There is a vector space E , which is also finite-dimensional, and it contains subspaces V1 and V2. I need help proving that:
1. ( V1 V2)0 = V10 + V20
2. ( V1+ V2)0 = V10 V20

Thanks!

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– user9413 Jun 13 '11 at 11:00
What did you try for yourself? – Asaf Karagila Jun 13 '11 at 11:54

I would try to prove first the second equality by element chasing, then one inclusion of the first (the inclusion of the space in the right hand side into the one on the left hand side to be more precise), and then finish computing dimensions.

For instance, as for the second equality,

$$v \in (V_1 + V_2)^\circ \ \Longrightarrow \ v\cdot w = 0 \qquad \text{for all } \qquad w \in V_1 + V_2 \ .$$

But vectors in this sum are of the form $w = w_1 + w_2$, for all possible $w_i \in V_i$, $i=1, 2$. Hence, $v\cdot w_1 = 0$ for all $w_1 \in V_1$ and $v\cdot w_2 =0$ for all $w_2 \in V_2$. Thus $v\in V_1^\circ \cap V_2^\circ$.

As for the first one, I would say

$$v \in V_1^\circ + V_2^\circ \ \Longrightarrow \ v = v_1 + v_2, \ \text{with}\ v_i \in V_i^\circ, \ i=1,2 \ .$$

And then compute what happens when you do $v\cdot w$ for such a $v$ and $w \in V_1 \cap V_2$.

$$\mathrm{dim} (V_1 \cap V_2)^\circ = \mathrm{dim} E - \mathrm{dim} (V_1 \cap V_2) \ .$$

And then continue with

$$\mathrm{dim} (V_1^\circ + V_2^\circ) = \mathrm{dim}V_1^\circ + \mathrm{dim}V_2^\circ - \mathrm{dim} (V_1^\circ \cap V_2^\circ ) = \dots$$

And, at a certain point, I would use the already proven equality (2).

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It seems my idea of proving this is wrong. – user9413 Jun 13 '11 at 12:00
@ Agustí, I think you need to mention why $(V_1 \cap V_2)^o \subseteq V^o_1 + V^o_2$ is true... but it is easy to see of your demonstration. – Hiperion Oct 22 '12 at 4:04