[Math] Dimension of the range of $T$ is equal to the codimension of $\ker T$

linear algebra

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.

Best Answer

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

  1. well-defined (What does well-defined mean in this context?),
  2. linear,
  3. bijective (i.e., surjective and injective).
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