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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.

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More generally, try to prove that $U / \ker T \cong \operatorname{range} T$. –  Srivatsan Oct 30 '11 at 18:18
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Hint: $f(x)=f(y) \Leftrightarrow x-y \in \ker(f)$. So how can you make $f$ injective? –  N. S. Oct 30 '11 at 18:19
    
@user9176. So is the $f$ in your comment the isomorphism between $U/kerT$ and range$T$? and if so how do I define it explicitly? –  aponga Oct 30 '11 at 18:49
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You define a new function $G: U/ker(t) \rightarrow range(T)$ by $G(x+ \ker(T))=T(x)$. You have to prove that $G$ is well defined, linear and isomorphism, but this is easy... The hard part was realizing that the question asks exactly what Sri said.... And $f$ in my comment was supposed to be $T$, typo :) –  N. S. Oct 30 '11 at 19:09

2 Answers 2

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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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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 $T$.

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Can you give more detail? –  aponga Oct 30 '11 at 18:20
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The problem basically asks them to prove the first isomorphism theorem for vector spaces, so using it doesn't seem the right approach... –  N. S. Oct 30 '11 at 18:24

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