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I proved that for subspaces $U_i$ of $V$ the inequality $\mathrm{dim}(U_1 + ...+U_m) \le \mathrm{dim}(U_1) + ... + \mathrm{dim}(U_m)$ holds.

I proved it as follows (would you please tell me if my proof is correct): For two subspaces $U,W$ we know that $\mathrm{dim}(U+W) = \mathrm{dim}U + \mathrm{dim}W - \mathrm{dim}(U \cap W)$ so that the claim is true for two subspaces. Then the claim follows because $U_1 + ... + U_{m-1}$ and $U_m$ are two subspaces hence $\mathrm{dim}(U_1 + ...+U_{m}) \le \mathrm{dim}(U_1 + ...+U_{m-1}) + \mathrm{dim}(U_m)$. Apply the argument again to get $ \mathrm{dim}(U_1 + ...+U_{m-1}) \le \mathrm{dim}(U_1 + ...+U_{m-2}) + \mathrm{dim}U_{m-1}$. Apply the argument $m$-times to obtain the desired inequality.

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  • $\begingroup$ You can look at it this way: Given $m$ bases for each subspace $B_1,\ldots,B_m$, the sum on the right hand side is $\sum |B_i|$. On the other hand, $\bigcup_i B_i$ generates the sum, but there is no guarantee that the set will be independent. Thus, upon eventual removal of one or more vectors, we will have an inequality. $\endgroup$
    – Pedro
    Aug 8, 2013 at 15:21
  • $\begingroup$ It seems to be correct for me. You can also argue by the basis elements, but it is not nicer. $\endgroup$
    – R.T.
    Aug 8, 2013 at 15:21
  • $\begingroup$ Ups, sorry @PeterTamaroff, I did not mean you... $\endgroup$
    – R.T.
    Aug 8, 2013 at 15:21
  • $\begingroup$ @Q.W. I know. ${}{}{}$ $\endgroup$
    – Pedro
    Aug 8, 2013 at 15:22

1 Answer 1

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It's correct. You can use an induction over $m$ argument if you want to avoid the "applying the argument $m$ times".

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