Isomorphism Transformation: Onto and One-to-One Concept Explanation

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Homework Statement



Is the following transformation an isomorphism:

a_0+bx+cx^{2}+dx^{3} \rightarrow \begin{bmatrix} a & b\\ c & d \end{bmatrix}

Homework Equations



A transformation is an isomorphism if:

1. The transformation is one-to-one
2. The transformation is onto

The Attempt at a Solution



I took an accelerated Linear Algebra course over the summer, and in the last lecture my professor barely touched the concepts of "onto" and "one-to-one"

I know a transformation is onto if T:V→W it maps to every vector in W (essentially no restrictions)
and that a transformation is one to one if for each vector in V maps to only one vector in W.

I have working through a bunch of problems by finding the transformation matrix and then taking its determinant to see if its one to one (i.e if det(T) ≠ 0) or finding the kernel (if Ker(T)=0)
I was testing if the transformation was onto by finding RREF and seeing if any restrictions popped up (if not then it was onto, I think this was right)

This problem seems extremely simple but I'm not sure where to start since I don't know what T looks like and haven't had much guidance in this type of problem yet (none actually)

Any help is appreciated, thanks!
 
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FeynmanIsCool said:

Homework Statement



Is the following transformation an isomorphism:

a_0+bx+cx^{2}+dx^{3} \rightarrow \begin{bmatrix} a & b\\ c & d \end{bmatrix}
You need to define this transformation more carefully. What exactly is its domain?

FeynmanIsCool said:

Homework Equations



A transformation is an isomorphism if:

1. The transformation is one-to-one
2. The transformation is onto
A linear transformation is an isomorphism (to be more specific, a vector space isomorphism) if 1 and 2 hold. So if V and W are vector spaces, and you suspect that a function ##T:V\to W## may be an isomorphism, you should usually start by verifying that it's linear.

FeynmanIsCool said:
I took an accelerated Linear Algebra course over the summer, and in the last lecture my professor barely touched the concepts of "onto" and "one-to-one"
Your descriptions of these terms are accurate, but I will still show you the definitions I use.

The terms "injective" and "surjective" are more popular. A function ##f:A\to B## is injective if the implication ##f(x)=f(y)\Rightarrow x=y## holds for all ##x,y\in A##, and surjective if ##f(A)=B##, i.e. if the range is equal to the codomain (rather than a proper subset of it). So a proof of injectivity should start with "Let x and y be arbitrary elements of A such that f(x)=f(y)". Then you prove that x=y. A proof of surjectivity should usually start with "let y be an arbitrary element of B". Then you find an x in A such that f(x)=y.
 
FeynmanIsCool said:

Homework Statement



Is the following transformation an isomorphism:

a_0+bx+cx^{2}+dx^{3} \rightarrow \begin{bmatrix} a & b\\ c & d \end{bmatrix}

Homework Equations



A transformation is an isomorphism if:

1. The transformation is one-to-one
2. The transformation is onto

The Attempt at a Solution



I took an accelerated Linear Algebra course over the summer, and in the last lecture my professor barely touched the concepts of "onto" and "one-to-one"

I know a transformation is onto if T:V→W it maps to every vector in W (essentially no restrictions)
and that a transformation is one to one if for each vector in V maps to only one vector in W.

I have working through a bunch of problems by finding the transformation matrix and then taking its determinant to see if its one to one (i.e if det(T) ≠ 0) or finding the kernel (if Ker(T)=0)
I was testing if the transformation was onto by finding RREF and seeing if any restrictions popped up (if not then it was onto, I think this was right)

This problem seems extremely simple but I'm not sure where to start since I don't know what T looks like and haven't had much guidance in this type of problem yet (none actually)

Any help is appreciated, thanks!

Let's call ##P_3## the linear space of 3rd degree polynomials and ##M## the 2 by 2 matrices. You don't need any representation other than what you are given to see if the given transformation, call it ##T##, is 1-1 and onto.

If ##T(p_1) = T(p_2)## can you show ##p_1 = p_2## using the usual properties of polynomials and matrices? Similarly, if ##A\in M## can you find ##p\in P_3## such that ##T(p)= A##?
 
Fredrik said:
You need to define this transformation more carefully. What exactly is its domain?

