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Let $V$ be a finite dimensional vector space over a field $K$, and let $W_1$, $W_2$ and $W_3$ be subspaces of $V$. By analogy with the inclusion-exclusion principle for sets, and taking into account the dimension formula for a sum of 2 subspaces, we can ask whether the following equality holds:

$$\dim(W_1 + W_2 + W_3) = \dim(W_1) + \dim(W_2) + \dim(W_3) − \dim(W_1 \cap W_2) − \dim(W_2 \cap W_3) − \dim(W_3 \cap W_1) + \dim(W_1 \cap W_2 \cap W_3)$$ (†) it is false for the sum of 3 subspaces.

This formula does not always hold: for example take three distinct lines in $\mathbb R^2$ as $U$, $V$, $W$. All intersections have 0 dimensions. The LHS is $2$, the RHS is $3$.

I would like to state general assumptions on the subspaces of $W_1$, $W_2$, $W_3$ which guarantee that the formula does hold and prove it under these assumptions.

But for which triples of (finite-dimensional) subspaces does this hold? Do you have any reference for reading? I studied a few books but could not seem to find anything. To sum up, my question I am trying to prove that the formula is True even though it is False.

Many thanks

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    $\begingroup$ You can find also some posts on Mathematics where a different formula was suggested, such as: Dimension of the sum of subspaces and The dimension of the sum of subspaces $(U_1,\ldots,U_n)$ And there is also this related post on MathOverflow: Is there a version of inclusion/exclusion for vector spaces? $\endgroup$
    – Martin Sleziak
    Commented Nov 20, 2021 at 7:50
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    $\begingroup$ @MartinSleziak Thank you for your sources. However, I would like to explore an assumption to prove this. (even if the formula does not always hold). $\endgroup$
    – user468543
    Commented Nov 20, 2021 at 8:16
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    $\begingroup$ One possible answer would be: If there is a basis such that each $W_i$ is generated by a subset of this basis then the formula holds. $\endgroup$
    – Friedrich Knop
    Commented Nov 20, 2021 at 8:54
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    $\begingroup$ I think you can answer this question in the spirit of indecomposable representations of quivers/posets. Any datum $(V; W_1, W_2, W_3)$ of three subspaces $W_i$ of a f.d. vector space $V$ is isomorphic to the sum of five indecomposable things: $(K;0,0,0)$, $(K; K, 0, 0)$, $(K; 0, K, 0)$, $(K; 0, 0, K)$ and $(K^2; K(1,0), K(0,1), K(1,1))$. Your formula holds if and only your datum doesn't contain the last indecomposable example. $\endgroup$
    – PseudoNeo
    Commented Nov 20, 2021 at 9:33
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    $\begingroup$ I have rolled back the yesterday’s anonymous suggested edit that claimed to “improved formatting and spelling”, but in reality just deleted two paragraphs that are setting the context for the question. $\endgroup$
    – Emil Jeřábek
    Commented Nov 22, 2021 at 15:32

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The idea of PseudoNeo's comment settles the converse of my statement modulo that he missed four cases.

According to the theory of indecomposable modules of the Dynkin quiver $D_4$ with all arrows pointing to the central node there are $12$ cases corresponding to the $12$ positive roots of the Lie algebra of type $D_4$. These cases for $(V;W_1,W_2,W_3)$ are

  1. $(0;K,0,0), (0;0,K,0), (0;0,0,K)$,
  2. $(K;0,0,0), (K;K,0,0), (K;0,K,0), (K;0,0,K), (K;K,K,0), (K;K,0,K),(K;0,K,K),(K;K,K,K)$
  3. $(K^2,K,K,K)$

All maps $W_i\to V$ in the first group are not injective. So, they are irrelevant. The second group consists precisely of those cases, where there is a basis of $V$ such that all $W_i$ are spanned by a subset of the basis. The last is case is causing the problems. In this case the difference $\mathrm{LHS}-\mathrm{RHS}=-1$. So we arrive at the following statement:

In the formula, one always has $\mathrm{LHS}\le \mathrm{RHS}$ with equality if and only if $V$ has a basis such that all $W_i$ are spanned by a subset of that basis.

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