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Let $X$ be a finite dimensional vector space over $K$, where $K=R$ or $K=C$, let $P=\operatorname{End}_K X$ be the ring of all endomorphisms of the space $X$, and let $I$ be a left ideal of $P$. Is it true that for each $f, g \in I$ there exists $h \in I$ such that $\operatorname{Ker} h=\operatorname{Ker} f \cap \operatorname{Ker} g$ ?

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Your question doesn't seem to use $I$ at all. Are you sure you have phrased it correctly? –  Zhen Lin Aug 9 '11 at 10:14
    
Thank you, I have just corrected. –  Richard Aug 9 '11 at 10:51
    
I have a very strong feeling that taking $h = \alpha f + \beta g$ for some $\alpha$ and $\beta$ will work, since $\mathrm{Ker(\alpha f + \beta g)} \supseteq \mathrm{Ker}(f) \cap \mathrm{Ker}(g)$, but I cannot seem to prove it. Choosing $\alpha$ and $\beta$ when given examples is quite trivial since there are "many" choices... maybe I just couldn't find a pathological counter-example though. It would suffice to find a canonical choice of $\alpha$ and $\beta$ for which the inclusion is always equality. –  Patrick Da Silva Aug 9 '11 at 11:51
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For each subspace $Y$ of $X$ there is the left ideal $I_Y$ of endomorphisms that annihilate all of $Y$. I think that all the left ideals of $P$ are of this form, and that is the key to this exercise. –  Jyrki Lahtonen Aug 9 '11 at 12:28
    
Yes, all the left ideals os $P$ are of the form $I=\{\phi \in P: \phi|_V =0\}$. But my question concerns just the proof of that theorem ( from "Algebra I" by Kostrikin and Shafarevich, $\S$ 8, example 13). It is the part of that proof which I don't understand. –  Richard Aug 9 '11 at 13:04
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1 Answer

up vote 2 down vote accepted

For each subspace $Y$ of $X$ there is the left ideal $I_Y$ of endomorphisms that annihilate all of $Y$. Actually all the left ideals of $P$ are of this form, and that could be the key to this exercise, but may also be the goal, so let's not use that :-)

Let $u_1,\ldots,u_r,\ldots,u_\ell$ be a basis for $\ker f$ such that $u_{r+1},\ldots,u_\ell$ is a basis for $\ker f \cap \ker g$. Extend the latter to a basis of $\ker g$ by adding $u_{\ell+1},\ldots,u_k$, and in the end to a basis of $X$ by adding vectors $u_{k+1},\ldots,u_n$. The vectors $g(u_1),\ldots g(u_r),g(u_{k+1}),\ldots,g(u_n)$ are then linearly independent, so for all $i$ outside the range from $r+1$ to $k$, there exists an endomorphism $h_i$ such that $h_i(g(u_i))=u_i$ and $h_i(g(u_j))=0$ for all $j\neq i$. So $h_i\circ g\in I$ maps $u_i$ to itself and the other basis vectors to zero.

Similarly composing $f$ from the left by suitably defined endomorphisms $h_{\ell+1},\ldots,h_k$ we get elements of $h_j\circ f\in I$ mapping $u_j, j=\ell+1,\ldots,k,$ to itself and the other basis vectors to zero.

The mapping $$h=h_1\circ g+\cdots+h_r\circ g + (h_{\ell+1}\circ f+\cdots +h_k\circ f)+h_{k+1}\circ g+\cdots+h_n\circ g $$ is in the left ideal $I$, sends the basis elements of $\ker f\cap \ker g$ to zero, and the other basis vectors to themselves. So this $h$ works.

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Great answer. I didn't notice that the vectors $g(u_1),...,g(u_n)$ were linearly independent when I tried to take the same path as you did, so I was kind of stuck when trying to find a linear combination... that is why my first guess was just to take elements of $K$ as coefficients. You wrote it out quite well. +1 and by the way there's a u missing on $h_i(g(u_i)) = u_i$ half-way in there. Close to that I think you meant $j \neq i$. –  Patrick Da Silva Aug 9 '11 at 14:47
    
@Patrick: Thanks for spotting the typos. Much appreciated. –  Jyrki Lahtonen Aug 9 '11 at 15:41
    
Thank you very much for solution. –  Richard Aug 9 '11 at 15:54
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