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A vector space W over the real numbers is the set of all 2 x 2

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data
    A vector space W over the real numbers is the set of all 2 x 2 Hermitian matrices. Show that the map T defined as:

    T(x,y,z,t) =
    [t+x y+iz]
    [y-iz t-x]

    from R4 to W is an isomorphism.

    2. Relevant equations



    3. The attempt at a solution
    I know that for the map to be isomorphic it has to have the following properties:
    - one to one (injective) and onto (surjective)
    - ker(T) = {0} and range(T) = W
    - The inverse map T^-1 has to exist
    - dimension of both R4 and W has to be the same.

    I don't really know how to apply these properties in showing that the map is isomorphic. Any help would be appreciated. Thanks in advanced!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 22, 2012 #2

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Isomorphism

    An isomorphism is by definition a linear bijection, and a bijection is by definition a map that's both injective and surjective. So there are three things you must verify:

    1. T is linear.
    T(ax+by)=...(use what you know about T here)...aTx+bTy.​
    2. T is injective.
    Show that Tx=Ty implies x=y.​
    3. T is surjective.
    Let z be an arbitrary 2×2 hermitian matrix. Prove that there's an x in ℝ4 such that Tx=z.​
     
  4. Feb 25, 2012 #3

    Deveno

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    Re: Isomorphism

    a hint, sort of:

    can you find linearly independent (over R, not C) 2x2 matrices in Mat2x2(C), A,B,C,D such that:

    T(x,y,z,t) = xA + yB + zC + tD?

    linearity is easy to prove, if so, and so is bijectivity.

    (why? because if {A,B,C,D} is LI, it's a basis of something, right?)

    another way to prove injectivity:

    show ker(T) = {(0,0,0,0)}.

    a pre-image of w in W should not be hard to find. figure out what y and z have to be, first, and then tackle x and t. you should get a system of two equations in x and t, easy to solve.
     
  5. Feb 25, 2012 #4

    HallsofIvy

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    Re: Isomorphism

    1) Your definition of "isomorphism" is incorrect. What you have defined is simply a one-to-one map from one set onto another. You have completely ignored the algebraic structure.

    2) And that means your sets must have an algebraic structure. Since you have defined W as a set of matrices, you already have a well defined addition and scalar product. But you are assuming (a, b, c, d)+ (u, v, w, x)= (a+u,b+ v, c+ w, d+ x), and x(a, b, c, d)= (ax, bx, cx, dx) and that really should have been stated.

    3) As Deveno said, in addition to showing that the function one to one and onto, you must show that it "preserves" the operations: T(u+ v)= T(u)+ T(v) and T(av)= aT(v).
     
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