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I need to prove that if $P_i:V\to V$ are projections ($P^2=P$) such that $\sum_{i=1}^kP_i=I$. Prove $P_iP_j=\delta_{ij}P_j$ where $\delta_{ij}$ is $0$ if $i\neq j$ and $1$ if $i=j$.

The hint says to use the fact that $\dim(\mathrm{Im}P_i)=\mathrm{tr}(P_i)$ but I don't really see how that helps me.

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  • $\begingroup$ Can we use the following as a fact?: "Let $W_1,\dotsc, W_k$ be subspaces of $V$. Then, $\mathrm{dim}(W_1)+\dotsb\mathrm{dim}(W_k)=\mathrm{dim}(V)$ if and only if $V=W_1\oplus\dotsb\oplus W_k$ (direct sum)". $\endgroup$
    – shall.i.am
    Jan 17 '17 at 5:43
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Let $W_i$ be the image of $P_i$. The linear transformation

$$\bigoplus\limits_{i=1}^k W_i \rightarrow V$$ $$(w_1, ... , w_k) \mapsto w_1 + \cdots + w_k$$

is surjective by hypothesis. These vector spaces have the same dimension, since

$$\sum\limits_{i=1}^k \textrm{Dim } W_i = \sum\limits_{i=1}^k \textrm{Trace } P_i = n$$

So this must be an isomorphism, i.e. $V$ is the internal direct sum of the $W_i$.

This means that for $v \in V$, the expression of $v$ as a sum $w_1 + \cdots + w_k$, with $w_i \in W_i$ is unique. Each element $w_i$ can be explicitly described as $P_i(v)$.

Let's show that $P_j \circ P_1 = 0$ for $j = 2, ... , n$. If $v \in V$, then

$$P_1(v) = P_1 \circ P_1(v) + P_2 \circ P_1(v) + \cdots + P_n \circ P_1(v)$$

But we can also express the element $P_1(v)$ just by using the first term in that sum. By the uniqueness we just mentioned, we must have $P_j \circ P_1(v) = 0$ for all $j \neq 1$.

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