MARTINA BALAGOVIC: Hi.
Welcome to recitation.
Today's problem is about change of basis.
It says the vector space of polynomials
in x of degree up to 2 has a basis 1, x, and x squared.
That's the obvious basis that you would
write for that vector space.
But today we're going to consider another basis, w_1,
w_2, and w_3.
And we don't know what w_1, w_2, and w_3 are explicitly.
What we know is that their values at x
equals minus 1, 0, and 1 are given by this table here.
So they are 1, 0, 0; 0, 1, 0; and 0, 0, 1.
We're asked to do the following.
We're asked to express this polynomial-- so y
of x is minus x plus 5-- in this basis, w_1, w_2, w_3.
We're asked to find the change of basis
matrices between these two bases, 1, x, x squared,
and w_1, w_2, w_3.
And finally, we're asked to find the matrix of taking
derivatives, which is a linear map on this space,
in both of these basis.
And let me give you an extra level of challenge, which
is to try to do as much of this as possible without explicitly
finding w_1, w_2, and w_3.
I'll let you think about the problem
and then you can come back and compare your solution
with mine.
Hi.
Welcome back.
So to start with the problem a, we
need to find coefficients alpha, beta, and gamma so that y of x
is expressed through, with this coefficient,
in this new basis, w_1, w_2, and w_3.
now, one way to do that would be to look
at this table of values, explicitly
find w_1, w_2, and w_3, so a quadratic polynomial is--
all the information we need about it
is in values at three points.
So say w_1 is a*1 plus b times x plus c times x squared.
Find a, b, and c.
Find w_1, w_2, w_3 explicitly, and then go back to this system
and try to find alpha, beta and gamma.
However, there's a trick.
Let's try to see if we can do it without finding
w_1, w_2, and w_3 explicitly.
So let me try to see what are the values of y
at these points.
So y is minus x plus 5.
So the values are 6, 5, and 4.
And let me, instead of considering this equation,
let me evaluate it at x is minus 1, x is 0, and x is 1.
What I get through this is that w at minus 1-- actually,
let me write this.
I get w at minus 1, which is a number,
equals alpha times w_1 at minus 1,
which is a number, plus beta times w_2 at minus 1,
plus gamma times w_3 at minus 1.
And similarly at 0, and similarly at 1.
And now, let me think of this as a linear system that
has unknowns alpha, beta, and gamma
coefficients, these values here at minus 1, 0, and 1.
And the right-hand side, well, what's
written here on the left-hand side. y at minus 1, y at 0,
and y at 1.
If you write this in a matrix and read off coefficients
from there, you get the following system.
So this is the matrix of the system read off from here.
These are the unknowns.
And these are the values of the right-hand side.
And I hope you'll agree that this is a very easy system
to solve.
We just get alpha is 6, beta is 5, and gamma is 4.
So the solution to the first part is y equals 6*w_1 plus
5*w_2 plus 4*w_3.
And let's notice another thing.
No matter what values we put here,
this matrix is always going to stay the same.
It's only the right-hand side that's going to change.
So if we're given any other polynomial
now to express in a basis w_1, w_2, and w_3,
we don't have to do any thinking.
We don't have to do any computations.
What we do is go back to our table at the beginning
and just read off-- let's go back to the table,
and just read these values.
So in this case, y is 6 times w_1, 5 times w_2,
and 3 times w_3.
And that's already a hint to solving
the b part, which is find the change of basis
matrices between 1, x, x squared and w_1, w_2, w_3.
Change of basis matrices means expressing
one basis in terms of another.
So as a part of the problem, we will
have to express 1, x, and x squared
in terms of w_1, w_2, w_3.
So let's just find their values at these three points.
One is a constant, it just takes value 1 everywhere.
x takes value minus 1 at minus 1, 0 at 0, and 1 at 1.
And x squared takes values 1, 0, and 1 at minus 1, 0, and 1.
And from this we can already conclude the part b here,
we can conclude that 1 equals w_1 plus w_2 plus w_3.
That x equals minus w_1 plus w_3.
And that x squared equals w_1 plus w_3.
And from this, we can immediately
write one change of basis matrix.
Namely, since we know how to express 1, x,
and x squared in terms of w_1, w_2, and w_3,
we can just copy this information over
to this matrix, getting 1, 1, 1; minus 1, 0, 1; and 1, 0, 1.
So which matrix is this?
This is a matrix-- so we have 1, x,
and x squared expressed in terms of w_1, w_2, and w_3.
So if we feed this matrix something expressed
in the basis 1, x, and x squared, say a, b,
and c, what it's going to throw out
is the same polynomial expressed in this basis here, w_1, w_2,
and w_3.
So I'm going to just write that this is a matrix of this basis
change.
How do we get the other one?
Well, very easy.
We know it's just the inverse of A.
So this is going to be the matrix that
takes something written in this basis
and transfers it to this basis.
I'm not going to calculate the inverse
of a matrix in front of you.
Instead I'm going to consult my oracle.
Sorry about that.
And my oracle says that the inverse
should be 0, 1, 0; minus 1/2, 0, 1/2; and 1/2, minus 1, 1/2.
And that solves the b part.
Let's go into the c part.
The c part required us to find a matrix of taking derivatives,
which is a linear map in the space of polynomials,
in both of these basis.
So let's first do the 1, x, x squared basis
because that one's easier.
I'm going to call it D_x.
So I'm going to work in basis 1, x, x squared.
And what I want to express is the transformation
of taking derivatives.
So here I'm going to write the vector
to which taking derivatives maps the polynomial 1, which is 0.
And that's this expressed in the basis 1, x, x squared.
In the second column I'm going to write x prime,
the vector to which D_x sends to vector
x, and that's equal to 1, which expressed in this basis
is 1, 0, 0.
And here I'm going to write x squared
prime, which is 2x, which, expressed in this basis,
is just this.
That one was easy.
For the other one, well we could calculate
w_1, w_2, w_3 explicitly, take the derivatives,
go back to the table and repeat the procedure
that we did already.
So expressing these derivatives in terms of w_1, w_2, w_3,
and that's a lot of work.
But we pretty much already did most of this work.
So we know how to take derivatives in this basis,
and we know how to go between these two basis.
So if we want to take a derivative of something written
in the basis w_1, w_2, w_3, well, let's first
write this something in basis 1, x, x squared.
Then let's take a derivative of it.
And then let's write it back in the original basis
that we want.
So it's multiplication of three matrices.
We have all three-- matrix multiplication is easy.
And my oracle, again, says that this should be minus 3
over 2, 2, minus 1 over 2; minus 1 over 2, 0, 1 over 2;
and 1 over 2, minus 2, 3 over 2.
And that solves the problem.
Now, one thing that I would want to discuss in the end
is how did you do with respect to my challenge, which
was let's try to do as much of it as possible
without finding w_1, w_2, and w_3 explicitly.
And it seems like we did really well.
There's nowhere on the board written w_1 equals,
w_2 equals, w_3 equals.
But is it really so?
It's not.
We calculated the matrix of A inverse here.
And what this really means is that w_1, w_2, and w_3, written
in the basis 1, x, and x squared, are as follows.
w_1 is minus 1/2 x plus 1/2 x squared.
w_2 is 1 minus x squared.
And w_3 is 1/2 x plus 1/2 x squared.
So you can check your work with the help of this matrix
in case you did find w_1, w_2, and w_3 explicitly.
And that's all I wanted to say today.