Under what conditions does a finite collection of subspaces of a finite-dimensional vector space have a simultaneous complement?

Question

Came across this question.

Under what conditions does a finite collection of subspaces of a finite-dimensional vector space have a simultaneous complement?

Definition: If $M$ is a subspace of a vector space $V$, a complement of $M$ is another subspace $V$ such that $M \oplus N = V$ and $M \cap N = \{0\}$

The hints suggest "it is easy for several subspaces to have a "simultaneous" complement, meaning a complement in common. It’s easy enough, but that doesn’t mean that it always happens"

I must be missing something basic. If $V = R^5$, $M$ is the subspace spanned by $\{1,0,0,0,0\}$, $\{0,1,0,0,0\}$, $\{0,0,1,0,0\}$, then its only supplement is the subspace spanned by $\{0,0,0,1,0\}$, $\{0,0,0,0,1\}$? How could there be another subspace other than $M$ that has the same complement?

so I can't even understand the hints, let alone figuring out the original question...

Answer 1

It is necessary and sufficient that they all have the same dimension.

Since the sum of the dimensions of a subspace and a complement of the subspace is the dimension of the entire space, subspaces with a common complement must have the same dimension.

Conversely, if the subspaces all have the same dimension, we can independently uniformly randomly choose as many vectors in the space as we need for the dimension of the complement. In each step and for each subspace, the space spanned by the subspace and the vectors chosen so far is a set of measure zero, so the probability that we choose a vector that lies in one of these spaces is $0$. Thus with probability $1$ we choose vectors that span a complement to each of the subspaces. Since something that can be chosen with probability $1$ must exist, it is sufficient that the subspaces all have the same dimension.

Answer 2

Assume that the field $K$ is infinite (else, it is easy to find a counterexample), but not necessarily equal to $\Bbb R$.

The subspaces $V_1,\dots,V_n$ must of course have the same codimension $d$. Let us prove that this necessary condition is sufficient, by induction on $d$. If $d=0$, the simultaneous complement is $N=\{0\}$. Assume now that $d>0$, and that the property holds for every codimension less than $d$. Since $K$ is infinite and the subspaces $V_1,\dots,V_n$ are proper, they cannot cover $V$. Choose $v\in V\setminus\cup_{i=1}^nV_i$ and let $W_i:=V_i\oplus Kv$ (of codimension $d-1$). By induction hypothesis, the $W_i$s have a simultaneous complement $N$, and then $N\oplus Kv$ is a simultaneous complement for the $V_i$s.