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Let $V$ be vector space of dimension 6 over field $Z/7Z$. Let $W_1$ and $W_2$ be two subspaces of $V$. Let $\dim(W_1)=4$, $\dim(W_2)=3$. what about dimension of $W_1 \cap W_2 $? I know the formula $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$. I tried with this formula but could not find the proper answer. I am confused with, is dimension ($W_1 \cap W_2$) one? or greater than equal to one or anything else?

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    $\begingroup$ Of course, the dimension can be as big as 3, if $W_2\subset W_1$. It can't be bigger; it can be as small as 1, but no smaller; and it can be anything in between, that is, it can be 2. $\endgroup$ – Gerry Myerson May 12 '16 at 12:32
  • $\begingroup$ @ Gerry Myerson what is the use of field Z/6Z here? $\endgroup$ – Arun Sharma May 12 '16 at 12:38
  • $\begingroup$ Is $W_1 \bigcup W_2 = V $ true? $\endgroup$ – Arun Sharma May 12 '16 at 12:47
  • $\begingroup$ Z / 6 Z isn't a field, nor does it appear in the question, Arun, and the union of two subspaces is never a vector space, unless one of the subspaces contains the other. $\endgroup$ – Gerry Myerson May 15 '16 at 12:46
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Note that $W_1 \subseteq W_1 + W_2 \subseteq V$. So $4 = \dim(W_1) \leq \dim(W_1 + W_2) \leq \dim(V) = 6$. Putting in the information we know into your formula gives that:

$$\dim(W_1 + W_2) = 7 - \dim(W_1 \cap W_2).$$

Using our inequality with this gives:

$$4 \leq 7 - \dim(W_1\cap W_2) \leq 6.$$

This inequality simplifies to:

$$ 1 \leq \dim(W_1 \cap W_2) \leq 3.$$

Without more information, we can't say more.

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  • $\begingroup$ can we say $(W_1 \bigcup W_2)= V$ by the above information? $\endgroup$ – Arun Sharma May 12 '16 at 17:10
  • $\begingroup$ $W_1 \cup W_2$ is not a subspace unless either $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$. $\endgroup$ – Ken Duna May 12 '16 at 17:11
  • $\begingroup$ So in fact, $W_1 \cup W_2$ cannot be $V$. $\endgroup$ – Ken Duna May 12 '16 at 17:12
  • $\begingroup$ Yes true...Thank you $\endgroup$ – Arun Sharma May 12 '16 at 17:13

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