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Let $U,W$ be subspaces of a vector space $V$. Show that $$\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$$ Hint: Show that the map given by $L:U×W\to V$ given by $L(u,w)=u-w$ is linear.

I can show that $L:U×W\to V$ given by $L(u,w)=u-w$ is a linear map. I also know that the dimension of $U×W$ is $\dim(U)+\dim(W)$. What do I do next? Any hints?

Parcly Taxel
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hasExams
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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: http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem – Julien 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 $L\cap W$. – Julien Jan 29 '13 at 19:16
  • thanks!! i see ... the reason why $u-w$ was given instead of $u+w$ – hasExams Jan 29 '13 at 19:18
  • Note that the map $(u,w)\longmapsto u+w$ would have worked too. – Julien Jan 29 '13 at 19:20
  • @julien try adding your comment as answer. – hasExams Jan 29 '13 at 19:21
  • I'm positive this is a duplicate but I can't find the duplicate :( – rschwieb Jan 29 '13 at 19:34
  • If one knows the formula for the dimension of the quotient space, all one has to do, is to show that the canonical map $T : U \longrightarrow \frac {U + W} {W}$ defined by $x \mapsto x + W,$ is a linear surjection with kernel $U \cap W.$ Once we show that we are done by the virtue of first isomorphism theorem. – Anil Bagchi. Jun 02 '23 at 07:01

4 Answers4

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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..

Julien
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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.

guest196883
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M.H
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Does the following suffice to prove it?

If we proceed to show the symmetric differences:

$dim(U \times W) = [ dim(U \setminus W) + dim(U \cap W) ] + [ dim(W \setminus U) + dim(U \cap W) ] $,

And we have:

$dim(U \oplus W) = dim(U \setminus W) + dim(W \setminus U) + dim(U \cap W)$

Thus,

$dim(U \times W) = dim(U \oplus W) + dim(U \cap W)$

freehumorist
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By the second isomorphism theorem for vector spaces, $U/(U\cap W)\cong U+W/W$. For any finite dimensional vector space U with a subspace W, $\dim U/W =\dim U-\dim W$, which is clear if one takes a basis for U: $\{u_1,...,u_r,u_{r+1},...,u_n\}$ where $u_1,...,u_r$ form a basis of W. Then $\dim U-\dim (U\cap W)=\dim(U+W)-\dim W,$ which implies $\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$.

Divide1918
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