# Math Help - differentiation operator basis

1. ## differentiation operator basis

Let V be the vector space

$f(x)=\sum^2_{k=1}(a_Kcoskx+b_ksinkx)$

Find the matrix $[D]_B$ of the differentiation operator D on V in the basis $\{ 1,sinx, cosx,sin2x, cos2x \}$.

So I don't understand how they got that matrix for D! To find the matrix for D, what I did was to compute the images of the standard unit vectors under the transformation. These are

$T(e_1)=(1,0,0,0,0)=(0, cos0, -sin0, 2cos0, 2sin0)$

And so on for

$D(e_2)=(0,1,0,0,0)$
$D(e_1)=(0,0,1,0,0)$
$D(e_1)=(0,0,0,1,0)$
$D(e_1)=(0,0,0,0,1)$

And then form the matrix $D_B=[D(e_1), D(e_2), D(e_3), D(e_4), D(e_5)]$. But the resulting matrxix I got was very different from the answer. So what do I need to do?

2. Originally Posted by demode
Let V be the vector space

$f(x)=\sum^2_{k=1}(a_Kcoskx+b_ksinkx)$

Find the matrix $[D]_B$ of the differentiation operator D on V in the basis $\{ 1,sinx, cosx,sin2x, cos2x \}$.

So I don't understand how they got that matrix for D! To find the matrix for D, what I did was to compute the images of the standard unit vectors under the transformation. These are

$T(e_1)=(1,0,0,0,0)=(0, cos0, -sin0, 2cos0, 2sin0)$

And so on for

$D(e_2)=(0,1,0,0,0)$
$D(e_1)=(0,0,1,0,0)$
$D(e_1)=(0,0,0,1,0)$
$D(e_1)=(0,0,0,0,1)$

And then form the matrix $D_B=[D(e_1), D(e_2), D(e_3), D(e_4), D(e_5)]$. But the resulting matrxix I got was very different from the answer. So what do I need to do?

They must be using the definition that 99% of all linear algebraists use: the matrix for D is the transpose of the coefficients matrix of the system: for example, you got $D(\cos x)=-\sin x$, and since $\cos x$ is the 3rd vector in the given matrix and $\sin x$ the 2nd one , that means that the 3rd column of the matrix will have -1 in the 2nd position, which it indeed has.
BTW, the given matrix is the one for D wrt the given basis: the standard basis' vectors you use there have nothing to do with this problem since they don't belong to the space you're working with.

Tonio