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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.

user4593
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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 01 '13 at 02:30
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    @MaisamHedyelloo That formula is not true. See http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23501#23501 –  Feb 01 '13 at 03:12
  • 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 http://math.stackexchange.com/questions/289971/given-two-subspaces-u-w-of-vector-space-v-how-to-show-that-textdim/289978#289978 – M.H Feb 01 '13 at 03:21

3 Answers3

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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 $W_1 \cap (W_2+W_3) = 0$.

MichaelNgelo
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Laura Balzano
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  • I think this thread might be relevant here http://mathoverflow.net/questions/17740/is-there-a-version-of-inclusion-exclusion-for-vector-spaces – Andreas Caranti Feb 01 '13 at 07:49
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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.

MichaelNgelo
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The condition $$\dim(W_1 + W_2 + W_3) = \dim W_1 + \dim W_2 + \dim W_3$$ is equivalent to the condition that the sum $W_1 + W_2 + W_3$ is a direct sum. We know that $W_1 + W_2 + W_3$ is a direct sum if and only if $W_1 \cap W_2 = 0$ and $(W_1 + W_2) \cap W_3 = 0$.

penseur_32
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