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Linear transformation and polynomial function

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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----[tex]\int[/tex]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[tex]\in[/tex]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




Homework Equations





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 ?
 

Answers and Replies

  • #2
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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----[tex]\int[/tex]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[tex]\in[/tex]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).




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

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