Proving Linear Transformation of Polynomials: Let Px be union of all polynomials

['AQF]

"Let Px be union of all polynomials.
Choose a an element of R, and define ta : Px --> R by ta(f) = f(a)
Let T=ker(Ta). Prove that the map
p(x) |--> (x − a)p(x)
is a linear, one-to-one, and onto transformation Px --> T ."

Is the assertation in the problem correct?
If so, how do you prove it?

 

[benorin]

Proof

Choose a\in\mathbb{R} and define t_{a}:P_{x}\rightarrow\mathbb{R} \mbox{ by }t_{a}(f)=f(a).
Let T=\mbox{ker}(t_{a}). Prove that if \Lambda\mbox{ maps } p(x) \mapsto (x-a)p(x), then \Lambda is a linear, one-to-one transformation of P_{x}\mbox{ onto }T.
Proof: Let f,g\in P_{x}, and let b,c\in\mathbb{R}.
Then \Lambda(bf(x)+cg(x))=(x-a)(bf(x)+cg(x))=b(x-a)f(x)+c(x-a)g(x)=b\Lambda(f(x))+c\Lambda(g(x)), so \Lambda is linear.
Since \Lambda(f(x))=\Lambda(g(x))\Rightarrow (x-a)f(x)=(x-a)g(x)\Rightarrow f(x)=g(x), \Lambda is one-to-one.
To see that \Lambda maps P_{x}\mbox{ onto }T, consider that
\forall f\in P_{x}, f\in\mbox{ker}(t_{a})\Leftrightarrow f(a)=0\Leftrightarrow \exists q\in P_{x} \mbox{ such that } f(x)=(x-a)q(x)\Leftrightarrow \exists q\in P_{x} \mbox{ such that } f(x)=\Lambda(q(x)) by the definition of ker and by the fundamental theorem of algebra.

 

[HallsofIvy]

In less formal terms:

T_a maps each polynomial p(x) into the number p(a).
The "kernel" of any linear transformation is the set of vectors it maps into 0 so the kernel of T_a is the set of all p(x) that have value 0 at x=a: p(a)= 0.

Let \Lambda be the transformation that takes p(x) into (x-a)p(x).

1. Linear: \Lambda(mp(x)+ nq(x))= (x-a)(mp(x)+ nq(x)= m{(x-a)p(x)}+ n{(x-a)p(x)}= m \Lambdap(x)+ n\Lambdaq(x).

2. One-to-One. Suppose \Lambdap(x)= \Lambdaq(x). That is (x-a)p(x)= (x-a)q(x). For x not equal to a, we can divide both sides by x-a to get p(x)= q(x). Since p and q are both polynomials, and so continuous, their values at x=a must also be the same: p(x)= q(x).

3. Range of \Lambda is a subset of kernel of T.
If q(x) is in the range of \Lambda, q(x)= (x-a)p(x) for some polynomial p. Trivially, q(a)= (a-a)p(a)= 0. Therefore q(x) is in the kernel of T.

4. \Lambda is "onto".
Let q(x) be in the kernel of T. Then q(x) is a polynomial such that q(a)= 0 and therefore has a factor of the form (x- a). q(x)= (x-a)p(x)= \Lambdap(x).

 

[Hurkyl]

Translating a problem can often make it easier to solve:

We want to say:

p(x) --> (x-a)p(x)

is a one-to-one, onto transformation of Px -> T

Breaking it down into pieces:

(1) This is a map Px -> T
is the same as
For any p(x) in Px, (x-a)P(x) is in T
which is the same as
For any p(x) in Px, (a-a)P(a) = 0


(2) This is a one-to-one map
is the same as
If (x-a)p(x) = (x-a)q(x) then p(x) = q(x).

(3) This is an onto map
is the same as
If q(x) is in T, then there is a p(x) s.t. q(x) = (x-a)p(x).
which is the same as
If q(a) = 0, then there is a p(x) s.t. q(x) = (x-a)p(x)