Question

10. Prove the following claim: A homomorphism is onto if and only if its rank equals the dimension of the codomain.

Answer 1

Solution:

Suppose that V and W are vector spaces over a field F and T : V $\rightarrow$ W be a homomorphism.

Claim: T(V) is a subspace of W.
Proof of the claim: Observe that 0_W = T(0_V) $\in$ T(V) and thus, T(V) is a nonempty subset of W.
Let w,w' be in T(V) and c,d be in F. By definition of T(V), there exist v,v' in V such that w = T(v) and
w' = T(v') .

Then, cw+dw' = cT(v)+dT(v') = T(cv)+T(dv') = T(cv+dv') which is in T(V) as cv+dv' is in V(which is a vector
space over F).
Thus, T(V) is a subspace of W.

Hence, it follows from the claim that dim T(V) =dim W iff T(V) = W (as T(V) is a subspace of W).

Finally, observe that rank(T) = dim T(V).

Now, T is onto if and only if T(V) = W if and only if dim T(V) = dim W if and only if rank(T) = dim W (where W is the codomain or the target of T).