Given U And W are distinct four dimensional subspaces of a vector space V of dimension 6 . Find the possible dimension of U $ \cap $ W ..
Attempt : dim(U $\cap$ W) = 8 - dim(U + W)
Now Dimension of U and W are 4. At maximum dim(U + W) can be of 6 dimension.Other possibilities are 5 and 4 . But in textbook they have omitted answer corresponding to 4 .Which i don't seem to know why ? Thanks for help.
$\endgroup$3 Answers
$\begingroup$Notice that $U\cap W$ is a subspace of $U$ and $W$ so
$$\dim(U\cap W)\le 4$$ but since $\dim(U\cap V)=4\iff V=U\cap V=U $ hence $$\dim(U\cap W)\le 3$$
- if $\dim(U\cap W)=0$ then the sum $U+ W$ is direct but in this case $$\dim(U+W)=\dim U+\dim W>6$$ which's impossible.
- if $\dim(U\cap W)=1$. Let $U\cap W=\operatorname{span}(e_1)$ and let $(e_1,u_2,u_3,u_4)$ a basis for $U$ and $(e_1,w_2,w_3,w_4)$ a basis for $W$ hence we see easily that $\dim (U+W)=7>6$ which's impossible.
- The cases $\dim(U\cap W)=2$ or $\dim(U\cap W)=3$ are possible using the same method as on the last point.
Conclusion The possible values of $\dim(U\cap W)$ are $2$ and $3$.
$\endgroup$ $\begingroup$If they are distinct then there cannot be 4 since if the dimension is 4, $U\cap W=U=W$.
$\endgroup$ 1 $\begingroup$Note. If $\dim(U+W) = 4$, we would find $U = U+W = W$ but we were told $U\neq W$. We could also say that $\dim(U+W) = 4$ implies $U = U \cap W = W$ (as Suzu Hirose mentions).
Recall that $$\dim(U+W) = \dim U +\dim W - \dim(U\cap W).$$ The possible values for $\dim(U+W)$ are $5$ and $6$. Using that $$\dim(U\cap W) = 4 + 4 - \dim(U+W)$$ we see that the possible values are $$\dim(U\cap W) = 8 - 5 = 3$$ $$\dim(U\cap W) = 8 - 6 = 2$$
$\endgroup$ 0
