Left ideals of the ring of endomorphisms 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 $\operatorname{Ker} h=\operatorname{Ker} f \cap \operatorname{Ker} g$ ?
 A: 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.
