In this problem, we're going to examine a different coordinate system.
Most of the time when working with vectors in the linear algebra, we work with what's
called the standard basis.
So for instance, when we write down the vector, I don't know x equals 1, 2, 3, that means
we have 1 in the direction of 100.
The 2 in the second coordinate tells me we move two units along the 010 direction.
And then 3 and the third element would tell us that we move three units along the 001
direction.
So the standard basis that we tend to work with implicitly just has a 1 in each corresponding
coordinate of the vector.
Sometimes, we want to work in different coordinate systems.
So for example in this example, we're going to work with a basis B that looks like the
following.
It's not going to meet-- these are going to be length three vectors.
The standard basis would be 100, 010, 001.
In this problem, we're going to work with the basis represented by 224 and negative
103, and 111.
It turns out that these vectors are all linearly independent, so they form a basis for R3.
So any vector in R3, I can write as a linear combination of these vectors.
But obviously, any vector in R3 that I would write as a linear combination of these vectors,
those coordinates would be different than how I would write the linear combination for
the standard basis, which would be the vector 100, 010, 001.
So it's just a change of coordinate systems, so to speak.
And in this first example, we're going get practice writing a vector that has been provided
in this coordinate system.
And we're going to translate that back into what I would call these standard coordinate
system.
So this is the basis that we're working with in this problem.
And we're provided a vector.
And the notation we use is x sub B and, this tells us that the vector x has been written
in the basis B. So anytime you see this square bracket with a subscript with a letter, that
is basically just an indication that the vector has been written, not in the standard basis,
which is usually our implicit assumption there, but it has been written with respect to the
basis B. So this vector is 2, 1, negative 3.
So this is the representation of the vector in this basis.
And it really tells us to move two units along this direction, and then move one unit along
this direction, and minus three units in to this direction.
So the interpretation is still very much the same.
It tells us how many units we've moved in each direction.
But the directions aren't the standard directions, these directions specified by this new basis
B. That's a little explanation about some of the things we'd like to do in linear algebra.
In this particular problem, what we're going to do is we're going to find x.
We're going to find the coordinates of xB in the original standard basis.
So how are we going to do that?
The key thing is that xB tells us how to construct x from the vectors in B. So let's take a look
at this.
Let me go ahead write down x and B, and let's think about what each of these entries means.
2 tells us to move, like I said, 2 units in the B1 direction.
So this vector right here is a vector in the standard basis, 224, obviously.
And this first coordinate tells us to move two units in that direction.
The second coordinate tells us to move one unit in the B2 direction, where B2 is this
session column.
And the negative 3 tells us to move three units in the negative direction of B3, where
B3 is this third column of the basis B. So we can go ahead and form this linear combination
now to figure out what x is.
X is going to be 2 times this first column plus 1 times the second column plus negative
3 times the third column.
So if I go ahead and do my algebra there, 2 times 2 is 4 times 224 times 228, 1 times
the second vector, this gives me the second vector, and then a negative 3 times the third
vector gives me a negative 3's everywhere.
If I add all these coordinates up, 4 minus 1 minus 3 gives me a 0.
4 plus 0 minus 3 gives me 1, and 8 plus 3 minus 3 gives me 8.
So I get the vector 0, 1, 8.
So this vector right here is the vector x.
And this represents the vector translated into the standard basis.
So we've started with a vector representation in one basis.
And using this interpretation of what this means, we've actually rewritten it in terms
of the standard basis.
So in some ways, this is like a coordinate transformation problem.
We had a vector representation in one coordinate system, and we were able to write it equivalently
in a different coordinate system, where the second coordinate system is what we call these
standard basis.
It's the normal way that we tend to write vectors.
And anytime we write a vector like this, that's what we mean.
We imply that we're using the standard basis.
It's only when we move to other coordinate systems and other bases, we use this slightly
different notation to indicate this vector is not written in the standard basis, this
vector has been written with respect to the basis B. And the basis B in this problem was
this set of vectors.