# Direct sum of two subspaces of a vector space

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?

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

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 $\dim(U \cap W)$.