If we have three subspaces of a vector space and want to determine if they form a direct sum,how can we determine this? I thought if $U_1,U_2,U_3$ were subspaces whose sum equals the vector space that we cannot check that
$U_1 \cap U_2=\{0\}$,
$U_2 \cap U_3=\{0\}$ and
$U_1 \cap U_3=\{0\}$
and conclude these subspaces form a direct sum.
Best Answer
The conditions for $U_1$, $U_2$ and $U_3$ so that $\textsf{V} = U_1 \oplus U_2 \oplus U_3$ are the following: \begin{align} (1) \quad & \textsf{V} = U_1 + U_2 + U_3 \\ (2) \quad & U_1 \cap (U_2+U_3) = U_2 \cap (U_1+U_3) = U_3 \cap (U_1+U_2) = \{0\} \end{align} I hope we agree with the first condition. Now, writte the zero vector as $$0 = u_1 + u_2 + u_3 \tag{$*$}$$ with $u_i$ living in $U_i$. Observe that the preceding equation can be expressed as $-u_1 = u_2+u_3$, and since the left hand side is in $U_1$ and the right hand side in $U_2+U_3$, we have $$-u_1 = u_2+u_3 = 0$$ so, $(*)$ reduces to $u_2+u_3 = 0$. Similarly, $u_2 = -u_3 \in U_2 \cap (U_1+U_3)$ implies that $u_2 = u_3 = 0$, and therefore, the only way to write $0$ as in $(*)$ is taking each $u_i = 0$.
In general, if $U_1,\dots,U_n$ are subspaces of a vector space $\textsf{V}$, $$\textsf{V} = \bigoplus_{i=1}^n U_i$$ if and only if $$\textsf{V} = \sum_{i=1}^n U_i \quad \textrm{and} \quad U_i \cap \sum_{j\neq i} U_j = \{0\}.$$ You can prove this using mathematical induction!