Does direct sum imply trivial pairwise intersection?

Question

Suppose $$V_{1},V_{2},\ldots,V_{m}$$ are subspaces of V.

If $$V=V_{1}\oplus V_{2}\oplus\cdots\oplus V_{m}$$

Then is it true that for any two subspaces $V_{p}$, $V_{q}$ where $p$ and $q\in \{1,2,...,m\}$, we have $V_{p} ∩ V_{q} =\{0\}$?

If that is true, how can we formally prove that?

Answer 1

Yes it is true.

A direct sum means that there exists a unique decomposition:

$$v=\sum_{i=1}^m \lambda_i v_i\in V=V_1\oplus\cdots V_m$$

where $v_i\in V_i$ for all $i\in\{1,\ldots,m\}$.

So if $v\in V_p\cap V_q$, then $v=\lambda_p v_p=\lambda_q v_q$.

You conclude by unicity that $\lambda_p=\lambda_q=0$ so $v=0$.

Answer 2

For $p \ne q$ we have

$V_p \cap V_q \subseteq V_p \cap \sum _{i=1,...,m, i\ne q}V_i$.

Since $V_p \cap \sum _{i=1,...,m, i\ne q}V_i=\{0\}$, the result follows.