Linear Algebra: Kernel, Basis, Dimensions, injection, surjections

[WK95]

Homework Statement

The Attempt at a Solution

Can someone please check my work?

 

[micromass]

Some comments:

a) You have not given the entire kernel. You have just produced a basis (or at least a spanning set) of the kernel.

d) I don't see why you care about the range here.

h) You did some calculations to show the map is invertible. OK. But there is a much easier way, do you see which one?

 

[WK95]

For a), I don't know how to write out the full set of vectors. All I know is that the vectors are linear combinations of the basis of the kernel.

I'm not sure what you mean by range for d). Injection refers to one-to-one and a theorem states that if the rank is equal to the number of columns, then it is injective or one-to-one. So if the rank is not equal to the numbers, it is not injective.

For h), I can't think of any easier way or any other way for that matter. Does your method involve determinants? If so, I didn't learn those yet.

 

[micromass]

For a), I don't know how to write out the full set of vectors. All I know is that the vectors are linear combinations of the basis of the kernel.

Yes, just write that out:

$$\textrm{Ker}(L) = \{(-2\alpha-\beta,3\alpha,\alpha,\beta)~\vert~\alpha,\beta\in \mathbb{R}\}$$

I'm not sure what you mean by range for d). Injection refers to one-to-one and a theorem states that if the rank is equal to the number of columns, then it is injective or one-to-one. So if the rank is not equal to the numbers, it is not injective.

OK, this is right. I was thinking of $L$ being injective if and only if $\textrm{dim}(\textrm{Ker}(L)) = 0$.

For h), I can't think of any easier way or any other way for that matter. Does your method involve determinants? If so, I didn't learn those yet.

If a function is invertible, then it must be both injective and surjective. That's not the case here, is it?

 

[WK95]

If a function is invertible, then it must be both injective and surjective. That's not the case here, is it?

Thanks a lot!