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Proving a map is isomorphic

  1. May 11, 2009 #1
    1. The problem statement, all variables and given/known data
    For each of the following pairs of vectors, define an explicit isomorphism to establish that the spaces are isomorphic. Prove that your map is an isomorphism.
    a)P3 and R4
    b)P5 and M(2,3)


    2. Relevant equations
    None


    3. The attempt at a solution
    a)I know that P3 and R4 are isomorphic because if the entries in a given polynomial in P3 are written out in a single row (or column), the result is a vector in R4 because we have 4 coefficients in a polynomial of highest degree 3

    b)I also know that P5 and M(2,3) are isomorphic because if the entries in a given 2 by 3 matrix are written out in a single row (or column), the result is a vector in R6 , and, likewise, if the entries in a given polynomial in P5 are written out in a single row (or column), the result is a vector in R6 because we have 6 coefficients in a polynomial of highest degree 5

    However, I'm not sure exactly sure how the question wants me to write my answers. Could you help me refine my answer so that I am answering what the question wants me to?
     
  2. jcsd
  3. May 11, 2009 #2

    CompuChip

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    How about (for a):
    [tex]\phi(a x^3 + b x^2 + c x + d \in P_3) = [a, b, c, d] \in \mathbb{R}^4[/tex]
    You still need to show that it's an isomorphism though.

    For b you can write down an isomorphism between P5 and R6 and M(2, 3) and R6 and then use that inverses of isomorphisms and compositions of two isomorphisms are again isomorphisms (if you haven't seen that yet, it's easy to prove).
     
    Last edited: May 11, 2009
  4. May 11, 2009 #3
    \phi, what exactly is this notation? and how do I show that a) is an isomorphism?

    Could you elaborate some more on b). I'm not quite sure i understand what you mean
     
  5. May 12, 2009 #4

    CompuChip

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    Do you know what the definition of an isomorphism (between, say, A and B) is?
     
  6. May 12, 2009 #5
    If A and B are isomorphic,then they are structurally identical. I just know this definition from class. What I get from this is that A and B have the same number elements and they are a one to one mapping :S
     
  7. May 12, 2009 #6

    CompuChip

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    OK sounds like you seriously need a good textbook or something :)
    So I assume we are talking about an isomorphism of vector spaces (over R). In fact the notion of isomorphism makes sense in a lot of places in mathematics, and could also apply to groups or rings in this context, but I will assume the simplest case.

    Definition: Let f: A -> B be a map between vector spaces A and B. f is called an isomorphism if f is a bijective map such that f(x + y) = f(x) + f(y) and f(ax) = a f(x) for all a in R and x, y in A.

    Basically, that just says that not only they have the same number of elements, but that they also have the same structure (it doesn't matter if you add two vectors in A and then map them to B, or you first map them to B and add their images).

    Definition: Two vector spaces A and B are called isomorphic if there exists an isomorphism between them.

    Can you find that somewhere in your class notes or textbook?

    So I claim that if you define a map
    [tex]\phi: P_3 \to \mathbb{R}^4[/tex]
    by
    [tex]\phi(a x^3 + b x^2 + c x + d) = [a, b, c, d]^T[/tex]
    that is an isomorphism. For this you need to check that
    a) phi is bijective (i.e. injective and surjective)
    b) preserves structure (i.e. when p, q are polynomials of degree < 4, then phi(p + q) = phi(p) + phi(q))

    a) is easy: for every vectors [a, b, c, d] in R4 you can find a polynomial such that the image under phi is [a, b, c, d] (surjective) and if you have two polynomials which both map to [a, b, c, d] then clearly they must be the same polynomial. I'll leave b) for you to try and get used to the definitions.
     
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