How to construct a base compatible with two filtrations Suppose $V$ is a $m$-dimensional vector space over some field $k$. We define filtration to be a ascending chain of subspaces (for some $n$):
$$Fil:Fil_0=0\subset Fil_1\subsetneq \cdots\subsetneq Fil_n=V.$$
We call a basis $E:=\{e_1,\cdots,e_m\}$ compatible with $Fil$ if $Fil_i\cap E$ is a basis for $Fil_i$ for all $i$. That is equal to say: $Fil$ is constructed by adding elements in $E$ at each level of the ascending chain.
I want to prove that for any two filtrations $Fil,Fil'$ in $V$, we could find a basis compatible with both of them. Could you provide some ideas or share some references about it? Thanks a lot.
 A: Refining the filtrations if necessary, we may assume that both filtrations have $n=m$.  Now the idea is to pick basis elements for one filtration one-by-one while ensuring compatibility with the other one.  Having picked $e_1,\dots,e_i$ to be a basis for $Fil_i$, we want to pick one more element $e_{i+1}$ to get a basis for $Fil_{i+1}$ while maintaining compatibility with $Fil'$.  To do this, let $j$ be minimal such that $Fil'_j\cap Fil_{i+1}\not\subseteq Fil_i$, and pick $e_{i+1}$ to be some element of $(Fil'_j\cap Fil_{i+1})\setminus Fil_i$.  (To motivate this choice, note that it is forced on us, since we need $Fil'_j\cap Fil_{i+1}\supsetneq Fil'_j\cap Fil_i$ to be spanned by its intersection with $E$ so our new element of $E$ has to come from $Fil'_j$.)
I now claim that the basis $E$ you get from this process is compatible with both filtrations.  Clearly it is compatible with $Fil$.  For compatibility with $Fil'$, we show by induction on $i$ that $Fil_i\cap Fil'_j\cap E$ spans $Fil_i\cap Fil'_j$ for all $j$ (the case $i=m$ is then the desired result).  Supposing this is known for a given value of $i$, let $j$ be minimal such that $Fil'_j\cap Fil_{i+1}\not\subseteq Fil_i$, so $e_{i+1}$ was chosen to be in $(Fil'_j\cap Fil_{i+1})\setminus Fil_i$.  For $k<j$, then, $Fil'_k\cap Fil_{i+1}=Fil'_k\cap Fil_i$ is spanned by its intersection with $E$ by the induction hypothesis.  For $k\geq j$, $Fil'_k\cap Fil_{i+1}\cap E=(Fil'_k\cap Fil_i\cap E)\cup\{e_{i+1}\}$, so by the induction hypothesis it suffices to show that $Fil'_k\cap Fil_{i+1}$ is spanned by $Fil'_k\cap Fil_i$ together with $e_{i+1}$.  To prove this, suppose $x\in Fil'_k\cap Fil_{i+1}$.  Since $Fil_{i+1}$ is spanned by $Fil_i$ together with $e_{i+1}$, we may write $x=y+ae_{i+1}$ where $a$ is a scalar and $y\in Fil_i$.  Since $e_{i+1}$ and $x$ are both in $Fil'_k$, $y$ is in $Fil'_k$ as well, so $x=y+ae_i$ writes $x$ as a sum of an element of $Fil'_k\cap Fil_i$ and a multiple of $e_{i+1}$, as desired.
