# Homework Help: Linear Transformations

1. Jul 13, 2012

### greendays

write P for the vector space of all polynomials, a$_{0}$+a$_{1}$x+a$_{2}$x$^{2}$+...+a$_{n}$x$^{n}$, , a$_{0}$, a$_{1}$,...,a$_{n}$$\in$R, n=0,1,2...

1. Find a linear transformation P->P that is onto but not one-to one
2. Find such a linear transformation, that is one-to-one but not onto

I have been thinking about this question for a long time, but still could not come up with some thoughts, can someone give me some hints?

2. Jul 13, 2012

### greendays

So if it is onto, the linear combination should have infinitely many solutions?
If it is one-to-one, then the linear combination should have only one solution?

3. Jul 13, 2012

### vela

Staff Emeritus
No. First, it doesn't make sense to say that a linear combination has solutions. What I think you are referring to are the solutions to the equation $a_0+a_1 x+ a_2 x^2 + \cdots a_n x^n = 0$. But this has nothing to do with the problem at hand.

What you have is P, the set of all polynomials, and you're trying to find functions that map a polynomial f(x), which is an element of P, to another polynomial g(x), which is also an element of P. For example, say function A: P→P maps a polynomial f(x) to the polynomial f(x)+1. Then
\begin{align*}
A(1) &= 2 \\
A(10x-4) &= 10x-3 \\
A(x^2) &= x^2+1
\end{align*} for instance. You can show that A is one-to-one and onto, but it's not linear. You need to figure out some functions from P to P which will meet the specified criteria.

4. Jul 13, 2012

### greendays

Thanks very much for your input, vela.
But how did you get A(1)=2?

5. Jul 13, 2012

### greendays

Say function T:A-->B, if onto but not one-to-one, then every element from B from get mapped to from A. If one-to-one but not onto, then some or every element from B get mapped to from A? But how can I come up with the function then?

6. Jul 14, 2012

### vela

Staff Emeritus
1+1 = 2

Here is a simple example. Let f:ℤ→ℤ be the mapping that takes n to 2n. This function is one-to-one; however, it does not map onto ℤ because its image, f(ℤ), doesn't contain the odd integers.

7. Jul 16, 2012

### greendays

Hmmm, let f: P-->P be the mapping that takes x to x$^{2}$, this function is onto but not one-to-one? (Is this correct for this question?)

I still cannot think of a function that is one-to-one, but not onto

8. Jul 16, 2012

### vela

Staff Emeritus
I don't think you understood the point I made in my first reply to you. You need to get that down before you can go onto the rest of the problem.

9. Jul 16, 2012

### greendays

I thought I understood what u meant
For my question, say a function A: P→P maps a polynomial f(x) to the polynomial f(x). Then
A(x)=x$^{2}$
A(x+2)=(x+2)$^{2}$
So this function is onto, but not one-to-one?:uhh:

10. Jul 16, 2012

### vela

Staff Emeritus
Note that x and x2 are elements of P. So what you wrote above doesn't mean the same thing as what you wrote below.

How do you get that A is onto? For example, the function f(x)=x is in P, right? What polynomial when squared is equal to x? There isn't one, so A can't be surjective.

11. Jul 16, 2012

### greendays

Hi vela,

Can you give me hints on this question? I have been thinking about it over and over again, but still can not figure it out

12. Jul 16, 2012

### vela

Staff Emeritus
Let's step back for a second. Consider functions from ℝ2 to ℝ2. Can you give an example of a function that is 1-1 but not onto and one that is onto but not 1-1?

13. Jul 16, 2012

### greendays

f(x)=2x+1 is 1-1 but not onto?

14. Jul 16, 2012

### greendays

no, I don't think I got the idea, so confused

15. Jul 16, 2012

### vela

Staff Emeritus
What are the definitions of a one-to-one function and an onto function? It doesn't seem like you know those.

16. Jul 17, 2012

### greendays

Let T: V$\rightarrow$W be a linear transformation.

1. T is said to be onto if im T=W.

2. T is said to be one-to-one if T (v)=T(v$_{1}$) implies v=v$_{1}$.

I think I understand the definitions, it just seems so hard to apply them into questions

17. Jul 17, 2012

### vela

Staff Emeritus
OK, suppose f:ℝ→ℝ is given by f(x)=2x+1. This isn't a linear function, but that doesn't matter here. It turns out this function is both one-to-one and onto.

To show it's one-to-one, suppose f(x)=f(y). That means 2x+1=2y+1, which implies x=y.

To show it's onto, let $y \in \mathbb{R}$. You want to show there is an $x \in \mathbb{R}$ such that f(x)=y. That's easy enough to do. Just solve 2x+1=y for x.

Does this make sense?

18. Jul 17, 2012

### greendays

yes, but back to the original question, how do you approach that question? I mean I know the definitions, but how to apply them to that question?

19. Jul 17, 2012

### vela

Staff Emeritus

20. Jul 17, 2012

### greendays

I tried to answer it in post 13, as I mentioned, I know the definitions, but have no clue how to use them.

Last edited: Jul 17, 2012
21. Jul 17, 2012

### vela

Staff Emeritus
If you don't know how to use them, you don't really know the definitions. Post 17 is an example of how to use them. Just consider plain old functions from ℝ to ℝ. Give an example of each type of function and explain/prove why you think a function is 1-1 or onto.