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Let $V$ be a vector space so that $V = U + W$ and suppose $\dim V = \dim U + \dim W$. Show, carefully, that $V = U \oplus W $.

Attempt:

Well, since

$$ \dim( \underbrace{U + W}_{= V} ) = \underbrace{\dim U + \dim W}_{= \dim V \; \; \text{given}} - \dim ( U \cap W) $$

Then, $\dim( U \cap W) = 0$, which means that $U \cap W = \{ 0 \}$ thus the result follows.

IS there another way to solve this problem without recurring to the inclusion exclusion formula?

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  • $\begingroup$ I would say your solution is, in some sense, optimal. $\endgroup$
    – Pedro
    Oct 6, 2016 at 17:44

1 Answer 1

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First, suppose $U \cap W$ is 1-dimensional, generated by a nonzero vector $x$. Let $U_{x}$ be a complement of $\mbox{span$\{x\}$}$ in $U$, and $W_{x}$ a complement of $\mbox{span$\{x\}$}$ in $W$.

We would then have $$ V = U_x \oplus \mbox{span$\{x\}$} \oplus V_{x}, \quad \quad (1) $$ and yet $$ \dim U = 1 + \dim U_{x}, \quad \dim W = 1 + \dim W_{x}, $$ which yields $$ \dim V = \dim U + \dim V = \dim U_{x} + 2 + \dim V_{x}, $$ contradicting (1).

Now proceed by induction on $\dim(U \cap W)$.

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