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Let $V$ be a vector space and let $T \in \operatorname{End}(V)$. If $\operatorname{rank}(T)$ and $\operatorname{null}(T)$ are finite, prove that $\dim(V)$ is finite.

I cannot use the Rank-Nullity Theorem as it only applies to finite dimensional vector space and I don't know whether $V$ is finite or infinite dimensional.

Jendrik Stelzner
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Dheeraj Putta
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    Take basis of $nullspace(T)$ and pick preimages of basis of $image(T)$. Then show that these together generate $V$. – Lev Borisov Apr 12 '16 at 11:18

3 Answers3

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The lemma used to prove this theorem is, on the other hand, always valid (and thus much more important) : if $f:V\to W$ is a linear map and $H$ is a supplement of $\ker(f)$ in $V$ (meaning $V=H\oplus \ker(f)$) then $f$ induces an isomorphism $H\to Im(f)$.

Try to see that the proof carries over to infinite dimension, and use that in your case.

Captain Lama
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The rank-nullity theorem $$\dim V = \dim \ker T + \dim im T$$ also holds for infinite-dimensional $V$, see here. The formula implies the claim.

Community
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Dietrich Burde
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If you can use the general fact that

a vector space always admits a basis

you can prove the Rank-Nullity theorem without any restriction on the dimension. Indeed

  • take a basis $\cal B$ of ${\rm Im}(T)$,
  • take a basis $\cal K$ of $\ker(T)$,
  • for each $b\in \cal B$, choose a $b^\prime$ such that $T(b^\prime)=b$ and call $\cal B^\prime$ the set of such elements $b^\prime$,
  • prove that $\cal K\cup\cal B^\prime$ is a basis of $V$.

Bingo!

AdLibitum
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