I am studying an exercise in Nomizu's Fundamentals of Linear Algebra, and the following question:
Assume that an inner product space $V$ is the direct sum $V = V_1 \oplus \cdots \oplus V_k$, where the subspaces $V_i$ are orthogonal to each other. Show that $(V_i)^{\perp}$ is the direct sum of all $V_j$'s such that $j \neq i$. Let $P_i$ be the projection of $V$ onto $V_i$. Show that
- $I = P_1 + \cdots + P_k$
- $\langle P_i(\alpha),\beta \rangle = \langle \alpha, P_i(\beta) \rangle$ for all $\alpha, \beta \in V$.
- $P_i^2 = P_i$ for each $i$ and $P_i P_j = 0$ for $i \neq j$.
My question is: What happens if the subspaces are not orthogonal? Do these claims still hold true?
For example, since $V$ is still a direct sum of $V_1, \ldots, V_k$, then for each vector $\alpha \in V$ I can write it uniquely as a sum of some $v_1 \in V_1, \ldots, v_k \in V_k$: $$\alpha = \sum_{i=1}^k v_i$$ Then if I define the projections $P_1, \ldots, P_k$ by: $$P_j \alpha = v_j$$ i.e. the map which simply extracts the $j^{\textrm{th}}$ component, do these projections not satisfy conditions (1), (2) and (3) stated above?