Proving a Linear Transformation is Onto

[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)...

 

[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".

 

[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.

 

[HallsofIvy]

Yes, thank you for the correction.

 

[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?

 

[WannabeNewton]

Yes.

 

[Fredrik]

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.

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

 

[WannabeNewton]

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

Thanks broski =D

 

[HallsofIvy]

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

Yes, not one of my better days is it!

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?