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Let $K$ be a field and $V$ be a vector space on $K$, $U$ and $W$ be two subspaces of $V$. Then if $\operatorname{codim}U$ and $\operatorname{codim}W$ are both finite, then prove the formula: $$ \operatorname{codim}(U\cap W)+\operatorname{codim}(U+W)= \operatorname{codim}U +\operatorname{codim}W. $$

Here the dimension of the space $V$ is without restriction, that's to say, the dimension of space $V$ could be infinite or finite.

(I think if we think about the quotient space, then the problem will be easy, but I still can not solve it.)

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Yes, passing to a suitable quotient is the key.

The hypothesis means that $V/U$ and $V/W$ are finite dimensional.

Now you can consider the map $$ T\colon V\to (V/U)\oplus(V/W) \qquad v\mapsto (v+U,v+W) $$ which is linear and has kernel $U\cap W$. Therefore $V/(U\cap W)$ is isomorphic to a subspace of $(V/U)\oplus(V/W)$, so it is finite dimensional.

Thus you can use Grassmann's formula in $V'=V/(U\cap W)$ for $U'=U/(U\cap W)$ and $W'=W/(U\cap W)$ and $U'+W'=(U+W)/(U\cap W)$.

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  • $\begingroup$ Thanks for you help, and could you please show me that how to deal with dim( U' ∩ W' ) ? $\endgroup$
    – painday
    Apr 9, 2017 at 1:55
  • $\begingroup$ @painday It's zero $\endgroup$
    – egreg
    Apr 9, 2017 at 7:29
  • $\begingroup$ OK, I understand it now. At first I try to find the relation between the bases of the four quotient spaces, and only prove an inequality. Your idea is better than mine. $\endgroup$
    – painday
    Apr 9, 2017 at 8:38

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