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?

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.