All right, in our last video in this playlist, we talked about computing a matrix to a power.
And we found that if we could diagonalize the matrix and write it in a diagonalization
form, being able to compute that matrix power was much simpler.
So in this video, we're actually going to look at, how do you get a matrix written in
a diagonal form like that?
So here's the matrix A that we're going to work with.
It's a 2 by 2 matrix.
And if possible, we're going to diagonalize the matrix.
In other words, we're going to try to write it as P times the diagonal matrix D times
P inverse.
So we're going to see if we can get it written in this form.
So here's something that's very interesting.
It turns out that we can always do this if A has unique eigenvalues.
If it has unique eigenvalues, then we can construct D by making the eigenvalues the
diagonal entries of that matrix.
And then the matrix P consists of the eigenvectors associated with those eigenvalues.
So any matrix that you have, if it has unique eigenvalues, you can always write in this
framework.
And these equations right here tell us how to construct those matrices.
D is always a diagonal matrix whose elements are the eigenvalues on the diagonal.
And P is a matrix whose columns are the eigenvectors.
So really what we need to do is we need to figure out what the eigenvalues and eigenvectors
are of A. Note that A is a lower triangular matrix.
All of the non-zero values are on or below the diagonal.
That means the eigenvalues are just the diagonal entries.
That means lambda 1 is 2 and lambda 2 is 3.
So we know what the diagonal elements are for our D matrix now.
We know that D should be this 2 by 2 matrix right here.
All right, that part's done.
What do we need to do now?
We need to find the eigenvectors V1 and V2.
So on the next chart, we'll actually work through those eigenvector computations.
Our eigenvalue lambda 1 is 2.
I need to find the solutions to this equation.
This is what it means to be an eigenvector.
So if we plug that in, we get 2 0 -1 3.
That's our matrix A. We need to subtract off 2 times the identity
matrix because I'm working with the eigenvalue 2.
And I need to find all solutions to this system of equations.
So I need to find all x1 and x2 that solve this.
If we simplify this just a little bit, we end up with this.
And if I look at that second set equation right here, this equation says that minus
x1 plus x2 is equal to 0, which if I rearrange says that x1 equals x2.
And remember, this always happens when we do eigenvector computations.
You always end up with three variables and an infinite number of solutions.
Any solution that has x1 and x2 will be an eigenvector.
So I'm going to set x1 equal to 1 just for simplicity.
And since x1 is 1, x2 has to be 1.
So this is my eigenvector, 1 1.
Similar computation for lambda to equal to 3.
I need to solve this system of equations right here.
So I have A. I need to subtract off 3 times the identity matrix.
And this is the system of equations I need to solve.
If I simplify just a little, this is the system of equations I need to solve.
And looking at this, equation 1 says that negative x1 equals 0.
That means x1 must be 0.
And the second column here all being zeros means that x2 is a free variable.
So x1 has to be 0.
And x2 can be anything I want.
So for simplicity, I'll just make it 1, so these are my eigenvectors.
I have one here.
And I have this one here, which corresponds to lambda 2 equal 3.
So I now know my eigenvectors, and I can construct the matrix P. So this is my matrix P right
here.
Each column is an eigenvector.
Let's go ahead compute P inverse, too.
We'll need that here in a minute.
Doing the inverse of a 2 by 2 is pretty easy.
I have other videos on that.
So that is the inverse of P. And let's go ahead and check.
Let's go ahead and compute P times D times P inverse.
If you multiply this out, you do indeed get 2 0 -1 3, which is equal to A.
We have been able to diagonalize A using the approach that we proposed.
This approach always works as long as A has unique eigenvalues.
Construct P as the eigenvectors.
Construct D as a diagonal matrix whose elements on the diagonal are the eigenvalues.