# Bijective versus Linear

1. Sep 26, 2010

### Anonymous217

Does a linear mapping imply that it is also bijective? I would assume this is not true because there wouldn't be a subcategory of linear mappings called bijective linear mappings then (isomorphisms, etc.).
Can someone give me an example of a linear mapping that is not bijective? I keep thinking in terms of R squared and how a line obviously shows it's one-to-one and onto, and I can't think of an example where a linear mapping isn't bijective. I'm probably missing something obvious.

2. Sep 26, 2010

### Office_Shredder

Staff Emeritus
The function L(x,y)=(x,0). Or even just L(x,y)=(0,0)

3. Sep 26, 2010

### Anonymous217

Thanks! That really was inherently obvious, especially that last example. I guess I keep thinking of linear mappings in relation to geometric lines.

4. Sep 27, 2010

### Fredrik

Staff Emeritus
It's also useful to know that if V is a finite-dimensional vector space, and T:V→V is linear, all of these claims are equivalent:

T is injective
T is surjective
T is bijective
T is invertible

This is theorem 3.21 in Axler.

5. Sep 29, 2010

### Landau

To prevent possible confusion, "invertible" here means "invertible in the category of vector spaces", i.e. there exists a linear inverse.

6. Sep 29, 2010

### Anonymous217

I had a quick question but I didn't want to make another topic about it because I feel like that would be a waste of space so I'll just cram it here.
What's the dimension of [PLAIN]http://data.artofproblemsolving.com/aops20/latex/texer/893a2ae84b20d51f3beecf662ec7aab02ee073b1.png [Broken] [\img]? (Sorry, I can't get [tex] tags to work here and I don't know how to do the Reals symbol in latex. I also seem to have problems displaying images with tags.)
It's just the xy-plane and I know it should be 2-dimensional, but then again it has 3 coordinates. Also, I know R^3/E_1 is isomorphic to E_12. Therefore, dim(R^3) - dim(E_1) = dim(E_{12}) and so it should be 2=2. So in general, my question is probably this: is the dimension only based on nonzero coordinates (since you could expand E_{12} to have more 0 coordinates I guess)?

Last edited by a moderator: May 5, 2017
7. Sep 29, 2010

### gerben

The dimension is the number of basis vectors you need to span the whole space. You can get anywhere in E_12, with multiples of basis vectors: a=(1,0) and b=(0,1). You cannot do it with less than two basis vectors, you can choose others for example (1,1) and (1,-1), but you need at least two. So, the dimension of E_12 is 2.

If you think of E_12 as being embedded in some higher dimensional space you might write the two basis vectors that span E_12 as a=(1,0,0,0,0,0) and b=(0,1,0,0,0,0), any linear combination of a and b will be of the form (x,y,0,0,0,0). Showing that linear combinations of a and b can never get out of E_12.

8. Sep 29, 2010

### Anonymous217

I completely forgot about doing it by basis vectors. Thanks!

9. Sep 29, 2010

### Fredrik

Staff Emeritus
The LaTeX code for $\mathbb R$ is \mathbb R. And img tags end with /img, not \img.