Isomorphism of P4 and R5 in a given inner product space

In summary, the inner product between two vectors in ℝ5 is determined by their values at the given values of tk, and each polynomial can be represented as a vector in P4. If the inner products are equal, then the vectors are in the same coordinate system.
  • #1
freddyfish
57
0
The isomorphism of ℝ5 and P4 is obvious for the "standard" inner product space.

The following question arise from an example in my course literature for a course in linear algebra. The example itself is not very difficult, but there is a statement without any proof, that if the inner product is determined by:

<p(t), q(t)>=Ʃp(ti)q(ti)

where tk=-2, -1, 0, 1, 2, respectively, (1≤k≤t)

then the polynomials in P4 are uniquely determined by their values at the given values of tk and each polynomial can be represented as a vector in ℝ5 where the entries of that the vector (from the top) are the polynomials value for each tk (same order as above).

Intuition confirmes this, but as far as I'm concerned intuition won't prove neither the uniqueness nor why the corresponding vectors in ℝ5 is formed from each polynomial's value for each given tk. Can anyone help me to prove this or at least present some idea that might be useful?

I would appreciate it //Freddy
 
Physics news on Phys.org
  • #2
Hey freddyfish and welcome to the forums.

Can you construct a way to show a co-ordinate transformation from P^5 to R^5?

If you can construct your metric tensor you can show that if the two inner products are equal, then you're done.
 
  • #3
Thanks! I will take a closer look at it, but also, I will soon get my share of tensors, since the autumn offers a course in tensor analysis for me. I'm not "supposed" to know about co-ordinate transformations just yet, but that's what summer vacations are for. To study interesting spaces and messed up co-ordinates, right? :p
 
  • #4
It's basically taking a lot of stuff you did with going from 2D polar to 2D cartesian and generalizing it in the context of general co-ordinate systems.

Some systems can be really messed up but that depends on your point of view: A lot of the initial examples will be pretty standard like cylindrical, polar, cartesian, parabolic and going between all of these.

Then you look at how you define metrics and inner products and that's the start of generalized co-ordinate system geometry where differential geometry enters the picture.
 
  • #5


Yes, the isomorphism between P4 and ℝ5 in this specific inner product space is indeed obvious. This is because the inner product in this space is defined by a sum of products of the polynomial values at specific points, which is essentially the same as the dot product in ℝ5. This means that the polynomials in P4 can be represented as vectors in ℝ5, with the same order of coefficients as the polynomial values at the given points. This representation is unique, as the values at the given points uniquely determine the polynomial. Therefore, an isomorphism exists between P4 and ℝ5 in this inner product space.

To prove this, we can consider the general form of a polynomial in P4, which is p(t) = a0 + a1t + a2t^2 + a3t^3 + a4t^4. We can then define a vector in ℝ5 as v = (a0, a1, a2, a3, a4) and calculate the inner product <p(t), q(t)> as v•w, where w is the vector representation of q(t). This shows that the inner product in this space is equivalent to the dot product in ℝ5.

Furthermore, we can show that this representation is unique by considering the fact that the values of a polynomial at specific points uniquely determine the polynomial. Therefore, if two polynomials have the same values at the given points, they must be the same polynomial. This means that the vector representation of each polynomial in ℝ5 is also unique, as it is based on the same values at the given points.

In conclusion, the isomorphism between P4 and ℝ5 in this inner product space is valid and can be proven by showing the equivalence of the inner product and the dot product, as well as the uniqueness of the vector representation of polynomials in ℝ5 based on their values at the given points. I hope this helps clarify the concept for you.
 

What is isomorphism in a given inner product space?

Isomorphism is a mathematical concept that refers to a one-to-one mapping between two mathematical structures. In the context of inner product spaces, isomorphism means that there exists a linear transformation between two spaces that preserves the inner product structure.

What is P4 and R5 in a given inner product space?

P4 and R5 refer to two different vector spaces. P4 is a space of polynomials of degree 4 or less, while R5 is a space of vectors with 5 components. In the context of isomorphism, P4 and R5 can be considered as two different inner product spaces that may or may not be isomorphic to each other.

How can we determine if P4 and R5 are isomorphic in a given inner product space?

To determine if two inner product spaces, such as P4 and R5, are isomorphic, we need to check if there exists a linear transformation between the two spaces that preserves the inner product structure. This means that the inner products of any two vectors in P4 should be equal to the inner products of their corresponding transformed vectors in R5.

What are the benefits of isomorphism in the context of inner product spaces?

Isomorphism in inner product spaces allows us to establish a one-to-one correspondence between two spaces, making it easier to study and understand the properties and structures of these spaces. It also allows us to apply theorems and techniques from one space to the other, which can be useful in solving problems or proving mathematical statements.

Can P4 and R5 be isomorphic in a given inner product space?

Yes, P4 and R5 can be isomorphic in a given inner product space. However, it is not always the case. For two spaces to be isomorphic, they must have the same dimension and their inner product structures must be preserved by a linear transformation. Thus, it is possible for P4 and R5 to be isomorphic in some cases, but not in others.

Similar threads

  • Linear and Abstract Algebra
Replies
8
Views
912
  • Calculus
Replies
4
Views
341
  • Linear and Abstract Algebra
Replies
8
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
2K
  • Linear and Abstract Algebra
Replies
6
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
2K
  • Linear and Abstract Algebra
Replies
13
Views
855
  • Linear and Abstract Algebra
Replies
2
Views
2K
  • Quantum Physics
Replies
8
Views
2K
  • Math POTW for University Students
Replies
2
Views
817
Back
Top