Dimension of the range of $T$ is equal to the codimension of $\ker T$ Given a linear transformation, $T:U\rightarrow V$ , I am asked to show that the dimension of the range of $T$ is the same as the codimension of the kernel of $T$. I am told that $U$ is not necessarily a finite dimensional vector space so I cannot assume that the dimension theory holds. As a matter of fact, I know nothing about codimensions and so I have no idea how to go about this question. I need some help. 
 A: A proof sketch. The codimension of $\ker T$ is just the dimension of the quotient space $U / \ker T$, so in order to the prove the claim, it suffices to show that $\newcommand{\range}{\mathop{\operatorname{range}}}$ $U / \ker T \cong \range T$. To this end, try to demonstrate an explicit isomorphism $S : U/\ker T \to \range T$.
Hint. Any element of $U/\ker T$ is a coset of the form $x + \ker T$ for some $x \in U$. Do you see a natural way to define $S(x + \ker T)$? Once you define $S$, to complete the proof, you will need to show that the map $S$ is


*

*well-defined (What does well-defined mean in this context?),

*linear, 

*bijective (i.e., surjective and injective). 

A: Hint: Use the first isomorphism theorem. The first isomorphim theorem states that: If you have a $T:U\longrightarrow V$ is a linear transformation then $U/\ker T\cong im T$. But the codimension of the kernel is the dimension of $U/ \ker T$ and 'cause $U/\ker T\cong im T$, $U/ \ker T$ have the same dimension of the range of $T$.
