# Isomorphism proof

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.

## 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.

Mentor
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.

Mentor
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

Mentor
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.