# Linear injective mapping

1. Oct 9, 2009

### Servarus

Hey guys, new to the forum but hoping you can help.

How do you prove that vector spaces V and U have a linear injective map given V is finite dimensional. I got the linear part but cannot really figure out the injectivity part, although I am thinking that it has to do with the kernel.

2. Oct 9, 2009

### HallsofIvy

Staff Emeritus
"V and U have a linear injective map" makes no sense. Two spaces do not have a "linear injective map". Do you mean "there exist a linear injective map from V to U"? That is true if and only if the dimension of V is less than or equal to the dimension of U. And you prove it by "associating" a unique basis vector of U to every basis vector of V and ignoring any that are left over. Map each basis vector of V into its associated basis vector or V.

3. Oct 9, 2009

### Tac-Tics

Like Halls said, be careful with your terminology.

Do you understand what an injective map (sometimes called a one-to-one map) is to begin with? You simply need to show that no two vectors map to the same place. Or in symbols, that given v1 and v2 are vectors in V, if v1 /= v2, then f(v1) /= f(v2).

The key ideas in your proof will be the following:

The map you're looking for is linear. That means it is fully defined simply by finding f(b1), f(b2), ..., f(bN) for any basis {b1, b2, ..., bN} for V. This means that f(b1), f(b2), ..., f(bN) are N vectors in U. Ask yourself: are these vectors eligible to be a basis for U? Are they enough of them? Can I choose f so that they are all linearly independent?

4. Oct 9, 2009

### Servarus

I apologize, yes I did mean there exists a linear injective map from V to U. And what I am really trying to figure out is how you prove that there does exist a linear injective map when there are no matrices or sets that have to do with either of vector space.

5. Oct 9, 2009

### Tac-Tics

Again, be careful with your terminology =-) What do you mean when you say there are no matrices or sets that "have to do" with either vector space?

As soon as you choose an ordered basis for both spaces, you have a column-vector representation for every vector in the space and a matrix representation for every linear map between the two spaces.

Of course, it all depends on what your professor expects of you. If he or she doesn't want you to talk about matrix representations, then you might have to jump through hoops, even if you already have an otherwise legitimate proof.

6. Oct 9, 2009

### HallsofIvy

Staff Emeritus
As I said before, in order that there exist a linear injective map from V to U, V must have dimension less than or equal to the dimension of U. In that case, let $\{v_i\}$, i= 1 to m, be a basis for V and let $\{u_i\}$, i= 1 to n, be a basis for U. Of course, m must be less than or equal to n.

Now define A:V->U by: $A(v_i)= u_i$ for i from 1 to m. For any v in V, $A(v)= A(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_mv_m)$$= a_1A(v_1)+ a_2A(v_2)+ \cdot\cdot\cdot+ a_mA(v_m)$$= a_1u_1+ a_2u_2+ \cdot\cdot\cdot+ a_mu_m$. That maps V onto the m dimensional subspace in U having $\{u_i\}$ for i= 1 to m as basis.

Since you mentioned "kernel" in your first post, a linear map is injective if and only if its kernel is the "trivial" subspace consisting of only the 0 vector.

Last edited: Oct 9, 2009
7. Oct 11, 2009

### Servarus

Alright, I finally understand.

Thank you both so much for all the help. It has been quite invaluable.