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Hope this isn't a duplicate.

I was trying to solve the following problem :

Let V and W be two vector subspaces of the vector space $\Bbb R^{10}$ over $\Bbb R$ of dimension 7 and 9 respectively. Then what can be said about the dimension of $V \cap W$?

(a) is 7

(b) is 6

(c) lies between 6 and 7

(d) is less than 6

I could not figure out anything on the problem. Thanks in advance for help.

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    Have you seen/covered a formula relating the dimensions $\dim V_1$, $\dim V_2$, $\dim (V_1\cap V_2)$ and $\dim (V_1+V_2)$ where $V_1$, $V_2$ are two subspaces of a given space? That will come in handy here. – Jyrki Lahtonen Feb 10 '18 at 05:15
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    Is it possible under some condition on $V,W$ to have $V\cap W=V$? If so,(b) and (d) are immediately ruled out. Is that condition on $V,W$ always fulfilled? If not, (a) is ruled out as well. – Hagen von Eitzen Feb 10 '18 at 05:20

2 Answers2

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$dim V +dim W=dim (V\cap W)+ dim( V+ W)$, where $V+W$ is the smallest subspace containing $V$ and $W$.

So, $16=dim(V\cap W)+dim(V+W)$.

But $9\le dim(V+W)\le 10$...

So, $6\le dim(V\cap W)\le7$.

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$V\cap W\subset V \Rightarrow dim (V\cap W) \le dim(V) = 7$.

From $dim(V)+dim(W)=dim(U+V)+dim(V\cap W)$, we have
$7+9=dim(V+W)+dim(V\cap W)$.

How $dim(V+W)\le 10$, so
$dim(V\cap W) \ge 7+9-10 = 6$

Therefore $dim(V\cap W)\in\lbrace{6, 7\rbrace}$.

If we take $V\subset W$, we have $dim(V\cap W)=dim(V)=7$.

If we take $V= span\lbrace e_1, \ldots, e_7\rbrace$ and $W= span\lbrace e_2, \ldots, e_{10}\rbrace$, we have
$V\cap W = span\lbrace e_2, \ldots, e_7\rbrace \Rightarrow dim(V\cap W)=6$.

Therefore $dim(V\cap W)$ lies between 6 and 7

Tiago Emilio Siller
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