Proving Isomorphism between Vector Space and Polynomial Space

In summary, the transformation T is defined as mapping sequences in V (vector space consisting of finite sequences in F) to polynomials in W (polynomials with coefficients from field F) by summing the sequence's terms multiplied by x raised to the corresponding indices. It is proven to be an isomorphism by defining another transformation T2 that maps polynomials back to sequences and showing that T and T2 are invertible. This shows that T is onto and one-to-one, making it an isomorphism.
  • #1
eckiller
44
0
Let V denote the vector space that consists of all sequences {a_n} in F (field) that have only a finite number of nonzero terms a_n. Let W = P(F) (all polynomials with coefficients from field F). Define,

T: V --> W by T(s) = sum(s(i)*x^i, 0, n)

where n is the largest integer s.t. s(n) != 0. Prive that T is an isomorphism.


I see how the transformation is mapping sequences to polynomials, but I don't even see how this is onto. Based on the sequence description, there comes a time where the remaining terms of every sequence is 0:

s_n = (s1, s2, ..., sn, 0, 0, ...).

So I don't see how that will "hit" every polynomial since the polynomials given in the problem don't have the "zero after finite many terms" restriction.
 
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  • #2
gosh this is so obvious, what to say?

you are just blocked from seeing it by some late night demon.

Recall, what is the definition of a polynomial?


In particular, what is the degree of a polynomial?

or just try to produce ontoness directly by actually writing down say,

the finite sequence that maps to the polynomial 5x^7 - x^6 +459x - 2.

then ask yourself what prevents you from doing this for any polynomial.
 
Last edited:
  • #3
Is this right?

For any p in P(F), p = c0 + c1 x + ... + ck x^k

Then define T2 : W --> V by T2( c0 + c1 x + ... + ck x^k ) = (c0, c1, ...,
ck, 0, 0, ...).

Then,

T2( T( (s0, s1, ..., sn, 0, 0, ...) ) )
= T2( s0 + s1 x + ... + sn x^n )
= (s0, s1, ..., sn, 0, 0, ...)

and,

T( T2( s0 + s1 x + ... + sn x^n ) )
= T( (s0, s1, ..., sn, 0, 0, ...) )
= s0 + s1 x + ... + sn x^n

which implies T is invertible, and hence an isomorphism?
 

1. What is an isomorphism proof?

An isomorphism proof is a mathematical technique used to show that two mathematical structures are essentially the same. It involves finding a one-to-one correspondence between the elements of the two structures and demonstrating that this correspondence preserves the structure and operations of the two structures.

2. How do I know when to use an isomorphism proof?

Isomorphism proofs are typically used when comparing two structures that have similar properties or structures but are defined differently. For example, if you are trying to show that two groups are essentially the same, you may use an isomorphism proof to demonstrate this.

3. What are the steps for conducting an isomorphism proof?

The steps for conducting an isomorphism proof may vary depending on the specific structures being compared, but generally involve establishing a one-to-one correspondence between the elements of the two structures, showing that the correspondence preserves the structure and operations, and providing a clear explanation of the proof.

4. What are some common mistakes to avoid when conducting an isomorphism proof?

Some common mistakes to avoid when conducting an isomorphism proof include assuming the isomorphism instead of proving it, not clearly explaining the correspondence between the elements, and not verifying that the operations are preserved. It is also important to be careful with notation and to clearly define any new symbols used in the proof.

5. Are there any tips or tricks for making isomorphism proofs easier?

Isomorphism proofs can be challenging, but there are some tips that can make them easier. These include starting with a clear understanding of the structures being compared, breaking the proof into smaller steps, and checking your work as you go. It can also be helpful to look at examples of isomorphism proofs for similar structures to get a better understanding of the process.

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