Prove Isomorphism When Columns of C are Linearly Independent

[LosTacos]

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.

Homework Equations

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.

 

[mfb]

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.

 

[LosTacos]

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.

 

[mfb]

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.

 

[LosTacos]

so how do you go about showing it is surjective

 

[LosTacos]

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

 

[mfb]

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.