Any left ideal of $M_n(\mathbb{F})$ is principal

The main result is the following:

Let $V$ be a finitely dimensional vector space. Then every left ideal in $\operatorname{End}(V)$ is of the form $I(W) \colon =\{T \in \operatorname{End}(V) \mid T = 0 \textrm{ on } W\}$.

Indeed, let $I$ be a left ideal of $\operatorname{End}(V)$ and $$W = Z(I) \colon= \{ w \in V \mid T(w) = 0 \textrm{ for all } T \in I\}.$$

Let us show that $I = I(W)$, or, in other words $$I = I(Z(I)).$$ for every left ideal $I$. Note that by definition $$Z(I) = \bigcap_{T \in I} \ker(T).$$ Since $V$ is a finite dimensional space there exist finitely many $T_1$, $\ldots $, $T_m \in I$ so that $$W=Z(I) = \bigcap_{i=1}^m\ker(T_i).$$ Consider the operator $\tilde T= (T_1, \ldots, T_m)$ from $V$ to $V^m$, with kernel $\bigcap_{i=1}^m\ker(T_i) = W$.

Let now $S \in\operatorname{End}(V)$ that is $0$ on $W$. It follows (by a standard universality result) that there exists $L:\operatorname{Im}(\tilde T) \to V$ so that $$S = L \circ \tilde T.$$ Now $L$ can be extended to the full $V^m$. We know the form of linear maps from $V^m$ to $V$. They are given by $L = (L_1, \ldots , L_m)$ with $L_i \in\operatorname{End}(V)$. Therefore we have $$S = \sum_{i=1}^m L_i T_i,$$ so $S \in I$.

${\bf Added.}$ Let again $I$ be a left ideal, $W = Z(I)$. We know from the above that $I = I_{W}$. From the proof above we see that any $T_i$ with $\cap_{i=1}^m\ker(T_i) = W$ are a system of generators. So take $T$ so that $\ker T= W$. Then $T \in I$ and moreover, $T$ generates $I$. Therefore $I$ is a principal ideal.

Obs: Similarly (by duality say) one shows that every right ideal $J$ of $\operatorname{End}(V)$ is also of the form $$J = J_W = \{ T \in \operatorname{End}(V) \mid\operatorname{Im} T \subset W\}$$ Moreover, any family $T_i$ such that $\sum_{i=1}^m \operatorname{Im} T_i = \sum_{T \in J} \operatorname{Im} T$ generates $J$. Similarly, any right ideal is principal.

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 $J = I_V$.