In this problem, we'll also be working with the null space of a matrix.
It's a little bit different from the first example problem we worked, though.
In the first example problem, we were just wanting to figure out, is this vector in the
null space.
In this example problem, we're going to more fully characterize all vectors in the null
space of a given matrix.
So the matrix we're going to work with in this problem is a 3 by 4 matrix A, and here
is the matrix.
And what we would like to do is we would like to find a full description of it's null space.
So we would like to be able to characterize all the elements of the null space of A. And
we're going to do that by finding a set of vectors that span the null space.
So we would like to find the set of vectors that span the null space and that will be
our full description of every possible vector that can be in the null space.
Because remember, the span of a set of vectors is all in your combinations.
So we're going to go ahead and find a compact way to write this infinite collection of vectors
that comprise the null space of A. So remember what the null space means.
I'm trying to find the set of vectors such that Ax equals 0.
So that's what I need to find and to find all the vectors x that satisfy this equation.
So what I can do is I can just go ahead and set up an augmented matrix.
So there's my matrix A and I've augmented it with the vector 0, 0, 0, because I'm trying
to find all solutions of this system of equations.
Now, what I'm going to do, I'm going to go ahead and just do very basic row operations
to actually solve this augmented matrix.
So if I that equation 2 equal equation 2 plus 3 times equation 1, and equation 3, I'll replace
with equation 3 minus 2 times equation 1, I can simplify things a little bit.
Equation 1 doesn't change at all, but equation 2 simplifies to this and equation 3 simplifies
to this.
So this is just the standard row operation stuff that we had videos on earlier in the
playlist.
And then let's go ahead and replace E3 with 2E3 plus E2.
And equations 1 and 2 won't change, but now, equation 3 looks like this.
And again, all I'm trying to do is get down to this reduced form, where I can pick off
the solution easily.
If I replace E2 with 1/6E2, I can turn that into a 1, which was something I would like
to do.
So equations 1 and 3 don't change, but equation 2 now transforms into this.
And now, I like this form.
I can look at this.
I have some pivots right here, obviously, in the first two columns.
So those are my basic variables.
X1 and x2 our basic variables.
But x3 and x4, since they don't have pivots, those are going to be free variables.
The system of equations is very much over determined.
I only have three equations but four unknowns.
We knew something like this would happen.
I'd end up with some free variables.
So since I have three variables, I'm sorry free variables, X3 and X4, I want to write
my solution in terms of those.
So if I look at that top row, that top row really says that x1 plus 2x3, I'm sorry, x1
equals X3 6x 4.
And the second row says that x2 equals a negative 2x4.
And then x3 and x4 are free, because they are the free variables in this system of equations.
So going from the matrix notation back to the equation notation so to speak is nice
at this point, because it tells me very explicitly what each coordinate of my solution is going
to be equal to.
Writing it this way is nice, too, because now I can go ahead and think about how I might
write the general solution in terms of my free variables.
The general solution, which I'll go ahead denote as the vector x will be equal to x3
times 2010 plus x4 times a negative 6 negative 201.
So any solution to this augmented matrix is always going to have this form.
And you can see what I've done.
I've written it in terms of the free variables x3 and x4 just by looking at this side of
the equation.
X1 has two x threes, that's why it has a two up here.
And it has minus 6x x4's, which is why there is a minus 6 right here.
And you just do the same thing for the other variables.
So I've been able to write the null space of my matrix, the general solution for all
the vectors in the null space as a linear combination of two vectors.
So what that really means is that my null space is really the span of those two vectors
2010 minus 6 201.
Because remember, this is a linear combination of these vectors.
It consists of all in your combinations, because x3 and x for free so that is the same thing
as the span of those vectors.
So I've been able to find a full characterization of my null space.
My null space of this matrix is really the span of these two vectors and the span is
an infinite collection.
It's all possible linear combinations of these two things.
Since my null space can be written as the span of two vectors, the dimension of my null
space is of size 2.
So we say this is a two-dimensional null space, because there's really two vectors that characterize
it by the span.