# Polynomial Proof

1. Oct 24, 2005

### '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?

2. Oct 25, 2005

### 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.

Last edited: Oct 25, 2005
3. Oct 25, 2005

### HallsofIvy

Staff Emeritus
In less formal terms:

Ta 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 Ta 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 $\Lambda$p(x)+ n$\Lambda$q(x).

2. One-to-One. Suppose $\Lambda$p(x)= $\Lambda$q(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)= $\Lambda$p(x).

Last edited: Oct 25, 2005
4. Oct 25, 2005

### Hurkyl

Staff Emeritus
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)

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?