which is defined right here.
It's a 3 by 3 matrix.
And we're going to do two things.
First, we're going to find what is called the characteristic polynomial of the matrix.
And then once we found the characteristic polynomial by setting it equal to zero and
finding the roots of the characteristic polynomial, we can find the eigenvalues.
So first, let's do the computation to find the characteristic polynomial.
By definition, the characteristic polynomial of any square matrix is always the determinant
of that matrix minus lambda times the identity matrix.
So that capital I there is the identity matrix.
So for our particular problem, that means we need to compute the determinant of the
matrix A, which was given in the problem, obviously, minus lambda times the appropriately
sized identity matrix.
Since A is a 3 by 3 matrix, that means that I needs to be a 3 by 3 matrix with zeros everywhere,
except ones down the diagonal.
Once I've multiplied the lambda times that matrix, then I will have lambdas down the
diagonal.
So the lambda times I quantity looks like this right here, and that gets subtracted
off of A, all right?
So if we go ahead and do the simplification to simplify this difference of matrices, that
obviously is 1 minus lambda.
2 minus 0 is 2.
9 minus 0 is 9.
12 minus 0 is 12.
11 minus lambda is 11 minus lambda.
2 minus 0 is 2, et cetera.
And now we need to compute the determinant of this matrix.
So for determinants, I have lots of examples on my YouTube channel on how to do that.
If you go back to some of the previous videos in the playlist, the easiest way to do a determinant
computation is via cofactor expansion.
So I'll go ahead and do cofactor expansion about this first row and column.
So we'll have 1 minus lambda times this 2 by 2 matrix right here.
And then we will have minus 2 times the sub matrix form from these two sections right
here.
And then finally, plus 9 times the 2 by 2 matrix right here.
So we basically did cofactor expansion about this first row, all right?
Now, 2 by 2 determinants are pretty easy to compute given a 2 by 2 matrix.
It's just this times this minus this times this, so kind of that cross pattern that you've
seen before in our previous determinant computation examples.
So if we keep working through this, I'll go ahead and rewrite what we had on the previous
chart.
We can go ahead and work through and write down these 2 by 2 determinants.
So there's the determinant of the first 2 by 2 matrix minus 2 times the determinant
of the second one.
So 12 times quantity 4 minus lambda minus 2 times 0 is 0.
And then plus 9 times, I'm going to get 0 minus 0, because 12 times 0 is 0.
Minus quantity 11 minus lambda times 0 is also 0.
Let's do a little simplification.
Obviously, subtracting off 0 doesn't matter right here, so we can simplify that.
Minus 2 times 12 is 24.
24, 4 minus lambda, it simplifies, then this is obviously 0.
If we keep going with some of our algebra here-- we'll go through this quickly.
This is just basic high school algebra at this point, just multiplying everything out.
24 times 4 is 96.
Minus minus makes it a positive 24 lambda.
Keep it going here.
Go ahead and distribute all these quantities.
I combined those into the minus 12.
And then if you go ahead and multiply this out, you'll get all of these terms right here.
Kind of ugly, but not hard.
And then from this final expression here, if we collect all like terms, I have a single
lambda cubed term, so just one of those.
And then, if I collect all of the lambda squared terms, there's some here, and there's some
here.
There's 16 of them in fact, so that's why I have 16 here.
And then in terms of the lambda terms, there's some there, and some there, and some there.
If you add all of those up, you get minus 35.
And then similarly, the constants, I have one constant there and some there.
If I combine those, I get the minus 52.
So when we're all said and done, having done all this algebra, we have gotten this polynomial
here, and that is what we call the characteristic polynomial of the system.
And that's why I've put it in this green box.
And that was our goal.
That first goal was to compute the characteristic polynomials.
So we've now done that.
So we kind of check that box.
The next part was find the eigenvalues.
Given a characteristic polynomial, we can easily find the eigenvalues of a matrix by
simply setting this equal to 0 and finding the roots of that polynomial.
So if I take the characteristic polynomial and actually set it equal to 0, I can solve
for the eigenvalues.
So that's what I did.
Right here, I took the characteristic polynomial.
I set it equal to 0.
Now, if you go to something like Wolfram or Matlab, you can easily find the roots of any
polynomial, and that's what I did.
It turns out that this factors very nicely, factors into this quantity right here.
Lambda minus 13 times quantity of lambda minus 4 times quantity lambda plus 1.
So looking at this, it's very easy to see where we have roots.
When lambda is 13, 13 minus 13 is 0.
So lambda equals 13 is a root of this polynomial.
That means one of our eigenvalues is lambda 1, we'll call it, equal to 13.
Looking at this term, we can see that another one of our eigenvalues is lambda equals 4.
So lambda 2 is equal to 4.
And then looking at this term, if lambda is negative 1 and negative 1 plus 1 is 0, this
term goes to 0, which satisfies the equation.
So we must have lambda 3 equals negative 1.
So those are the eigenvalues that we were able to determine by finding the roots and
the characteristic polynomial.
And that concludes this example.