T:P3→M2,2


Fredrik said:
linear transformation is an isomorphism (to be more specific, a vector space isomorphism) if 1 and 2 hold. So if V and W are vector spaces, and you suspect that a function ##T:V\to W## may be an isomorphism, you should usually start by verifying that it's linear.

Ok,
I do know that its linear if: k(Tu)=T(ku) and T(u+v)=T(u)+T(v), that's simple enough.



Fredrik said:
Your descriptions of these terms are accurate, but I will still show you the definitions I use.

The terms "injective" and "surjective" are more popular. A function ##f:A\to B## is injective if the implication ##f(x)=f(y)\Rightarrow x=y## holds for all ##x,y\in A##, and surjective if ##f(A)=B##, i.e. if the range is equal to the codomain (rather than a proper subset of it). So a proof of injectivity should start with "Let x and y be arbitrary elements of A such that f(x)=f(y)". Then you prove that x=y. A proof of surjectivity should usually start with "let y be an arbitrary element of B". Then you find an x in A such that f(x)=y.

Nice,
I like these definitions better than the ones I was working with.
 
FeynmanIsCool said:
T:P3→M2,2
Right, so you should show that for all ##u,v\in P_3## and all ##a,b\in\mathbb R##, we have ##T(au+bv)=aTu+bTv##. Or you can break it up in two separate statements, as you did here:

FeynmanIsCool said:
Ok,
I do know that its linear if: k(Tu)=T(ku) and T(u+v)=T(u)+T(v), that's simple enough.
 
LCKurtz said:
If ##T(p_1) = T(p_2)## can you show ##p_1 = p_2## using the usual properties of polynomials and matrices? Similarly, if ##A\in M## can you find ##p\in P_3## such that ##T(p)= A##?

I have the tools to but I am not really sure where to start.
" If ##T(p_1) = T(p_2)## can you show ##p_1 = p_2## using the usual properties of polynomials and matrices?" I know this is the test for 1-1

and

" Similarly, if ##A\in M## can you find ##p\in P_3## such that ##T(p)= A##?" is the test for onto, but I am not sure where to start. What properties of polynomials/matrices to I use?
 
Fredrik said:
Right, so you should show that for all ##u,v\in P_3## and all ##a,b\in\mathbb R##, we have ##T(au+bv)=aTu+bTv##. Or you can break it up in two separate statements, as you did here:
Right, and it turns out T is linear. Since the two statements hold. I'm just confused on how to show 1-1 and onto. Before in other problems, I could find T (or was given T). Now, since I don't know T, I am a little confused on how to test its "onto-ness" or "one to one-ness".
 
This is what I mean:
A simple problem that popped up with the same question was:

c_{0}+c_{1}x\rightarrow (c_{0}-c_{1}, c_{1})<br />

with T:P1→R2

This case is simple since I can find T, its just \begin{bmatrix} 1 &-1 \\ 0& 1 \end{bmatrix}
I could easily find its det (test for 1-1) and RREF restrictions (onto test)

On this problem, it just seems more vague...
 
FeynmanIsCool said:
I have the tools to but I am not really sure where to start.
" If ##T(p_1) = T(p_2)## can you show ##p_1 = p_2## using the usual properties of polynomials and matrices?" I know this is the test for 1-1

and

" Similarly, if ##A\in M## can you find ##p\in P_3## such that ##T(p)= A##?" is the test for onto, but I am not sure where to start. What properties of polynomials/matrices to I use?

You are making this much harder than it is. Say you are given$$
A = \begin{bmatrix}a & b \\ c & d\end{bmatrix} \in M$$Can you write down a polynomial in ##P_3## that maps to it? That would show ##T## is onto.
 
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  • #10
LCKurtz said:
You are making this much harder than it is. Say you are given$$
A = \begin{bmatrix}a & b \\ c & d\end{bmatrix} \in M$$Can you write down a polynomial in ##P_3## that maps to it? That would show ##T## is onto.

sure, wouldn't It just be: a_{0}+bx+cx^{2}+dx^{3}?

edit: ahh...so that it? If the polynomial given was say: a_{0}+bx+cx^{2}+d(x+1)^{3} then it wouldnt?
 
  • #11
Yes. Simple, isn't it. Now show ##T## is 1-1.
 
  • #12
I got it from here, Thanks LCKurtz and Fredrik.Im making it way harder than it it. Ohh well, now I know!

Thanks!
 

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