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Linear transformation, isomorphic

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data

    Let B be an invertible n x n matrix. Prove that the linear transformation L: Mn,n [tex]\rightarrow[/tex] Mn,n given by L(A) = AB, is an isomorphism.

    3. The attempt at a solution

    I know to show it is an isomorphism i need to show that L is both onto and one-to-one.

    By the theorem that says:

    Let T:V[tex]\rightarrow[/tex]W be a linear transformation with vector spaces V and W both of dimension n. Then T is one-to-one if and only if it is onto.

    To prove both conditions needed for an isomorphism i can just prove it is one-to-one as in this case, 'V' and 'W' are the same dimension and so proving L is one-to-one also proves it is onto.

    To prove it is one-to-one, i need to determine the kernel of L and show that it is {0}. To do this i need to use the fact that B is an invertible n x n matrix and L(A)=AB.

    I need some guidance on how to use these features to show the kernel of L os {0}???
    Thanks in advance
     
  2. jcsd
  3. Apr 27, 2008 #2
    well suppose L(A) = L(C), so AB = CB, then ...some stuff... implies A = C, so L is 1-1, you can fill in the missing step

    your way, suppose L(A) = 0, so AB = 0, so ....some stuff... implies A = 0, so kerL = {0} so it's 1-1, again you can fill in the missing step


    note the missing step is the same in both approaches
     
    Last edited: Apr 27, 2008
  4. Apr 27, 2008 #3
    Does it have something to do with the fact that the rows and columns of the invertible matrices are linearly independant and so the kernel must be 0 because of this? Or do i have to use notation to do with the linear transformation conditions? Any more help would be greatly appreciated. Thanks
     
  5. Apr 27, 2008 #4

    Hurkyl

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    You're overthinking it. In this case, I think it would be easier to directly prove it's an isomorphism, rather than use that indirect method.
     
    Last edited: Apr 27, 2008
  6. Apr 27, 2008 #5

    HallsofIvy

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    If Ax= 0 and A is invertible, then x= ?
     
  7. Apr 27, 2008 #6
    I assume that x =0, but what is the significance of A being invertible?
     
  8. Apr 27, 2008 #7
    you don't assume that x=0. remember, you need to show that IF Ax = 0 THEN x=0. the significance of A being invertible is that, in general, a matrix times a non-zero vector could be the zero vector. for example, if B = [itex] \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)[/itex] and [itex]x = \left(\begin{array}{c} 0 \\ 1 \end{array}\right)[/itex] then Bx = 0. notice, of course, that B is not invertible though.
     
  9. Apr 28, 2008 #8

    HallsofIvy

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    If Ax= 0, and A is invertible, how would you solve the equation? If you don't know what A being invertible has to do with the solution to this equation, then you need to go back and review the basics of linear transformations.
     
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