Show that $V$ is the internal direct sum of $U$ and $W$

Question

Let $V$ be finite-dimensional and let $U$ and $W$ be subspaces of $V$. Suppose $U \cap W = \{ 0\}$ and $\dim(U) + \dim(W) = \dim(V)$. I want to show that $V$ is the internal direct sum of $U$ and $W$.

I'm told that $V$ is the internal direct sum of $U$ and $W$ if $V=U+W$ and $U \cap W=\{0\}$.

I'm already told that $U \cap W =\{0\}$ so I'm left to show that $V=U+W$.

I know that $$\dim(V)=\dim(U)+\dim(W)=\dim(U)+\dim(W)-\dim(U \cap W)=\dim(U+W)$$

Thus $V$ is isomorphic to $U+W$.

Is this enough to show that any $v \in V$ can be written as $u+w$ or do I need to do more?

Answer 1

Since $U+W$ is a subspace of $V$ and $$ \dim(U+W)=\dim V $$ then $U+W=V$: they aren't just isomorphic, but actually equal. If you need more reasons: a basis for $U+W$ has the same number of elements as a basis of $V$, hence it is a basis for $V$.

Answer 2

As $U\subset V$ and $W\subset V$ you have $U+V\subset V,$ so it is indeed enough to show that any $v\in V$ can be written as $u+w$ with $u\in U$ and $w\in W.$