# Linear transformation and polynomial function

• ak123456
For the "determine" part, you can use the standard basis for Pn, which is {1, x, x^2, ..., x^n}, to find a basis for the image of I. To find the matrix M, you can use the basis for Pn and Pn+1 mentioned in the problem.

## Homework Statement

from calculus we know that ,for any polynomial function f : R-R of degree <= n,the fuction of I(f) :R-R ,s----$$\int$$f(u) du is a polynomial function of degree <=n+1
show that the map I: Pn--Pn+1 , f--I(f) is an injective linear transformation, determine a basis of the image of I and find the matrix M$$\in$$M(n+2)*(n+1)(R) that represents I with respect to the basis 1,t,...t^n of Pn and the basis 1,t,...t^(n+1) of Pn+1

## The Attempt at a Solution

can i use L(x+y)=L(x)+L(y) aL(x)=L(ax) to show linear transformation ? but for what value i can choose for x, y
for ''determine ...'' part i have no idea for it ,any help ?

ak123456 said:

## Homework Statement

from calculus we know that ,for any polynomial function f : R-R of degree <= n,the fuction of I(f) :R-R ,s----$$\int$$f(u) du is a polynomial function of degree <=n+1
show that the map I: Pn--Pn+1 , f--I(f) is an injective linear transformation, determine a basis of the image of I and find the matrix M$$\in$$M(n+2)*(n+1)(R) that represents I with respect to the basis 1,t,...t^n of Pn and the basis 1,t,...t^(n+1) of Pn+1

## The Attempt at a Solution

can i use L(x+y)=L(x)+L(y) aL(x)=L(ax) to show linear transformation ?
No, you can't use these -- you have to show that they hold for this transformation. For x and y, use polynomial functions of degree <= n. For example, you could let f(x) = anxn + an-1xn-1 + an-2xn-2 + ... + a1x + a0, and g(x) = bnxn + bn-1xn-1 + bn-2xn-2 + ... + b1x + b0.
Then show that L(f + g) = L(f) + L(g) and that aL(f) = L(ax).

ak123456 said:
but for what value i can choose for x, y
for ''determine ...'' part i have no idea for it ,any help ?