# Proving a Linear Transformation is Onto

1. Mar 8, 2013

### BrainHurts

There's this theorem:

A linear map T: V→W is one-to-one iff Ker(T) = 0

I'm wondering if there's an analog for showing that T is onto? If so could you provide a proof?

I'm thinking it has something to do with the rank(T)...

2. Mar 8, 2013

### HallsofIvy

The definition of "onto" is b
"f: U-> V is said to be "onto" V if and only if f(U)= V"

Now, for linear transformations, f(U) is always a subspace of V so f is "onto" V if and only if range(f)= dim(V).

Do you know that "range-nullity" theorem? For linear transformation f:U-> V nullity(f)+ range(f)= dim(V). In particular, if f is one-to-one, so that nullity(f)= 0, we must have range(f)= dim(V) so that f is also "onto" V. Conversely, if f is "onto" V, then it is also one-to-one. That is, a linear transformation is "one-to-one" if and only if it is "onto".

3. Mar 8, 2013

### WannabeNewton

Halls I believe you left something out. A linear map $T:U\rightarrow V$ is injective if and only if it is surjective provided that $dimU = dimV$. In particular, as you said if $T$ is injective then $nullity(T) = 0$ thus $rank(T) = dim(T(U)) = dimU$ by rank nullity but we then need that $dimU = dimV$ because $dim(T(U)) = dimV,T(U)\subseteq V\Rightarrow T(U) = V$ and similarly for the converse. I believe you had a typo because it should be $rank(T) + nullity(T) = dimU$ as opposed to $dimV$.

4. Mar 8, 2013

### HallsofIvy

Yes, thank you for the correction.

5. Mar 8, 2013

### BrainHurts

HallsofIvy, just to be sure, you mean that rank(f) because range(f) is a set, and if f is onto then range(f) = V

I am familiar with the rank-nullity theorem. Let f:U-->V is a linear map, rank(f) + nullity(f) = dimU. I think that's what you meant because I am familiar with the statement that T (A linear map) is 1-1 iff nullity(T) = 0

WannabeNewton: You guys both made a similar statement that a linear map T: U-->V is 1-1 iff range(T) = V and that dim(U)=dim(V)

So T is an isomorphism if dim(U)=dim(V) and if T is either 1-1 or onto?

6. Mar 9, 2013

Yes.

7. Mar 9, 2013

### Fredrik

Staff Emeritus
Just a little LaTeX tip: Use something like \operatorname or \mathrm to get $\operatorname{dim}V$ instead of $dim V$.

8. Mar 9, 2013

### WannabeNewton

Thanks broski =D

9. Mar 9, 2013

### HallsofIvy

Yes, not one of my better days is it!