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Theorem from Hoffman and Kunze

  1. Sep 25, 2009 #1
    Theorem:
    Let V and W be n-dimensional vector spaces over the field F of complex/real numbers. Then the space of linear transformations L(V,W) is finite dimensional and has dimension mn.

    Proof:
    Let B = {[tex]\alpha 1, \alpha 2 ... , \alpha n[/tex]} and B' = {[tex]\beta 1, \beta 2,... \beta m[/tex]} be ordered bases for V and W respectively. For each pair of integers (p,q) with 1[tex]\leq[/tex] p [tex]\leq[/tex] m and 1 [tex]\leq[/tex] q [tex]\leq[/tex] n, we define a linear transformation E(p,q) from V into W by

    E(p,q)([tex]\apha i[/tex]) = 0, if i[tex]\neq[/tex] q
    =[tex]\beta p[/tex], if i = q


    =[tex]\delta[/tex](i,q)[tex]\beta[/tex]p.

    According to theorem, there is a unique linear transformation from V into W satisfying these conditions. The claim is that the mn transformations E(p,q) form a basis for L(V,W).

    Let T be a linear transformation from V into W. For each j, 1 [tex]\leq[/tex] j [tex]\leq[/tex] n, let A(i,j),....,A(m,j) be the coordinates of the vector T[tex]\alpha i[/tex] in the ordered basis B', i.e.,

    T[tex]\alpha j [/tex] = [tex]\sum^{m}_{p=1}[/tex]A(p,j) [tex]\beta p[/tex].

    We wish to show that
    T = [tex]\sum^{m}_{p=1} \sum^{n}_{q=1}[/tex] A(p,q) E(p,q) .... (1)

    Let U be the linear transformation in the right hand member of (1). Then for each j

    U[tex]\alpha[/tex]j = [tex]\sum_{p} \sum_{q} A(p,q) E(p,q)(\alpha[/tex]j)

    = [tex]\sum_{p} \sum_{q} A(p,q) \delta[/tex](j,q)([tex]\beta[/tex]p)

    = [tex]\sum^{m}_{p=1}[/tex]A(p,j) [tex]\beta p[/tex]

    = T[tex]\alpha[/tex]j

    and consequently U = T. Now (1) shows that the E(p,q) span L(V,W); we must prove that they are independent [ THIS IS THE PART THAT I DON'T UNDERSTAND. I COULD FOLLOW UP TO HERE]. But this is clear from what we did above; for, if the linear transformation

    U = [tex]\sum_{p} \sum_{q}[/tex] A(p,q) E(p,q)

    is the zero transformation, then U[tex]\alpha[/tex]j = 0 for each j, so

    [tex]\sum^{m}_{p=1}[/tex]A(p,j) [tex]\beta p[/tex] = 0

    and the independence of the [tex]\beta[/tex]p implies that A(p,j) = 0 for every p and j.

    ------END OF PROOF IN TEXT

    Now let me explain a little more clearly what I don't understand with a rather simple example.

    Let S be the set of ordered pairs (a,1) with 1[tex]\leq[/tex] a [tex]\leq[/tex] n, a is an integer, and F be the set of real numbers.

    Now let me define a function f(i,j), f: S [tex]\rightarrow[/tex] F, such that

    f (i,j) [(a,1)] = [tex]\delta(j,a)[/tex]

    This could be represented as a space of nx1 column matrices with 1s in the jth position.

    What I am trying to point out is that f(1,1) maps to the matrix [1 0 0 0..... 0], but so does f(1,2). If both of them map to the same fellow, how the heck are the two linearly independent?
     
    Last edited: Sep 26, 2009
  2. jcsd
  3. Sep 26, 2009 #2
    Okay, wait. I see a flaw in my argument. It doesn't matter if two f(i,j) map to the same vector in F. That ought to be a good thing actually since that just says we can even construct linear transformations which aren't 1:1. So that's that. But I still don't understand how the mn linear transformations are linearly independent. I fully understand how the span the space of linear transformations. I just can't connect the dots.
     
  4. Sep 26, 2009 #3
    DONE. You can delete this thread now. I was able to prove it right after posting it here as usual.
     
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