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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

  1. $I = P_1 + \cdots + P_k$
  2. $\langle P_i(\alpha),\beta \rangle = \langle \alpha, P_i(\beta) \rangle$ for all $\alpha, \beta \in V$.
  3. $P_i^2 = P_i$ 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?

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Good question! The answer is: no, claim #2 doesn't hold any more. (In addition, the assertion "Show that $(V_i)^\perp$ ..." is also no longer true.)

Let's just take each $V_i$ to be one-dimensional, spanned by fixed vectors $e_i$. Consider claim #2 with $\alpha=e_1$, $\beta=e_2$ and $i=1$: $$ \langle P_1(v_1),v_2 \rangle = \langle v_1,v_2 \rangle \quad\text{and}\quad \langle v_1,P_1(v_2) \rangle = \langle v_1,0 \rangle = 0, $$ so the two quantities can be equal only if $v_1$ and $v_2$ are orthogonal. Similarly, for $(V_1)^\perp$ to equal the direct sum $V_2 \oplus \cdots \oplus V_k$, it's necessary for each $v_j$ to actually be in $(V_1)^\perp$, which again requires $v_1$ to be orthogonal to $v_2$ etc.

What's happening is while the functions you define are perfectly good linear transformations, they're simply not orthogonal projections (so calling them "projections" at all is a little iffy). Consider $v_1=(1,0)$ and $v_2=(1,1)$ in $\Bbb R^2$, for example. You could make them orthogonal projections by changing the inner product, but that changes the inner product space into a different one (despite the fact that the underlying sets are the same).

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