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Suppose $U$ and $W$ are distinct four dimensional subspaces of a vector space $V$, where $dim\ V=6$. Find the possible dimensions of subspace $U \cap W$
asked Dec 22, 2017 by randomisation (1,920 points)  

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$dim\ U = 4 \\ dim\ W = 4 \\ dim V = 6 $

As $U$ and $W$ are distinct,

$dim\ (U+W) > dim\ U, dim\ W \\ \implies dim\ (U+W) > 4$

Also, as $U + W$ is a subspace of $V$,

$dim\ (U+W) \leq dim\ V \implies dim\ (U+W) \leq 6$

Thus, we get $dim\ (U+V) = 5$ or $6$

Now, $dim\ (U \cap W)$

                 $= dim U + dim W - dim (U+W) \\ = 4 + 4 - 5 \text{ or } 4+4-6 \\ = 3 \text{ or } 2.$
answered Dec 22, 2017 by randomisation (1,920 points)  
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