All right.
In this example, we're going to work through the kind of computational mechanics of computing
eigenvalues of a matrix.
Obviously with Matlab and other software, it's very easy just to plug-in that matrix
you have and get out the eigenvalues numerically, but it's also nice to be able to do this by
hand and know how to do these types of computations all with pen and paper, so to speak.
So here's the matrix we're going to work with.
It's just a 2 by 2 matrix, Matrix A. And we are going to find two things-- first,
we're going to find what we call the characteristic polynomial of the matrix and then from the
characteristic polynomial, if we set that polynomial equal to 0, the roots of that polynomial
give us the eigenvalues of the matrix.
So those are kind of the two key steps.
So first, let's go ahead and compute the characteristic polynomial.
By definition the characteristic polynomial of a matrix is the determinant of the matrix
minus lambda times the identity matrix, I.
So we're need to compute the determinant of A minus lambda I.
So that determinant of A, that's our given matrix A, minus lambda times the identity
matrix.
So in this case, we're dealing with 2 by 2 matrices.
So it's going to be lambda times a 2 by 2 matrix I. I do a little algebra here.
Those 2 by 2 matrices after subtracting, I have negative 1 minus lambda.
6 minus 0 is 6, 2 minus 0 is 2, and 3 minus lambda is 3 minus lambda.
So I need to take the determinant of this matrix.
If you recall, that's pretty easy.
That's just this times this minus this times this.
So those things kind of criss-crossed.
So I'm going to get minus 1, minus lambda times 3 minus lambda, minus 6 times 2.
So that is what I need to simplify a little bit algebraically.
If we multiply this out, and negative times 3 is a negative 3.
A negative 1 times negative lambda is plus lambda.
A negative lambda times 3 is a negative 3 lambda.
And then a negative lambda times a negative lambda lambda squared.
And then minus 12 for this term right here.
And if we simplified just a little bit and combined some like terms, like lambda squared
minus 2 lambda minus 15.
So this is what we call the characteristic polynomial of the matrix, and by definition,
it's always equal to the determinant of A minus lambda I.
So this is the characteristic polynomial that we were looking for.
Now we can find the eigenvalues.
By definition, the eigenvalues of a matrix are the roots of the characteristic polynomial.
So if we set the characteristic polynomial equal to 0 and find the lambda that now satisfy
this equation, that by definition, is the eigenvalues of the matrix.
So let's do that.
Turns out this factors pretty nicely.
It factors into lambda minus 5 times lambda plus 3.
Now there's no reason why the roots have to be integers.
You'll often see when people do examples, that you get integers for the eigenvalues.
That just kind of what people tend to do when making problems for pen and paper.
But in general, when you have a matrix and compute the eigenvalues, you can get any type
of real valued number with decimals and things like that.
So don't be tricked into thinking that eigenvalues are always integers.
They just often are for these kind of contrived problems that people do as examples.
All right, so we were able to factor the polynomial like this.
So that tells me that the roots are two of them.
Lambda 1 is equal to 5.
When lambda is equal to 5, we get 5 minus 5 is 0.
And then looking at this one, if lambda is negative 3, the same thing happens.
So my roots are 5 and negative 3.
So this right here is the collection of eigenvalues.
We had a 2 by 2 matrix, so there should be two eigenvalues.
And they are in fact equal to 5 and negative 3.
So that conclude this example, showing you how to, by hand, compute the characteristic
polynomial.
By setting that equal to 0 and finding the roots, that by definition yields the eigenvalues
of the matrix.