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I have proof of this but not getting please help me. If $End_{K}(V)$ is a simple ring then $V$ is finite dimensional vector space. proof:- Let us assume that $V$ be not a finite dimensional vector space over field $K$. Define $I=\{f \in End_{K}(V)\mid\dim_{K}f(V)<\infty\}$ $0 \in I$ and $1\notin I$ $(0)\subseteq I \neq End_{K}(V)$ Now to show that I is both sided ideal we get contradiction. i am not getting it from my text book

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Only the basic verification stands between you and the answer:

If the image of $f$ and $g$ are finite dimensional, show that the images of $-f$ and $f+g$ are finite dimensional also.

For any $h\in End_K(V)$, give a reason why $fh$ has finite dimensional image.

Finally, why should the image of $hf$ be finite dimensional?

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  • $\begingroup$ How we can prove if $V$ is finite dimensional then $End(V)$ is simple ring ? $\endgroup$ – user197636 Mar 18 '15 at 20:11
  • $\begingroup$ @user197636 You can show it's isomorphic to the full ring of square matrices $M_n(F)$ for some $n$, and then it's simple for this reason. $\endgroup$ – rschwieb Mar 19 '15 at 1:07
  • $\begingroup$ 's thanks alot now i will try this. $\endgroup$ – user197636 Mar 19 '15 at 3:22

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