# Injective linear transformation

1. Aug 27, 2009

### beetle2

1. The problem statement, all variables and given/known data

We regard each polynomial p(t) an element of R(t) as defining a function

$p:R\rightarrow R, x \rightarrow p(x)$

prove that

$g:R[t]\rightarrow R[t], p(t) \rightarrow \int_{0}^{t}p(x)dx$

defines an injective linear transformation.
2. Relevant equations

3. The attempt at a solution

As the function is only defined for $t \geq 0$ is fair to say that when t = 0, g(T) = 0 = ker(g)
is the only vector in the kernel. and as the functions domain is well defined

then function is injective?

2. Aug 27, 2009

### slider142

First, the function p output by the functional g is defined for all real t, as the integral is defined for all real t. Secondly, g is not a linear transformation of t, it is a linear transformation of p. Properties of g are not related to variations in t, but variations in p (Ie., what polynomials, if any, are mapped to the 0 polynomial?)

3. Aug 27, 2009

### beetle2

Are they the polynomials with zero coefficient?

4. Aug 28, 2009

### boneill3

Do I need to find out all the elements p(t) of the set of polynomials [t] that after integration map to the zero polynomial?

5. Aug 28, 2009

### nirax

where r u stuck with this ??? ... linear part is trivial. to prove the injective part observe that any polynomial cannot be zero in the whole of (0,1) as it can atmost have finitely many zeros.

i can help u more only if you show evidence of your effort. you seem to be stuck at just the question, marvelling at what to prove.

6. Aug 28, 2009

### boneill3

I am stuck with identifying the function G.

I thought a function is injective IF and only IF Ker{T}= 0.

I'm not sure what you mean by "any polynomial cannot be zero in the whole of (0,1) "

7. Aug 28, 2009

### nirax

g is defined ... what r u identifying ? g is injective iff g(p) = 0 => p =0.

which is the same as your definition in this case (why ?)

> I'm not sure what you mean by "any polynomial cannot be zero in the whole of (0,1) "

it means that only zero polynomial (one which is identically zero) can take the value zero at every point between 0 and 1.

but like i said, you are stuck with a benumbed mind. come out of it. there is nothing so terrifying that you freeze.

8. Aug 28, 2009

### boneill3

g is injective iff g(p) = 0 => p =0.

Normally polynomials are not injective because f(x^2) = 9 could be x = 3 or -3

If it is injective we assume there must be an left inverse function f of g = idx

so take two elements of G x and y

g(x) => =0 = g(y) => 0

If g(x) = g(y), then x = g(f(x)) = g(f(y)) = y, so that x = y = idx

9. Aug 28, 2009

### nirax

who said polynomials are injective ?

g maps {set of all polynomials} to {set of all polynomials}. you r supposed to prove that g is injective not elements of {set of all polynomials}. and what is this supposed to mean (along with whatever you wrote) --

g(x) => =0 = g(y) => 0

10. Aug 28, 2009

### slider142

The zero polynomial is the polynomial that is the zero vector in the vector space R[t]. Use the vector space axioms to identify this element. Intuitively, it should be the polynomial p(t) = 0.

11. Aug 28, 2009

### boneill3

Hi guy's
Thanks for your help but I think I better go back to the books because I'm not sure what the question asks

the first part says:

$p:R\rightarrow R, x \rightarrow p(x)$

A function explicitly maps an variable x in R to p(x) in R

I'm confused about the function G

$g:R[t]\rightarrow R[t], p(t) \rightarrow \int_{0}^{t}p(x)dx$

how they tie together

12. Aug 28, 2009

### HallsofIvy

Staff Emeritus
What "G" are you talking about? If you mean "g" then then it is, just as it says, the function that maps each polynomial into its integral (with the stipulation that its value at 0 is 0).

in particular
$$g(a+ bt+ ct^2)= \int_0^t (a+ bx+ cx^2)dx= at+ \frac{1}{2}bt^2+ \frac{1}{3}ct^3$$.

13. Aug 28, 2009

### boneill3

So to identify the zero vector..

we prove $\lambda x = 0v$ if and only if $\lambda = 0$ or $x = 0v$

take $x \in R[t]$
Suppose that $\lambda x = 0v$ and $\lambda \neq 0$
than:

$x = 1x$

$= (\frac{1}{\lambda}\lambda)x$

$= \frac{1}{\lambda}(\lambda x)$

$= \frac{1}{\lambda}0v$

$= 0v\\$

take $x,y,z \in R[t]$

We need to prove $x+y = 0v$ if and only if $y = -x$

Suppose $x+y = x+z$

Than:

$-x+(x+y) = -x+(x+z)$

$(-x+x)+y = (-x+x)+z$

$0v+y = 0v+z$

$y = z$