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Isomorphisms have the same dimensions

  1. Jan 30, 2013 #1
    Advanced Calculus of Several Variables, 5.6:

    Two vector spaces V and W are called isomorphic if and only if there exist linear mappings [itex] S : V \to W[/itex] and [itex]T : W \to V[/itex] such that [itex]S \circ T[/itex] and [itex] T \circ S [/itex] are the identity mappings of [itex]S[/itex] and [itex]W[/itex] respectively. Prove that two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

    I'm having a hard time proving the problem formally. I'm also having a hard time proving sufficiency of the criterion. In terms of necessity, I have this much:

    It's clear that if one space, say V, has more dimensions than the other, then V which all map to the same value of W. For instance, [itex] v \in V : <v,w> = 0 [/itex] for all [itex]w \in W [/itex] all map to 0 in W (by the positivity property of inner products). There is then no function W -> V that can reclaim the lost information.

    Formally, for all [itex] S : V -> W, S(v) = 0 [/itex] when v is orthogonal to every vector in W.

    Consider two vectors [itex]v_{1}[/itex] and [itex]v_{2}[/itex] which are both orthogonal to every vector in W.

    Then, for example, [itex]S(v_{1}) = S(v_{2}) = 0 \in W[/itex], and we have [itex] (T \circ S) (v_{1}) = (T(S(v_{1})) = T(S(v_{2})) = T(0). [/itex]

    In order for the isomorphism to exist, [itex] (T \circ S) (v_{1}) = v_{1} [/itex] and [itex] (T \circ S) (v_{2}) = v_{2} [/itex]. But this is impossible since [itex](T \circ S) (v_{2}) = (T \circ S) (v_{1})[/itex], and [itex]v_{1} ≠ v_{2}[/itex]

    Though I am satisfied by this, I have the feeling that I'm supposed to show this using matrix multiplication. The formalism that is eluding me is hiding in the phrase "The identity mappings of S and W respectively". What does that mean? [itex]S \circ T = I_{w}[/itex]? What does that mean?
     
    Last edited: Jan 30, 2013
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  3. Jan 30, 2013 #2

    HallsofIvy

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    Are you allowed to assume a given inner product on the spaces? That does not appear in the statement.

    I would think that the simplest way to do this would be to show that if [itex]\{v_1, v_2, ..., v_n\}[/itex] is a basis for V then [itex]\{S(v_1), S(v_2), ..., S(v_n)\}[/itex] is a basis for W.
     
  4. Jan 30, 2013 #3
    Earlier in the chapater, a theorem is given that:

    "Let [itex] L : V \to W [/itex] be linear, with V being n-dimensional. If Ker L = 0, then L is one-to-one, and Im L is an n-dimensional subspace of W."

    I had a hard time with the proof of this theorem, and the exact place where I had the difficulty is relevant to this proof.
    The author writes that

    "To show that the subspace Im L is n-dimensional, start with a basis [itex]v_{1}, ... v_{n}[/itex] for V. Since it is clear, (by the linearly of L) that the vectors [itex]L(v_{1}), ..., L(v_{n}) [/itex] generate a basis for Im L, it suffices to prove that they are linearly independent..."

    To me, it isn't clear at all how the linearity of L makes [itex]L(v_{1}), ..., L(v_{n}) [/itex] a basis for Im L. Well, intuitively, I'm not surprised. It seems pretty clear that an orthonormal basis becomes a basis for any linear transformation, but I can't provide a formalism for that. And moreover, I can't see how *any* basis does so, especially not formally. One of my goals in this reading is to get better at providing formal arguments for all of my proofs. I actually spent a few minutes trying to convince myself of the fact the first time I read this "proof" but never was able to do so. And now I've just spent another hour trying again. Could you explain it to me?
     
  5. Jan 30, 2013 #4

    Dick

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    Any element of Im L can be written as L(v) for some v in V. Since v1,v2,...,vn is a basis of V, v can be written as a1*v1+a2*v2+...+an*vn. If you work out L(a1*v1+a2*v2+...+an*vn) you should be able to show L(v1),L(v2),...,L(vn) must span Im L. Next, to finish showing it's a basis you need to show L(v1),L(v2),...,L(vn) are also linearly independent. Use the definition of linearly independent. It will be easiest to show that if you assume they are linearly dependent then that would contradict Ker L={0}. Try it!
     
  6. Jan 31, 2013 #5
    Thanks Dick, your remark that "Any element of Im L can be written as L(v) for some v in V" was what I was missing to prove that [itex]L(v_{1}), ... , L(v_{n})[/itex] spans W if [itex]v_{1}, ... , v_{n} [/itex] spans V. Because any vector in v can be written as [itex]a_{1}v_{1} + ... a_{n}v_{n}[/itex], then any vector in W can be written as [itex]L(a_{1}v_{1} + ... + a_{n}v_{n})[/itex]. By linearity of L, then, any vector in W can be written as [itex]a_{1}L(v_{1}) + ... a_{n}L(v_{n})[/itex], proving that [itex]L(v_{1}), ... , L(v_{n})[/itex] spans W.

    Now if we have that [itex]v_{1}, ... , v_{n} [/itex] are a basis and Ker L = 0, it's easy to prove that [itex]L(v_{1}), ... , L(v_{n})[/itex] are also linearly independent and thus constitute a basis: but do I even have to do this for 5.6? I've shown that dim V = dim W is necessary in my first post and sufficient in this post. Why is it being a basis necessary?

    PS: Ivy, you are correct in pointing out that an inner product is not defined in this problem statement, but we have that dim Ker V = dim V - dim Im V, so we know without an inner product that there will be infinitely many zeroes of V if dim V - dim Im V > 0, so the proof is easily mended.
     
    Last edited: Jan 31, 2013
  7. Jan 31, 2013 #6

    Dick

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    To show 5.6 using this theorem you don't have to repeat the whole thing. If you can show TS=ST=identity means S and T are one-to-one then it's easy from there.
     
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