is there a case where the intersection of two distinct subspaces of a direct sum is not {0}.

Question

is there a case where the intersection of two distinct subspaces of a direct sum is not {0}.

ie) If U1, U2, . . . , Uk are subspaces of a vector space V and V = U1 ⊕ U2 ⊕ · · · ⊕ Uk, then Ui ∩ Uj = {0} for any two distinct i, j ∈ {1, 2, . . . , k}.

I know that the sum of the subspaces intersected with some Uj is {0}, but I am not sure if Ui ∩ Uj = {0} for any distinct i and j

Answer 1

Yes, this is true: If $V= U_1 \oplus U_2 \oplus \dots \oplus U_k$ then by definition for any $i>1$ we have $$(U_1 + \dots U_{i-1}) \cap U_i = \{0\}$$ From that it follow for any $i \ne j$, let's say w.l.o.g. that $j<i$ that $$(U_1 + \dots + U_j + \dots + U_{i-1}) \cap U_i = \{0\}$$ but $(U_1 + \dots + U_j + \dots + U_{i-1}) \supset U_j$ because the elements $0 + \dots + 0 + v_j + 0 + \dots +0$ with the $v_j \in U_j$ are elements of $(U_1 + \dots + U_j + \dots + U_{i-1})$. Therefore also $U_j \cap U_i = \{0\}$.

Answer 2

Yes this is clearly true, but the pairwise intersection condition is rather weaker than the direct sum condition. The direct sum condition says that every vector in the sum is so in a unique manner: there is only one way to write it as a sum of vectors each in the respective subspace. But if $v$ were a nonzero vector in both $U_i$ and $U_j$, then you could either write is as a sum of vectors that are $v$ in $U_i$ and zero on other components, or the same but with $v$ on the $U_j$ component, and this shows that the sum in this case cannot be direct.