Recently, I want to prove a structure theorem of submodule of finitely generated module over PID. By several process, it reduces to the following problem in linear algebra.
Given a linear space $V$, and a family of subspaces $$\begin{array}{ccccccccc} V^1_1 & \to & V^1_2 & \to & \ldots & \to & V^1_n\\ \downarrow & & \downarrow & & \vdots && \downarrow\\ V^2_1 & \to & V^2_2 & \to & \ldots & \to & V^2_n \\ \downarrow & & \downarrow & & \vdots && \downarrow\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \downarrow & & \downarrow & & \vdots && \downarrow\\ V^m_1 & \to & V^m_2 & \to & \ldots & \to & V^m_n & = & V \end{array}$$ where all the arrows are inclusion. Can we pick up a set of basis of $V$, say $\mathfrak{B}$, such that $$V_i^j=\operatorname{span}(V_i^j\cap \mathfrak{B})$$ That is, $V_i^j\cap \mathfrak{B}$ is exact a set of basis for $V_i^j$.
After careful contemplation, I find it is a very subtle problem. Here are some observations.
If $m=1$, the problem can be easily solved by basis expansion. More exactly, pick a set of basis of $V_1^1$, expand it to a set of basis of $V_2^1$, expand it to a set of basis of $V_3^1$, etc.
If $m=2,n=2$, the problem can be solved by so-called modular formula, i.e. $$\dim (V_1\cap V_2)+\dim (V_1+V_2)=\dim V_1+\dim V_2$$ We first pick a set of basis of $V_1^1$, expand it to a set of basis of $V_1^2\cap V_2^1$, then expand it to $V_1^2$ and $V_2^1$ respectively. The union of them form a basis of $V_1+V_2$, then expand it to a set of basis to $V_2^2$.
Not all the shapes can be solved. Note that $$\begin{array}{ccccc} 0 & \to & 0 & \to & \operatorname{span}(e_1) \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & \operatorname{span}(e_2) & \to & \operatorname{span}(e_1,e_2) \\ \downarrow & & \downarrow & & \downarrow \\ \operatorname{span}(e_1+e_2) & \to & \operatorname{span}(e_1,e_2) & \to & \operatorname{span}(e_1,e_2) \end{array}$$ is an example.
Denote $\mathcal{V}=\{V_i^j\}$, let $\overline{\mathcal{V}}$ be the lattice generated by $\mathcal{V}$ (that is, the smallest family of linear subspace containing $\mathcal{V}$ and closed under $\cap$ and $+$). Then each member of $W\in \overline{\mathcal{V}}$ satisfies $$W=\operatorname{span}(W\cap \mathfrak{B})$$ Since $$\begin{cases} W_1=\operatorname{span}(W_1\cap \mathfrak{B})\\ W_2=\operatorname{span}(W_2\cap \mathfrak{B}) \end{cases}\Rightarrow\begin{cases} W_1\cap W_2=\operatorname{span}((W_1\cap W_2)\cap \mathfrak{B})\\ W_1+ W_2=\operatorname{span}((W_1+ W_2)\cap \mathfrak{B}) \end{cases}$$ If such set of basis $\mathfrak{B}$ exists, then $V\mapsto V\cap \mathfrak{B}$ gives a faithful lattice representation of $\overline{\mathcal{V}}$ to a distributive lattice $2^{\mathfrak{B}}$. As a result, a necessary condition of the existence of such basis is $\overline{\mathcal{V}}$ is distributive. Note that the counterexample in 3. is exact the smallest lattice non-distributive $M_5$ $$\begin{array}{ccccc} && \cdot && \\ & \diagup & \mid & \diagdown & \\\cdot & & \cdot && \cdot \\ & \diagdown & \mid & \diagup & \\ && \cdot &&\end{array}$$
(Due to amrsa, This is wrong)
If each $V_i^j=V_{i+1}^j\cap V_i^{j+1}$, it is likely to be sufficient. Firstly, we can prove $V_i^j=V_{i+k}^j\cap V_i^{j+h}$ for any $k,h\geq 0$. The cases of $k=0$ or $h=0$ are trivial. The direction of $\subseteq$ is easy. Note that $$V_{i+k}^j\cap V_i^{j+h}\subseteq V_{i+k}^{j+h-1}\cap V_{i+k-1}^{j+h}= V_{i+k-1}^{j+h-1}$$ Therefore, by induction, $$V_{i+k}^j\cap V_i^{j+h}=(V_{i+k}^j\cap V_{i+k-1}^{j+h-1})\cap (V_i^{j+h}\cap V_{i+k-1}^{j+h-1})=V_{i+k-1}^{j}\cap V_i^{j+h-1}=V_i^j$$ Then we can expand the basis from left-up corner, fulfilling the first column, and second column without nontrivial linear relations, etc.
If 6 is right, then $n=2$ is correct, by considering a new diagram $$\begin{array}{ccccccccc} V^1_1 & = & * & = & \ldots & = & *\\ \downarrow & & \downarrow & & \vdots && \downarrow\\ V^1_2 \cap V^2_1 & \to & V^1_2 & = & \ldots & = & * \\ \downarrow & & \downarrow & & \vdots && \downarrow\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \downarrow & & \downarrow & & \vdots && \downarrow\\ V^1_n\cap V^2_1 & \to & V^1_n\cap V^2_2 & \to & \ldots & \to & V^1_n \\ \downarrow & & \downarrow & & \vdots && \downarrow\\ V^2_1 & \to & V^2_2 & \to & \ldots & \to & V^2_n & = & V \end{array}$$
If 6 is right. We can prove the following conclusion. For two chains of subspace $\{V_1^j\subseteq V_2^j\subseteq \ldots \subseteq V_n^j\}_{j=1,2}$, we can pick a set of basis $\mathfrak{B}$ such that $$V_i^j=\operatorname{span}(V_i^j\cap \mathfrak{B})$$ By considering $$\begin{array}{ccccccccc} V^1_1\cap V^2_1 & \to & V^1_1\cap V^2_2 & \to & \ldots & \to & V_1^1\cap V^2_{n} & \to & V^1_1\\ \downarrow & & \downarrow & & \vdots && \downarrow && \downarrow\\ V^1_2\cap V^2_1 & \to & V^1_2\cap V^2_2 & \to & \ldots & \to & V_2^1 \cap V^2_{n} & \to & V^1_2 \\ \downarrow & & \downarrow & & \vdots && \downarrow&& \downarrow\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \downarrow & & \downarrow & & \vdots && \downarrow && \downarrow\\ V^1_n\cap V^2_1 & \to & V^1_n\cap V^2_2 & \to & \ldots & \to & V^1_n\cap V^2_{n} & \to & V^1_n \\ \downarrow & & \downarrow & & \vdots && \downarrow && \downarrow\\ V^2_1 & \to & V^2_2 & \to & \ldots & \to & V^2_n & \to & V \end{array}$$ But there are counterexample when there are three chains see 3.
If we consider more general problem, given a poset $X$, a order homomorphism $f:X\to \{\textrm{all the subspace of $V$}\}$, when can we pick a set of basis $\mathfrak{B}$ such that $$\forall x\in X, f(x)=\operatorname{span}(f(x)\cap \mathfrak{B})$$ I think the condition on $X$ can be solved by above discussion. If $X$ can be written as a union of two chains, then such set of basis always exists, otherwise, there exists counterexample. But actually, the condition of $f$ is still ambiguous.
So, after long explanation of my observation, my question is what condition should be made on $\{V_i^j\}_{ij}$, such set of basis exists.