# Transformation Matrix issue

1. Nov 18, 2009

### unfunf22

1. The problem statement, all variables and given/known data
Find m(DT), that is, find the matrix for the transformation DT where D is the derivative operator and T: V -> V , T(p(x)) = xp'(x).
The polynomial is of degree <= 3, and the basis for it is (1,x,x^2, x^3).

2. Relevant equations
Basic matrix multiplication needs to be understood.

3. The attempt at a solution
I have DT(1,x,x^2,x^3) -> D((0,x,2x^2,3x^3)) -> (0, 1, 4x, 9x^2).
So I form my matrix by solving the following:
M*[1, x, x^2, x^3] = [0, 1, 4x, 9x^2].
I get that the matrix is 4x4, of course, and is defined by row-vectors as follows:

[0,0,0,0], [0,1,0,0], [0,0,4,0], [0,0,0,9].
As you see, that matrix does take (1,x,x^2,x^3) -> (0,1,4x,9x^2).

The book; however, says that the transformation matrix is the following, defined by row vectors: [0,1,0,0], [0,0,4,0], [0,0,0,9], [0,0,0,0].
I can't see how that gives the right answer.

What am I doing wrong here?

2. Nov 18, 2009

### Hurkyl

Staff Emeritus
Your notation is... peculiar. I think all you need to do is pay more care to what you write really means.

DT(1,x,x^2,x^3) -> D((0,x,2x^2,3x^3)) -> (0, 1, 4x, 9x^2)​
What did you really mean by this? Is this just shorthand for the four equations
DT(1) = 0
DT(x) = 1
DT(x^2) = 4x
DT(x^3) = 9x^2​
or did you mean something else?

M*[1, x, x^2, x^3]​
This doesn't make sense -- it would be the matrix product of a 4x4 array of scalars with a 1x4 array of vectors....

3. Nov 18, 2009

### unfunf22

I meant that [1,x,x^2,x^3] is a column-vector, so it would look like
|1|
|x|
|x^2|
|x^3|

M multiplying that, so it would be 4x4 matrix times a 4x1 amatrix, to give out a 4x1 matrix.

4. Nov 18, 2009

### D H

Staff Emeritus
You shouldn't have any xs in those vectors. The basis for your space is 1, x, x2, and x3. The various powers of x come from the basis, not the vectors. For example, the vector that corresponds to the polynomial 1+x+x2+x3 is simply [1,1,1,1].

5. Nov 18, 2009

### unfunf22

So then I should be showing that D(T([1,1,1,1])) -> [0,1,4,9] ? Wouldn't this still have the same problem as before? Can I not consider these as column vectors? Maybe that's the issue.

6. Nov 18, 2009

### D H

Staff Emeritus
No, you should be showing that D(T([1,1,1,1])) -> [1,4,9,0].

7. Nov 18, 2009

### unfunf22

Oh, alright, well if I have DT([1,1,1,1]) = [1,4,9,0] then that would be

the matrix would be
[1,0,0,0]
[0,4,0,0]
[0,0,9,0]
[0,0,0,0]
As [1,0,0,0]*[1,1,1,1] = [1] (corresponding to 0 degree poly)
[0,4,0,0]*[1,1,1,1] = [4] (corresponding to x)
[0,0,9,0]*[1,1,1,1] = [9] (corresponding to x^2)
[0,0,0,0]*[1,1,1,1] = [0] (corresponding to x^3)

That matrix still isn't what is in the book. But the matrix that is in the book would work fine as well, so how would I decide on which matrix to use?

Is there a more natural way to go about this? (I've already done this homework and turned it in for a grade, I am just using this problem to study for a test, so presenting methods to solve it will not be giving me an unfair advantage over other students grade wise).

-Ian

8. Nov 18, 2009

### D H

Staff Emeritus
The problem here is you are not using a general third order polynomial. The vector corresponding to $a+bx+cx^2+dx^3$ is [a,b,c,d]. What is T(p(x))? D(T(p(x)))?