# What about dimension of $W_1 \cap W_2$?

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?

• 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. – Gerry Myerson May 12 '16 at 12:32
• @ Gerry Myerson what is the use of field Z/6Z here? – Arun Sharma May 12 '16 at 12:38
• Is $W_1 \bigcup W_2 = V$ true? – Arun Sharma May 12 '16 at 12:47
• 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. – 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:

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

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