All right.
In this example, we're going to work with eigenvectors and eigenspace and finding a
basis for the eigenspace of a matrix.
The matrix we're going to work with is this 3-by-3 matrix A right here.
And to save us some time, we're going to be told some of the eigenvalues of the matrix.
This 3-by-3 matrix obviously has three eigenvalues, and we're told that they are equal to 8, negative
16, and 40.
What we're going to do is we're going to essentially find the eigenvectors and find a basis for
those eigenvectors.
So more generally, we say that we're going to be finding a basis for the eigenspace.
So you can think of the eigenvectors as forming this vector space.
We're going to find a basis for that vector space.
And since it's associated with the eigenvectors, we call it an eigenspace.
All right.
So what do we need to do?
We need to find the null space of lambda I minus A. Remember, that's how you find an
eigenvector.
An eigenvector is the null space of lambda I minus A for each lambda.
So this equation right here, we're going to need to evaluate individually for each of
these lambdas.
For each of those lambdas, we'll evaluate it and then find the null space.
That's our eigenvector.
And then we'll find a basis for that eigenvector.
All right.
So number one, let's go ahead and start for the case where our value is equal to 8.
That was the first one we're provided, so that's the one that makes the most sense to
start with.
I need to compute 8I minus A. So that is the identity matrix scaled by 8.
And I need to subtract off the matrix A, my original starting matrix, which if I do the
arithmetic, gives me this matrix right here.
This is the matrix I'm going to work with, and I need to find the null space of that
matrix.
Remember, the null space means find the linear combinations of these rows that gives us the
all 0 vector.
So this right here, setting up this equation equal to all zeros, I'm going to try to find
x1 and x2 and x3 to solve this system of equations.
The points, x1, x2, and x3, in general, in solving a system of equations there's three
options.
You could have one solution, no solution, or an infinite number of solutions.
Since I'm finding the null space of lambda I minus A and dealing with eigenvectors, these
types of systems of equations are always going to have an infinite number of solutions when
working this particular type of problem.
All right.
So let's go ahead and solve this system of equations.
If I look at equation 2-- equation 2, which is this second row right here-- tells me that
a negative 32 times x2 has to equal 0.
So that means that x2 has to be 0.
And if I look at equation 1, that tells me that 24 minus 84 x2, which I just solved for
as having to be 0, that also has to equal 0.
So if that's true, that means that x3 also has to be equal to 0.
Excuse me, I misspoke there.
If you look at this equation right here, x3 can be anything since it's being multiplied
by 0.
It's x1 that must be equal to 0.
If you simplify this, you're going to get 24 x1 equals 0.
And solving for x1, obviously, gives 0, so both x1 and x2 have to be 0.
And then from looking at this column right here, we can see that x3 can actually be anything.
So x3 is actually a free variable because it can be any number.
And since it gets multiplied by zeros here, it still is going to satisfy this system of
equations.
So what we found is that my solution to this linear system of equations is, in fact, a
vector that has a 0 for x1, a 0 for x2, and then it has x3 can be anything.
Well, since x3 can be anything, I'm free to choose.
I'm going to go ahead and just choose the value 1, and this vector right here now serves
as a basis for all solutions.
Any multiple of this vector is also a solution, right, because I'd still have zeros in the
first elements.
And I would still have x3 as the final coordinate, but x3 could be any number.
So this vector right here serves as a basis for this first eigenvector.
All right.
Let's continue.
Number 2, lambda equals negative 16.
Same procedure, I need to solve the system of equations minus 16 times the identity matrix
minus A equals 0.
So, first, let's just compute what that quantity is.
There is my scaled identity.
I subtract off the matrix A. That yields this matrix here.
And for this matrix, we need to find the null space.
The null space is the x1, x2, x3 such that I get 0, so I need to solve this linear system
of equations.
Again, by inspection, I can kind of figure some things out pretty easily.
Looking at equation 1, we see that x2 has to be equal to 0.
Looking at equation 3, I have this system of equations because x2 is 0.
So if I solve for this, I see that x1 has to equal a negative x3.
So in terms of the solution to this system of equations, x2 always has to be equal to
0.
And this relationship between x1 and x3 has to hold.
So my solution is always going to have a 0 as the second coordinate.
It's always going to have some number in the third coordinate.
And whatever number is there, I have to have the negative of it in the x1 spot.
So, again, we're trying to find a basis for the solution.
There's an infinite number of solutions, but they're all going to have that form.
So I'm going to go ahead and choose x3 as 1, just to make things simple.
If x3 is 1, that means this is equal to 1 since I set it equal to 1.
But the value in my x1 coordinate has to be negative 1, has to have that form right there.
So this vector right here forms a basis for the solution to this system of equations.
This vector times any number is also a solution to this linear system of equations.
That's why it's the basis for it.
All right.
Let's keep going, one more.
So the last eigenvalue we need to deal with is 40.
I do my same computations.
40I minus A gives me this matrix right here.
And then I need to find the null space of that matrix, which means find all, x1, x2,
and x3, such that this matrix operating on them gives me the all 0 vector.
So if I look at this and do some row reductions-- we're going to skip that step; I have plenty
of other videos that show you how to do this step via row reduction-- you can simplify
it to this matrix right here, this linear system of equations.
And then looking at this, we see some interesting things.
From equation 2, we get that x3 has to be equal to 0.
If I look at equation 1, I have x1 minus 1.5.
x2 has to equal 0, which if I rearrange says that x1 has to be 1.5 times x2.
So now I can go ahead and solve for this basis.
x3 is always 0, so I'll put a 0 there.
And then as long as this relationship holds between the first two coordinates, I have
a solution.
So I'm going to go ahead and set x2 equal to 1.
If x2 is equal to 1, then by this equation here, that means x1 has to be 1.5.
So this is the final basis vector that I was looking for.
So we've found three vectors that serve as basis vectors for what we call the eigenspace
of the matrix.