# Dimension of sum of 3 subspaces

Let $W_1$, $W_2$ and $W_3$ be finite-dimensional subspaces of a vector space.

Show that it may happen that $W_i \cap W_j = 0$ for all $i \ne j$, but still $\dim(W_1 + W_2 + W_3) \ne \dim W_1 + \dim W_2 + \dim W_3$.

I have a counterexample of three lines that represent each subspace that intersect at 0, but I don't really understand why that works.

I know this property holds for two subspaces so I'm confused why it doesn't hold for three.

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What if $W_3$ contains a vector of the form $w_1+w_2$ with $w_1\in W_1$ and $w_2\in W_2$? – David Mitra Feb 1 '13 at 2:30
@MaisamHedyelloo That formula is not true. See mathoverflow.net/questions/23478/… – user38268 Feb 1 '13 at 3:12
i think we can define base for $w_1\cap w_2\cap w_3$ and expand it $for W_i \cap W_j$ and $w_i$ like my approach in this question math.stackexchange.com/questions/289971/… – Maisam Hedyelloo Feb 1 '13 at 3:21

We know that $\dim(W_1 + W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$. Therefore, for three subspaces, $\dim(W_1+W_2+W_3) = \dim(W_1)+\dim(W_2+W_3)-\dim(W_1 \cap (W_2+W_3))$. Just because $W_1 \cap W_2 = 0$ and $W_1 \cap W_3 = 0$ does not mean that $W_1 \cap (W_2+W_3) = 0$.
I try to give an intuitive answer. In the case with two subspaces, the condition $W_1 \cap W_2 = 0$ translates to the "linear independence of the two subspaces": \begin{align*} & \lambda_1 x_1 + \lambda_2 x_2 = 0 , \text{ where } x_1 \in W_1, x_2 \in W_2\\ \Longrightarrow & \lambda_1 x_1 = - \lambda_2 x_2 \in W_1 \cap W_2 = 0 \\ \Longrightarrow & \lambda_1 = \lambda_2 = 0 . \end{align*}
While in the case with three or more subspaces, the pairwise "linearly independence" condition $W_i \cap W_j = 0$ for $i \neq j$ is not strong enough to ensure that each $W_i$ would "generate new/indepedent directions". We may need conditions like $(W_1 + \cdots + W_{j} ) \cap W_{j+1} =0$ for all $j \geq 1$ to ensure that $W_{j+1}$ is really independent from the subspace generated by the first $j$ subspaces.