Linear transformation and polynomial function

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SUMMARY

The discussion focuses on demonstrating that the integral operator I, defined as I(f) = ∫f(u) du, is an injective linear transformation from the space of polynomial functions Pn of degree ≤ n to Pn+1 of degree ≤ n+1. Participants clarify that to show linearity, one must verify the properties L(x+y) = L(x) + L(y) and aL(x) = L(ax) using polynomial functions. The basis for Pn is {1, t, ..., t^n} and for Pn+1 is {1, t, ..., t^(n+1)}. The matrix M representing I can be derived from these bases.

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  • Understanding of polynomial functions and their degrees
  • Knowledge of linear transformations and their properties
  • Familiarity with integral calculus, specifically the integral of polynomial functions
  • Ability to construct and manipulate matrices
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  • Study the properties of linear transformations in vector spaces
  • Learn how to compute the integral of polynomial functions
  • Explore the concept of injective functions and their implications in linear algebra
  • Investigate matrix representation of linear transformations between different polynomial spaces
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Students and educators in calculus, particularly those focusing on linear algebra and polynomial functions, as well as mathematicians interested in the properties of integral operators.

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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----\intf(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\inM(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 ?
 
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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----\intf(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\inM(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 ?
 

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