# How can we make sure that for three direct sums, zero can be written uniquely as a sum?

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.

• What does it mean for a collection of subspaces to "form a direct sum"? – lulu Jan 5 at 21:20
• @lulu The only way to write $0$ as a sum $u_1+\dots +u_m$ where each $u_j \in U_j$ is by taking each $u_j$ equal to zero – user736276 Jan 5 at 21:25
• @lulu I guess what your'e going for is the union of all the sets is equal to the zero vector – user736276 Jan 5 at 21:30
• So, in $\mathbb R^3$, if one space is the $x-$axis and another is the $y-$axis, you'd say these "form a direct sum"? – lulu Jan 5 at 21:30
• Regardless of the terminology..cho0se a basis of each subset and then ask whether the union of those basis vectors is linearly independent. Your test is insufficient, as (for example) you can take three lines through the origin in the plane. These are pairwise independent but, clearly, the triple is not independent. – lulu Jan 5 at 21:32

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!