Determining the matrix of a linear transformation

[Jennifer1990]

1. The problem statement, all variables and given/known data
Let D : P3--> P2 be differentiation of polyonimals. Determine the matrix of D with
respect to the standard basis of P3.



2. Relevant equations
None


3. The attempt at a solution
I think D=[1 0 0; 0 1 0; 0 0 0]. This is from inspection though because I know that the resulting matrix must exclude the numbers in the last row of the matrix.
Could someone give me a hint on how to start this question?

[quasar987]

Well first, do you know what is means by "the standard basis of P3"?

[Jennifer1990]

The standard basis is [1 0 0; 0 1 0 ; 0 0 1]

[Hurkyl]

The standard basis is [1 0 0; 0 1 0 ; 0 0 1]
How can that be? None of those are even elements of P3.

[Jennifer1990]

ohh wait wait , I meant [ 1 0 0 0; 0 1 0 0 ; 0 0 1 0; 0 0 0 1]

[Hurkyl]

You have the dimension right, but those still aren't elements of P3.

[Jennifer1990]

um....then is it the set of different polynomials? {1, x, x^2, x^3}...i think

[Hurkyl]

{1, x, x^2, x^3}...i think
Yes, that's the standard basis for P3.

[AUMathTutor]

It's either that or the other way around. I think that P3, however, means polynomials of degree less than 3; so x^3 is excluded.

Now, what you do is you do the linear transformation on each of the basis vectors. This will produce a vector which is an element of P2. Say that U = P3 and V = P2, and D goes from U to V. Then L(u) = v, where u is in U and v is in V.

So you need to find: v1 = L(u1), v2 = L(u2), v3 = L(u3) (and v4 = L(u4), if you include x^3)

Then the matrix for D will have columns v1, v2, and v3 (, and v4 if you include x^3), in that order. Then if you take M, the matrix for D, and try

M*u = v, you'll find that v is a vector in P2 which represents the derivative of the polynomial u.

[Jennifer1990]

So
v1 = L(u1) = 0
v2 = L(u2) = 1
v3 = L(u3) = x
v4 = L(u4) = x^2

How am i suppose to rewrite this to represent the matrix D?.....because this is now the standard basis for P2

[AUMathTutor]

Well, the derivative of x^3 isn't really x^2... it's 3x^2.

And, like I said, I believe that P2, P3, etc. means "the set of all polynomials of degree less than 2, 3, etc.". I might be wrong. That's how it was presented to me, however.

I'll do one part of it for you - so here we go.

U = P3 = [1, x, x^2], V = P2 = [1, x]

D(x^2) = 2x = transform(0, 2) wrt V

So the first column of M will be this vector, namely,


M = 0 ? ?
2 ? ?


To fill in the other question marks, do the same thing with the other basis vectors and put them in. Then you'll see that things like

M * transform(3, 7, 2) = transform(7, 4).

Does that make more sense?

[Hurkyl]

So
v1 = L(u1) = 0
v2 = L(u2) = 1
v3 = L(u3) = x
v4 = L(u4) = x^2

How am i suppose to rewrite this to represent the matrix D?.....because this is now the standard basis for P2
What is L? You were supposed to be finding the coordinate matrix for D.


Do you know how to find the coordinates of a vector with respect to a basis?

[Jennifer1990]

To find the coordinates of a vector with respect to a basis, row reduce the vector with the new basis.

To:AUMathTutor
I think D works out to be [0 1 0 0 ; 0 0 2 0; 0 0 0 3] if we include polynomials of degree 3 in P3

[AUMathTutor]

I think that's right. I see I made a mistake in putting (0, 2) in the first column if x^2 comes last in my scheme.

I think that what you have is correct, if of course you allow polynomials of degree 3 in P3 and polynomials of degree 2 in P2.