Let $M\ge 1$, for any $l|M$, $l\ne M$ (so that $M=lm$) we define $V_l$ to be the subspace of $\mathbb{Q}^M$ satisfying the equations
$\begin{cases} x_1=x_{l+1}=x_{2l+1}=\dots=x_{(m-1)l+1}\\ x_2=x_{l+2}=x_{2l+2}=\dots=x_{(m-1)l+2}\\ \vdots\\ x_l=x_{2l}=x_{3l}=\dots=x_{ml} \end{cases}$
Notice that $\dim V_l=l$ and that if $l_1|l_2|M$, we have that $V_{l_1}\subseteq V_{l_2}$. Also, we have that $V_{l_1}\cap V_{l_2}=V_{GCD(l_1,l_2)}$. Some explicit computations led me to think that it is true that
$\dim(\sum_{l|M}V_l)=\sum_{l|M}\dim(V_l)-\sum_{l_1,l_2|M; l_1\ne l_2}\dim(V_{l_1}\cap V_{l_2})+\sum\dim(V_{l_1}\cap V_{l_2}\cap V_{l_3})-\dots+(-1)^{a}\dim(\bigcap V_l)$
where $a$ is the number of the spaces $V_l$ and $l$ is always different from $M$ in the sums. Is someone able to give a proof of this fact?
Comments:
Let $V_1,\dots,V_n$ be subspaces of $\mathbb{Q}^M$. It is known that, whereas $\dim(V_1+V_2)=\dim(V_1)+\dim(V_2)-\dim(V_1\cap V_2)$, there is not such an "inclusion-exclusion" formula for the sum of more than two subspaces. This is essentially because the operations of sum and intersection of vector spaces don't commute in general or equivalently because a specific chain complex is not always exact (see https://mathoverflow.net/questions/17740/is-there-a-version-of-inclusion-exclusion-for-vector-spaces). But of course inclusion-exclusion could work for specific families of subspaces!
For example, if $V$ is a vector space and $\{v_1,\dots,v_n\}$ is a basis of $V$, the inclusion-exclusion works for every family of subspaces of $V$ spanned by some elements of the basis $\{v_1,\dots,v_n\}$.