Hi, I'm working on the following problem and was wondering if anyone could check to see if I've solved it correctly (I can't help but wonder if my method is incorrect - it seems a little too simple).
Define by . Find the range of
My attempt is as follows:
Let (standard form for a polynomial of )
Then
so range
It seems a little crude to me for some reason...
Is it correct?
Thanks,
By that I meant it's a linear combination of the elements forming the "obvious" basis for . Those elements are: .
The range of is all possible linear combinations of its basis elements:
(These were obtained by applying the linear transformation to defined above)
Does this reasoning sound right?
I agree that form the obvious bases of . However, is in no way a general form or representative of all polynomials in that space; it's just a single particular polynomial. A general form would be for some real numbers .
I agree that your original conclusion "range T = span " means "The range of is all possible linear combinations of its basis elements: ." But how does this conclusion follow from the reasoning before that?
I don't think you showed how any linear combination of except was produced by T from that particular f(x).
We agree that
.
A typical proof of the fact that range(T) = A for some set A consists of two parts: the first one shows range(T) ⊆ A and the second one shows A ⊆ range(T). The first part fixes an arbitrary and shows that T(f) ∈ A. The second part fixes an arbitrary g ∈ A and finds some such that T(f) = g.
since T(f) always has an "x" in it (unless f'(x) = 0, the 0-polynomial), one suspects that the image of T is: A = {f(x) in P_{3}(R): f(0) = 0} (that is, polynomials with 0 constant term).
clearly T(f) is always in that set, because g(x) = x(f'(x)) always satisfies: g(0) = 0(f'(0)) = 0, no matter what f' happens to be. so we've found a containing set for T(f). T(f) ⊆ A.
now suppose g(x) = a_{1}x + a_{2}x^{2} + a_{3}x^{3} = x(a_{1} + a_{2}x + a_{3}x^{2}) is any such polynomial.
if we can find an f(x) with f'(x) = a_{1} + a_{2}x + a_{3}x^{2}, we're done, because that's a pre-image for g.
by ordinary techniques of first-year calculus, one sees that any anti-derivative of a_{1} + a_{2}x + a_{3}x^{2} will do, in particular:
f(x) = a_{1}x + (a_{2}/2)x^{2} + (a_{3}/3)x^{3} yields T(f) = g. so A ⊆ T(f).
geometrically, what is happening is this: differentiation "loses one dimension", so we go from P_{3}(R) to P_{2}(R), while "multiplication by x" acts like a shift-operator, moving the 3 coordinates we get for our derivative over one position to the right (which still leaves us with a 3-dimensional space).