# Polynomial

• Dec 2nd 2011, 04:00 PM
MathIsOhSoHard
Polynomial
f: P1(R) -> P1(R)

f(1+4x) = 1-2x
f(-2-9x) = 2+4x

Find the transformation matrix for f in the monomial basis.

|1 -2| |1 2|
|4 -9| |-2 4|

I tried to multiply these two matrices but that doesn't give me the correct facit.
• Dec 2nd 2011, 11:40 PM
FernandoRevilla
Re: Polynomial
Quote:

Originally Posted by MathIsOhSoHard
f: P1(R) -> P1(R) f(1+4x) = 1-2x f(-2-9x) = 2+4x Find the transformation matrix for f in the monomial basis.

Write $\displaystyle \begin{Bmatrix} f(1)+4f(x)=1-2x\\-2f(1)-9f(x)=2+4x\end{matrix}$ . Express, $\displaystyle f(1)=a+bx,\;f(x)=c+dx$ then, the solution is $\displaystyle \begin{bmatrix}{a}&{c}\\{b}&{d}\end{bmatrix}$ .
• Dec 3rd 2011, 06:21 AM
MathIsOhSoHard
Re: Polynomial
Quote:

Originally Posted by FernandoRevilla
Write $\displaystyle \begin{Bmatrix} f(1)+4f(x)=1-2x\\-2f(1)-9f(x)=2+4x\end{matrix}$ . Express, $\displaystyle f(1)=a+bx,\;f(x)=c+dx$ then, the solution is $\displaystyle \begin{bmatrix}{a}&{c}\\{b}&{d}\end{bmatrix}$ .

I'm not sure that I understand this entirely. :(
• Dec 4th 2011, 12:11 AM
FernandoRevilla
Re: Polynomial
Quote:

Originally Posted by MathIsOhSoHard
I'm not sure that I understand this entirely. :(

Have you studied the theory? If $\displaystyle T:V\to W$ is a linear map, $\displaystyle B_V=\{v_1,\ldots,v_m\}$ and $\displaystyle B_W=\{w_1,\ldots,w_n\}$ are basis of $\displaystyle V$ and $\displaystyle W$ respectively then, $\displaystyle [T]_V^W=[\;[T(v_1]_{B_W},\ldots,\;[T(v_n)]_{B_W}]$ .
• Dec 4th 2011, 02:36 AM
Deveno
Re: Polynomial
for $\displaystyle P_1(\mathbb{R})$ the monomial basis is $\displaystyle \{1,x\}$. so whatever matrix represents f in this basis, we have:

$\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}1\\4\end{bmatrix} = \begin{bmatrix}1\\-2\end{bmatrix}$

and

$\displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}-2\\-9\end{bmatrix} = \begin{bmatrix}2\\4\end{bmatrix}$

now the column (a,c) is f(1,0) = f(1+0x) = f(1). so let's find this first. remember, matrix multiplication is linear. this means that 9f(1,4) + 4f(-2,-9)
= f(9,36) + f(-8,-36) = f(1,0).

but f(1,4) = (1,-2), and f(-2,-9) = (2,4). so 9f(1,4) + 4f(-2,-9) = 9(1,-2) + 4(2,4) = (9,-18) + (8,16) = (17,-2). so a = 17, c = -2.

now let's find the second column (b,d), which is f(0,1) = f(0+1x) = f(x).

-2f(1,4) - f(-2,-9) = f(-2,-8) + f(2,9) = f(0,1). so f(0,1) = -2(1,-2) - (2,4) = (-2,4) + (-2,-4) = (-4,0). so b = -4, d = 0.

finally, let's check:

$\displaystyle \begin{bmatrix}17&-4\\-2&0\end{bmatrix} \begin{bmatrix}1\\4\end{bmatrix} = \begin{bmatrix}1\\-2\end{bmatrix}$

and

$\displaystyle \begin{bmatrix}17&-4\\-2&0\end{bmatrix} \begin{bmatrix}-2\\-9\end{bmatrix} = \begin{bmatrix}2\\4\end{bmatrix}$

so that is indeed the correct matrix.