It the following matrix isomorphic?

[soopo]

Is the following matrix isomorphic?

Homework Statement

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).

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.

 

[phreak]

The mapping is also linear, so L is surjection.

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?

 

[soopo]

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?

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.

 

[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.

 

[soopo]

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.

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.

 

[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.

 

[phreak]

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

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.

 

[soopo]

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.

Thank you!
Your answer is excellent.

 

[Dick]

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

 

[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.

 

[Dick]

What linear map are you talking about now?

 

[soopo]

What linear map are you talking about now?

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.

 

[HallsofIvy]

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).