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.

Printable View

- Dec 2nd 2011, 04:00 PMMathIsOhSoHardPolynomial
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 PMFernandoRevillaRe: Polynomial
- Dec 3rd 2011, 06:21 AMMathIsOhSoHardRe: Polynomial
- Dec 4th 2011, 12:11 AMFernandoRevillaRe: Polynomial
- Dec 4th 2011, 02:36 AMDevenoRe: Polynomial
for the monomial basis is . so whatever matrix represents f in this basis, we have:

and

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:

and

so that is indeed the correct matrix.