TITLE: Any left ideal of $M_n(\mathbb{F})$ is principal
QUESTION [11 upvotes]: I'm working on the following problem:

Let $A$ be the ring of $n \times n$ matrices over a field $\mathbb{F}$.
(a) Show that for any subspace $V$ of $\mathbb{F}^n$, the set $I_V$ of matrices whose kernel contains $V$ is a left ideal of $A$.
(b) Show that every left ideal of $A$ is principal.

I've done part $a)$, but would like to know if you can prove $b)$ directly from $a)$. It seems to me that given the left ideal $J$, it should be the case that if $V$ is the intersection of the kernels of matrices in $J$, then we should have $J = I_V$. I can show that $I_V$ is principal, and certainly $J$ is contained in $I_V$, but I can't show the other direction.
I think you can prove $b)$ by considering the subspace $W$ of $\mathbb{F}^n$ consisting of the rows of elements of $J$, which is of dimension $k \leq n$ say, and then showing that $J$ is generated by any matrix whose first $k$ rows are some basis for $W$ and whose final rows are all $0$. But it seems that we should be able to do the problem just using $a)$, and I'd like to know how to do it!

REPLY [2 votes]: Note that $I_V$ is principal left ideal of $M_n(\mathbb{F})$ whose generator can be taken to be any linear map $T \colon \mathbb{F}^n \rightarrow \mathbb{F}^n$ with $\ker(T) = V$. To see this, choose a basis $(w_1, \ldots, w_n)$ of $\mathbb{F}^n$ such that $(w_1, \ldots, w_k)$ is a basis of $\ker(T) = V$. Define $f_i = T(w_i)$ for $k + 1 \leq i \leq n$ and complete them to a basis $(f_1, \ldots, f_n)$ of $\mathbb{F}^n$.
Let $S \colon \mathbb{F}^n \rightarrow \mathbb{F}^n$ be a linear map with $V \subseteq \ker(S)$. We need to find a linear map $P \colon \mathbb{F}^n \rightarrow \mathbb{F}^n$ such that $PT = S$. Define $P$ by requiring that 
$$ P(f_i) = \begin{cases} 0 & 1 \leq i \leq k, \\
S(w_i) & k + 1 \leq i \leq n. \end{cases}$$
Then
$$ P(T(w_i)) = \begin{cases} P(0) = 0 = S(w_i) & 1 \leq i \leq k, \\
P(f_i) = S(w_i) & k + 1 \leq i \leq n \end{cases} $$
which shows that $PT = S$.
Given any left-sided ideal $J$, let $V = \cap_{T \in J} \ker(T)$. Note that there must exist a linear map $T \in J$ with $\ker(T) = V$. Since $J$ is a left-sided ideal, we have
$$ (M_n(\mathbb{F}))T = I_V \subseteq J \subseteq I_V $$
which shows that $J = I_V$.