Left ideal of $\text{M}_n(\mathbb{F})$, universality result which isn't clear.

Question

Let $V$ be a finite dimensional vector space. Then every left ideal in $\text{End}\,V$ is of the form$$I(W) := \{T \in \text{End}\,V : T = 0 \text{ on }W\}.$$

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 $$S = L \circ \tilde T.$$

---

To me, it's not clear why the bolded part is true. Could anybody expand on this standard universality result as to why its true, and perhaps provide a proof of it?

Answer 1

We want to define $L$ such that $$L\left(T_1\left(v\right),\dots,T_m\left(v\right)\right)=S\left(v\right)$$ so that's how we define $L$ (every element in $\mathrm{Im}\left(\tilde{T}\right)$ can be written in that way). It is well-defined because $\left.S\right|_W=0$ and also $\left.T_i\right|_W=0$. And of course it's linear.

I'm sure that there is some universal property here, but I'm not sure which.

Answer 2

We regard $\tilde{T}$ as an epimorphism $\tilde{T} \colon V \to \operatorname{im} \tilde{T}$.

Let $p \colon V \to V\!/W$ denote the canonical projection. Because $S|_W = 0$ we have $W \subseteq \ker S$, so by the universal property of the quotient the linear map $S$ factors through a unique linear map $\overline{S} \colon V\!/W \to V$ with $S = \overline{S} \circ p$.

By the first isomorphism theorem (and thus again by the universal property of the quotient) the linear map $\tilde{T}$ induces an isomorphism $$ \overline{T} \colon V/\ker \tilde{T} \to \operatorname{im} \tilde{T}, \quad \overline{v} \mapsto \tilde{T}(v). $$ Because $\ker \tilde{T} = W$ this is an isomorphism $\overline{T} \colon V\!/W \to \operatorname{im} \tilde{T}$. This isomorphism satisfies $\tilde{T} = \overline{T} \circ p$ and thus $p = \overline{T}^{-1} \circ \tilde{T}$.

Under the isomorphism $\overline{T} \colon V\!/W \to \operatorname{im} \tilde{T}$ the linear map $\overline{S} \colon V\!/W \to V$ corresponds to the linear map $L := \overline{S} \circ \overline{T}^{-1} \colon \operatorname{im} \tilde{T} \to V$, for which we have $$ L \circ \tilde{T} = \overline{S} \circ \overline{T}^{-1} \circ \tilde{T} = \overline{S} \circ p = S. $$

Here is a commutative diagram of the whole situation:

---

PS: One can also shorten the above by saying that $\tilde{T} \colon V \to \operatorname{im}(\tilde{T})$ is a cokernel of the inclusion $i \colon W \hookrightarrow V$; then the statement follows immediately from $S \circ i = 0$ and the universal property of the cokernel. (The underlying calculations stay the same, but are better hidden.)