Condition for Direct Sum of Two Subspaces

The next result gives a simple condition for testing whether a sum of two subspaces is a direct sum.

Direct sum of two subspaces

Suppose $U$ and $W$ are subspaces of $V$. Then

$$U+W\text{ is a direct sum}⟺U\cap V=\{0\}$$

The symbol $⟺$ means “if and only if ”; this symbol could also be read to mean “is equivalent to”.

Proof.

First, we show that if we assume $U+W$ is a direct sum, we must have $U\cap W=\{0\}$.

Suppose $U+W$ is a direct sum. Suppose some element $v$ is in both $U$ and $V$, or $v\in U\cap W$. Then, we can write the zero vector as $0=v+(-v)$, where $v\in U$ and $-v\in W$. Since $U+W$ is a direct sum, the only way for us express $0$ as a sum of vectors from $U$ and $W$ is if both vectors are $0$. Thus, we have $v=0$. Thus, $U\cap W=\{0\}$, completing the proof in one direction.

Now we want to prove the other direction; assuming that $U\cap W=\{0\}$, we want to show that $U+W$ is a direct sum.

Suppose $u\in U,w\in W$, and

$$0=u+w.$$

To complete the proof, we only need to show that $u=w=0$, which we showed above for the condition for direct sums. The equation above implies that $u=-w$. Since we are dealing with subspaces that are closed under scalar multiplication, $u=-w\in W$ means that $u\in W$ as well, which in turn means $u\in U\cap W$. Since our beginning assumption was $U\in W=\{0\}$, we must have $u=0$, which by the equation above implies that $w=0$, completing the proof.

The result above deals only with the case of two subspaces. When asking about a possible direct sum with more than two subspaces, it is not enough to test that each pair of the subspaces intersect only at $\{0\}$. To see this, consider the the non-direct sum example above. In that non-example of a direct sum, we have $V_1\cap V_2=V_1\cap V_3=V_2\cap V_3=\{0\}$.