Let's go ahead and work another problem, dealing with coordinate system representation.
This one will be a little bit different than the one we just worked in the playlist.
In this problem, we have a basis B and we have a vector x.
And notice this vector x, since we're not using the brackets with the subscript B, we
mean that this x is written with respect to the standard basis, just like we always do
when we write down the vector x.
The basis B is this set of three vectors, so we're working essentially in R3 here, and
the vector x, again written with respect to the standard basis, is the vector 2, 3, 4.
What we're going to do in this problem is, we are going to go the other direction.
In the previous example, we had a vector x sub B, which was written with respect to the
basis B. And we figured out how to write it in terms of the standard basis.
In this problem, we're going to start off with a vector written with respect to the
standard basis and we're going to figure out how to write its coordinates with respect
to the basis B. So what we're going to do is, we're going
to find x of B, the coordinates of x written with respect to the basis B that we're given
here in this problem.
So the question is, how can I write 2, 3, 4 as a linear combination of these basis?
That's really the problem that we need to solve.
I need to be able to write x as some number, c1 times this first column plus some other
number, I don't know, c2, times this second column plus some final unknown number, c3,
times the third column.
That should be a c3 right there, just fixed that.
If we can find these numbers, the c1, c2, and c3, then the vector x to B will simply
be c1, c2, and c3.
This by definition are the coordinates of x written with respect to the basis B. So
really what this problem boils down to is, can I find c1, c2, and c3?
So how do I write x as a linear combination of these three vectors?
Well, that's pretty easy to do.
I can do this by just solving a system of equations.
I'm trying to write the vector x as some unknown c1 times this vector plus c2 times this vector
plus c3 times the vector 1, 1, 1.
This is what I need to solve, which I can rewrite as a collection or an augmented matrix
essentially, right?
I can write this as the B vector kind of on the right side of the equation, times all
of these.
And what I'm solving for here in this augmented equation are the unknowns c1, c2, and c3.
So it really just goes back to all of the first videos on this playlist, with how do
you solve a linear system of equations.
So how do we solve this linear system of equations?
I'm going to do row reduction and if I do row reduction, we'll skip the steps here,
again, I have lots of videos that show you how to do row reduction, what we'll end up
with is the following matrix.
I'll end up with the matrix here, which has basically an identity matrix and the final
column will be equal to a negative 1, 1, 5.
So having done row reduction, this tells me that my c1 is equal to a negative 1 because
by looking at this equation, this really says 1 c1 is equal to a negative 1.
And the second row here tells me that 1 times c2 is equal to 1.
And this third row here tells me that 1 times c3 is equal to 5.
So I have solved for the coordinates c1, c2, and c3 . So we now know that, that is equal
to a negative 1, 1, . And 5 let's go and just do a little check.
That is the answer.
The answer is what we were trying to figure out.
Let's just go ahead and just do a little check just to make sure we did our math correctly.
So we have claimed that the coordinates of x with respect to the basis B are minus 1,
1, and 5, that's the vector.
So what that means is that if we take a negative 1 times B1 plus 1 times B2 plus 5 times B3,
we should get out x.
So let's just do that math real quick.
This is almost like a reverse check of the row reduction steps that we just did.
So a negative times that first vector gives me minus 2, minus 2, minus 4.
1 times the second vector does give me the second vector.
And then 5 times the vector of 1s gives me 5s everywhere.
If I add all these up, I get a negative 2 minus 1 plus 5 and minus 2 plus 0 plus 5,
minus 4 plus 3 plus 5, which gives me 2, 3, and 4, which is indeed equal to x, the vector
that we started with.
So again, this last part here is really just a check to make sure our row reduction solution
is legitimate, but we have done what we wanted to do.
Given a basis B and a vector x that was written with respect at the standard basis, we figured
out how to write x with respect to the basis B, which really boils down to solving a linear
system of equations.
You can pick off the coordinates of each dimension of the vectors that we're working with.