If $S_1,S_2,S_3$ have direct sum, then, $S_1 = (S_1+S_2) \cap (S_1+S_3)$?

Question

If $S_1,S_2,S_3$ have direct sum, then, $S_1 = (S_1+S_2) \cap (S_1+S_3)$?

I tried this way:

Call $\mathcal F_1$ a family of vectors that generates $S_1$.

Call $\mathcal F_2$ a family of vectors that generates $S_2$.

Call $\mathcal F_3$ a family of vectors that generates $S_3$.

Then we have $S_1$=$span \{\mathcal F_1,\mathcal F_2\} \cap span\{\mathcal F_1,\mathcal F_3\}$?

So we know that the subspaces have direct sum, so this mean that: $S_2 \cap S_3 = \{0_v\}$

Thus, this is true... or that's what I guess... How does it look?

Answer 1

Let $V=S_1\oplus S_2\oplus S_3$ be the direct sum. Let $v\in (S_1+S_2)\cap(S_1+S_3)$ and write $v=v_1+v_2+v_3$ with $v_i\in S_i$ uniquely determined. Since $v\in S_1+S_2$, we get $v_3=0$. On the other hand, $v\in S_1+S_3$ implies $v_2$. Consequently, $v=v_1\in S_1$.