Linear transformation, isomorphic

  • Thread starter karnten07
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  • #1
karnten07
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Homework Statement



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.

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
 

Answers and Replies

  • #2
ircdan
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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:
  • #3
karnten07
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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

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
 
  • #4
Hurkyl
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I know to show it is an isomorphism i need to show that L is both onto and one-to-one.
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.
 
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  • #5
HallsofIvy
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If Ax= 0 and A is invertible, then x= ?
 
  • #6
karnten07
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If Ax= 0 and A is invertible, then x= ?

I assume that x =0, but what is the significance of A being invertible?
 
  • #7
eok20
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I assume that x =0, but what is the significance of A being invertible?

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.
 
  • #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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