Left ideals of $\mathrm{End}_{k}(V)$ I'm trying to solve the following problem:
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Let V be a vector space over a field $K$ with $\dim_k V =n$ and consider the ring $R = \End_k(V)$. If $U$ is a subspace define $I_U = \{ f \in R : U \subset \ker f\}$. Show that $I_U$ is a left ideal of $R$ and that every left ideal has this form.
Showing $I_U$ is left ideal is straightforward. I'm having trouble with the second part. I started taking $I$ left ideal of $R$. And define
$m = \min \{\dim(\ker f) : f \in I\}$ and $U = \ker f $ where $f$ is any element of $I$ with $\dim(\ker f) = m$. Then I want to show that $I=I_U$.
If $g \in I_U$ I need to construct $h$ such that $g= h \circ f$. However, I'm having trouble defining $h$.
For the other inclusion, I don't know where to start.
 A: Following the comment by Alex Wertheim, we let $U=\cap \ker f$ over all $f\in I$. Let $v_i$ be a basis of $V$ complimentary to $U$. Then for each $i$ there is $f_i\in U$ such that $$f_i(v_i)\neq 0$$
We can then find $g_i$ such that $\dim \ker g_i=n-1$ and $$g_i\circ f_i(v_i)\neq 0$$ by assumption, $$h_i=g_i\circ f_i\in I$$
Now let $l\in I_U$ be any endomorphism. And let $m_i$ be any endomorphism such that $$m_i\circ h_i(v_i)= l(v_i)$$
Now it is easy to see that
$$l=\sum m_i\circ h_i$$
A: That’s a good idea, but there is still some work to be done. Indeed, what basically needs to be proved is that $U$ is effectively the smallest kernel of an element of $I$, not merely one of minimal dimension. For this, we’ll use the full ideal structure of $V$.
Show that, for any $g: V \rightarrow V$, the left ideal generated by $g$ is exactly the set of $h: V \rightarrow V$ such that $\ker{h} \supset \ker{g}$. In particular, with $g=f$, $I \supset I_U$.
Hint:

 Complete a basis of the kernel of $g$ into a basis of the kernel of $h$ into a basis of $V$. Use the fact that $g$ maps any supplement space of its kernel isomorphically to the image of $g$.

Now, let $f$ be as you defined and assume that there is some $g \in I$ be such that $\ker{g} \not\supset \ker{f}$ (in particular $f$ is not injective). Show that there is some $g\in I$ with the same property such that the image of $g$ is in direct sum with the image of $f$.
Hint:

 Use the right bases again and the previous step! The second $g$ is a certain endomorphism whose kernel will contain that of the first $g$.

Conclude that for such $f,g$, the kernel of $f+g$ is strictly contained in $U$ and deduce a contradiction. Conclude that $I \subset I_U$.
