Eigenvalues of $\sum_i A_i P_i$ with $(A_i)_{jk}=\delta_{ij}\delta_{ik}$ and $P_i$ orthogonal projections Assume $$C=\sum_{i=1}^nA_iP_i$$ where each $P_i$ is an $n\times n$ orthogonal projection matrix ($P_i=P_i^\top$ and $P_i=P^2_i$) and $A_i$ is an $n\times n$ matrix of zero elements except the $(i,i)$-th element equal to one. 
Can I say something about the eigenvalues of $C$?
 A: The matrix $A_iP_i$, is a matrix containing a lot of zeros, and its $i$-th row is equal to the $i$-th row of $P_i$. That is, the resulting matrix has rows all with Euclidean norm equal to $1$. This implies $\|C\|_2 \le \sqrt n$:
$$
\|Cx\|_2^2 =\sum_i (c_i^Tx)^2\le \sum_i \|c_i\|^2 \|x\|^2 = n \|x\|^2,
$$
where $c_i$ is the $i$-th row of $C$.
This bound is realized if all rows of $C$ are the same up to factors with absolute value $1$.
This shows that all eigenvalues of $C$ are in the circle $\{z\in \mathbb C:\ |z|\le \sqrt n\}$.
This bound is sharp as the matrix $C=(c_{ij})$ with $c_{ij}=\frac1{\sqrt n}$ shows.
Using Gershgorin theorem, one can prove inclusions for eigenvalues.
The radius of the Gershgorin circle $i$ can be estimated by 
$$
\sum_{j\ne i} |c_{ij}| \le \sqrt{n-1} \left(\sum_{j\ne i} |c_{ij}|^2 \right)^{1/2} = \sqrt{(n-1)(1-|c_{ii}|^2)}.
$$
Then the eigenvalues of $C$ are inside the circles with center $c_{ii}$ and radius as above. If more is known about the entries of $C$, the bound can be improved.
