# It the following matrix isomorphic?

1. Jan 21, 2009

### soopo

Is the following matrix isomorphic?

1. The problem statement, all variables and given/known data
a) Is the following matrix isomorphic?
$$L: \Re^{2} \rightarrow \Re^{2}$$

$$L(x) = (-x_{2}, 0), where x = (x_{1}, x_{2}) \in \Re^{2}.$$

b)
Define Ker(L) and Im(L).

3. The attempt at a solution
a) I need to show that the L is bijection. I know that it is injection,
because it can be presented as Ax = b. The mapping is also linear, so L is
surjection. Thus, L is bijection and isomorphic.

b)
$$Ker(L) = \Re^{2}$$
$$Im(L) = \Re^{2}$$

I am not sure is this enough.

Last edited: Jan 22, 2009
2. Jan 21, 2009

### phreak

This is incorrect. There are many examples of linear, non-surjective functions. For instance, the projective function. What is the definition of a surjective function? Moreover, what is the definition of an injective function?

3. Jan 21, 2009

### soopo

Injection: from one to one i.e. each element in the domain is determined once by the element at the codomain
Surjection: all elements in the codomain are determined by those at the domain

This seems to mean that L is not isomorphic. The reason is perhaps that L is not injection as x1 is not determined at the codomain by the domain. That is all elements of the codomain are not determined.

4. Jan 21, 2009

### Dick

L is neither an injection nor a surjection. Just cite an example for each. Find two elements in the domain that map to the same image point and find a point in the range that isn't hit by any domain point. Neither is at all hard.

5. Jan 21, 2009

### soopo

Let x2 = 0 and x1 = 0. This means that two points in the domain map to one codomain. So it is not an injection.

The other one is harder. I first thought to let x2 = 1 and x1 = 1. Then, I started to wonder that it is not even a funtion as no domain points hit points at the codomain.

Please, let me know whether I am at the right direction or not.

6. Jan 21, 2009

### Dick

You do mean (x1,x2)->(-x2,0), right? (1,1)->(-1,0), (0,1)->(-1,0), not injective. Nothing maps to (0,1), not surjective. Case closed.

7. Jan 21, 2009

### phreak

No, no. To prove that it's not an injection, you must find two points in the domain that map to the same point. The domain is $$\mathbb{R}^2$$. Hence, you must find two points x=(x1,x2) and y=(y1,y2) such that $$L(x) = L(x_1,x_2) = L(y_1,y_2) = L(y)$$. Since you seem to be stuck, I'll give you an example. Let x= (1,1) and y = (0,1). Then by definition of L, we have L(x) = (-x2,0) = (-1,0) = (1,0) and L(y) = (-y2,0) = (-1,0). This means that we have found two points in the domain of L that map to the same point. But this contradicts the definition of injectivity.

8. Jan 21, 2009

### soopo

Thank you!

9. Jan 21, 2009

### Dick

You WILL have an easier time with the next one, right? Don't overthink it. It's not all that abstract.

10. Feb 23, 2009

### soopo

b)
The image should be
Im(L) = ($$x_{1}, x_{2}$$)

The kernel should be got by solving the null space that is Ax = 0.
I am wondering how to put the domain and codomain to Ax = 0.
We should get two simple equations from the matrix.

I am keen on saying that the kernel is
Ker(L) = ($$x_{1} = 0, x_{2} = 0$$)
It seems to be the only solution - not sure though.

11. Feb 23, 2009

### Dick

What linear map are you talking about now?

12. Feb 24, 2009

### soopo

I will correct the mistake.

We get the kernel by solving the nullspace
Ker(L) = {$$(x_{1}, x_{2})^{T}: x_{1} = 0$$ and $$x_{2} = 0$$}

The image is the same as we apparently want it for the nullspace
Im(L) = {$$(x_{1}, x_{2})^{T}: x_{1} = 0$$ and $$x_{2} = 0$$}

It should now be correct.

13. Feb 24, 2009

### HallsofIvy

Staff Emeritus
Please answer Dick's question! What linear map are you talking about? Are you still talking about L(x1,x2)= (-x2, 0)? If so, then, no, the kernel is NOT {(0,0)}.

And it makes no sense to say of the image "apparently we want it for the nullspace". The image of L is set of all y such that L(x)= y for some x. It has little to do with the nullspace (= kernel).