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Let $ U, W $ be subspaces of vector space $ V $. Show that $$ \text{Dim}( U) + \text{Dim} (W) = \text{Dim}(U+W) + \text{Dim} ( U \cap W) $$

Given Hint: Show that map given by $ L:U \times W \to V $ given by $ L(u,w) = u-w $ is a linear map.

I can show that $ L:U \times W \to V $ given by $ L(u,w) = u-w $ is a linear map. Also I know that dimention of $U\times W$ is $ \text{Dim}( U) + \text{Dim} (W) $. What do I do next? Any hints?

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Show that $\text{Dim}(U+W)= \text{Dim}( U) + \text{Dim} (W) -\text{Dim} ( U \cap W)$. –  A.D Jan 29 '13 at 19:12
    
Isn't that same? I think Dimension of $u-w$(image of map), will be $\text{Dim}(U+W)$, will the dimension of kernel of transformation will have $\text{Dim}(U\cap W) $ ?? –  hasExams Jan 29 '13 at 19:13
    
You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem –  1015 Jan 29 '13 at 19:13
    
@julien am I right?? will they share the same null space? –  hasExams Jan 29 '13 at 19:14
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Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $L\cap W$. –  1015 Jan 29 '13 at 19:16
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2 Answers

up vote 4 down vote accepted

The range of the map $L$ is clearly $U+W$.

Now the nullspace is: $$ \mbox{Ker} \;L=\{(u,w)\;;\; u=w\in U\cap W\} $$ so it is isomorphic to $U\cap W$ via the map $v\longmapsto (v,v)$.

By the rank-nullity theorem applied to $L$, we find: $$ \mbox{rank}\;L+\mbox{null}\;L=\mbox{dim}\;(U\times W) $$ which yields the desired formula, which is sometimes called Grassmann formula..

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Hint: assume $\dim ( U \cap W)=k$ and let $B_{ ( U \cap W)}={[x_1,...x_k]}$ then expand this base for W and U let these base are $$B_U={[x_1,...x_k,y_1,..,y_n]}$$ $$B_w={[x_1,...x_k,z_1,..,z_m]}$$ then prove C={$x_1,...x_k$,$y_1$,..,$y_n$,$z_1$,..,$z_m$} is base for W+U you can show C generate W+U and C is independent.

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i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!! –  hasExams Jan 29 '13 at 19:23
    
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it. –  Sam DeHority Jan 29 '13 at 19:24
    
@DoctorBatmanGod context :P –  hasExams Jan 29 '13 at 19:25
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