# Linear Transformation P2 > R^2

1. Jun 19, 2008

### aredian

1. The problem statement, all variables and given/known data

If L( p(x) ) = [ integral (p(x)) dx , p(0) ]

find representation matrix A such that

L (a + bx) = A[a b]^T

2. Relevant equations

3. The attempt at a solution
I don't quite understand the question.
I think that:
if the base from p2 is {1, x} then any vector in p2 is of the form a + bx.
Then I can find L(a) = (a1 , a2) and L(bx) = (a3 , a4)
And use it to get A?

Can someone re phrase the question for me?

2. Jun 19, 2008

### benorin

Given: $$L(p(x)) = \left[\begin{array}{c}\int p(x)dx\\p(0)\end{array}\right]$$

Put $$L(ax+b) = \left[\begin{array}{c}\int (ax+b)dx\\b\end{array}\right]$$

keeping going... the goal is to come up with a matrix A such that $$A\left[\begin{array}{cc}a&b\end{array}\right]^T$$ equals the right-hand side of the last equation.

3. Jun 19, 2008

### aredian

$$L(ax+b) => \left[\begin{array}{cc}x&x^2 / 2 \\0&1\end{array}\right]$$ $$[\begin{array}{c}\ \alpha\\ \beta \end{array} \right]$$

4. Jun 20, 2008

### matt grime

I think you're missing something important: the map is supposed to go from P^2 to R^2, so you can't map ax+b as you claim as the integral in the first coordinate yields (infinitely many) polynomial expressions. Surely the integral should be over some interval (I'd suggest the integral from 0 to 1 as the most likely).

Your matrix in the last post can't make sense, since it implies that the polynomial variables are allowed to appear in the positions in vectors in R^2.

5. Jun 23, 2008

### aredian

yup that was it.. I missed the interval, so the xs turned out to be 1's

thanks