- #1

WK95

- 139

- 1

## Homework Statement

## The Attempt at a Solution

Can someone please check my work?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter WK95
- Start date

In summary, your homework statement does not have any clear explanations and it is difficult to follow.

- #1

WK95

- 139

- 1

Can someone please check my work?

Physics news on Phys.org

- #2

- 22,183

- 3,324

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?

- #3

WK95

- 139

- 1

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.

- #4

- 22,183

- 3,324

WK95 said: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:

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

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?

- #5

WK95

- 139

- 1

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

Thanks a lot!

The kernel of a linear transformation is the set of all vectors in the domain that map to the zero vector in the codomain. In other words, it is the set of all inputs that result in an output of zero.

A basis in linear algebra is a set of linearly independent vectors that can be used to represent any vector in a vector space. It is the smallest set of vectors that span the entire vector space.

The dimension of a vector space is the number of vectors in a basis for that vector space. In other words, it is the minimum number of linearly independent vectors needed to span the entire space.

A linear transformation is an injection if each distinct input has a distinct output. In other words, no two different inputs map to the same output.

A surjection is a linear transformation in which every element in the codomain has at least one preimage in the domain. In other words, every output in the codomain is mapped to by at least one input in the domain.

- Replies
- 15

- Views
- 1K

- Replies
- 1

- Views
- 859

- Replies
- 15

- Views
- 1K

- Replies
- 7

- Views
- 1K

- Replies
- 3

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Replies
- 4

- Views
- 2K

- Replies
- 9

- Views
- 2K

- Replies
- 4

- Views
- 1K

Share: