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.
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 $j \neq 1$.