Linear Transformations for Polynomials: Onto vs. One-to-One

[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?

Thanks in advance!

 

[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?

 

[vela]

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.

 

[greendays]

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.

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

 

[greendays]

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.

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?

 

[vela]

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

1+1 = 2

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?

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.

 

[greendays]

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.

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

 

[vela]

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.

 

[greendays]

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.

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?

 

[vela]

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?)

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.

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?

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.

 

[greendays]

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.

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

 

[vela]

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?

 

[greendays]

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

 

[greendays]

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?

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

 

[vela]

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

 

[greendays]

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

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

1. T is said to be onto if I am 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

 

[vela]

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?

 

[greendays]

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?

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?

 

[vela]

Can you answer the question I asked in post 12?

 

[greendays]

Can you answer the question I asked in post 12?

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

 

[vela]

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.