Isomorphism proof

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Homework Statement



Let L:R->R be a linear operator with matrix C. Prove if the columns of C are linearly independent, then L is an isomorphism.


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The Attempt at a Solution



Assume the columns of C are linearly independent. Then, the homogenous equation Cx=0 is the trivial solution. Need to show L is both 1-1 and onto. Assume that L(c1)=L(c2). Need to show that c1=c2. Well, since the matrix is linearly independent and Cx=0 is trivial solution, then each column represents its own solution. Therefore, if L(c1)=L(c2), then it must be that c1=c2. Now need to show L is onto. So for every d, there exists a c such that L(c)=d. Since C is linearly independent, no vector can be expressed as a linear combination of others. So, for each d, there will be a c where L(c)=d. Therefore, L is an isomorphism.
 

Answers and Replies

  • #2
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I think this should be L:Rn->Rn?
Then, the homogenous equation Cx=0 is the trivial solution.
An equation is not the same as a solution. The equation has the trivial solution only.

Well, since the matrix is linearly independent and Cx=0 is trivial solution, then each column represents its own solution.
How do you mean that?
Hint: Use linearity.

Now need to show L is onto.
"is onto"?

Since C is linearly independent, no vector can be expressed as a linear combination of others.
No vector of your matrix, right.
So, for each d, there will be a c where L(c)=d.
That does not follow from the previous statement.
 
  • #3
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Okay so the equation has the trivial solution. So that means that the only solution is the trivial one, which is represented by the columns themselves, therefore 1-1. And since C is linearly independent, no vector in the matrix can be expressed as a linear combination of otheres. Therefore, each output has an input where L maps to that.
 
  • #4
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So that means that the only solution is the trivial one, which is represented by the columns themselves, therefore 1-1.
The columns have no relation to the trivial solution. The trivial solution is x=0.


Consider the linear operator F:R->R2, F(x)=(x,0). It satisfies everything you list in your posts, but it is not surjective: there is no x such that F(x)=(1,1). As long as you do not use that L maps Rn on itself, the argument cannot work.
 
  • #5
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so how do you go about showing it is surjective
 
  • #6
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Because the columns of A are linearly independent, each column will have a unique solutions such that there will exist an x such that L(x)=b which implies Ax=B
 
  • #7
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Injective linear operators from R^n to R^n are always surjective. This follows from a more general result about the dimensions of the image and the kernel.
 

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