0

Let $V$ be a finite-dimensional vector space, $End_{K}(V)$ the ring of endomorphisms. Furthermore, let $I$ be an ideal of $End_{K}(V)$, $I \neq {0}$.

I want to show that for all $u, v \in V, u,v \neq 0$, there exists $f \in I$ so that $f(u) = v$.

If $f$ is injective, I believe we are done since that means all nonzero vectors are in the image of $f$, due to there only being a trivial kernel.

Now assume that $f$ is not injective, i.e. the kernel is not trivial. This is where I am lost, since I don't see how if $u \in N(f)$, $f(u) = v$. The hint given in the paper is to use the basis expansion theorem. Beyond that I am stuck.

Newbie1000
  • 351
  • 1
    What do you mean "if $f$ is injective"? You are the one producing the $f$. It is not a given. Maybe that is what is confusing you? – rschwieb Mar 07 '24 at 16:52
  • 1
    What you want to prove is false right now. What if $I=(0)$? Then there is no $f\in I$ with that property. – Arturo Magidin Mar 07 '24 at 16:55
  • Assume I is not equal to zero. I will add that in. – Newbie1000 Mar 07 '24 at 16:56
  • $\mathrm{End}_K(V)$ is well-known to be isomorphic to the ring of $n\times n$ matrices over $K$ (where $n=\dim(V)$). This ring is simple: the only ideals are $(0)$ and $\mathrm{End}_K(V)$. The existence of $f$ in the former is clearly false. The existence of $f$ in the latter is trivial. – Arturo Magidin Mar 07 '24 at 16:57
  • 1
    So this example is supposed to be part of a larger proof that $End_{K}(V)$ is simple. Consider $I$ any arbitrary nonzero ideal, not necessarily the entire ring. I should have mentioned that detail. – Newbie1000 Mar 07 '24 at 17:01
  • Any nonzero ideal is the entire ring. – Arturo Magidin Mar 07 '24 at 17:17
  • Find any nonzero linear transformation $f$ in $I$. Multiply on the right by an endomorphism that takes $u$ to a vector not in the kernel of $f$. Then multiply on the left by an endomorphism that takes the image of that vector to $v$. – Arturo Magidin Mar 07 '24 at 17:20

0 Answers0