$P=\frac{1}{n}\mathbf{1}\mathbf{1}^T$($\mathbf{1}=[1,1,\dots,1]^T$), $I\in R^n$ is a indentity matrix, $C$ is a positive definite symmetric matrix. Is $P+(I-P)C$ positive definite? How to prove that?
With a random positive definite symmetric matrix $C$, I did 10 million experiments. All eigenvalues of $P+(I-P)C$ are with positive real parts.
Note that even if the eigenvalues of this matrix is nonnegative, it will only be positive definite if $$ P + (I-P)C $$ is also symmetric. However, this is only the case when $P$ and $C$ commute, i.e., $[P,C] = 0$.
Here's a $2 \times 2$ counterexample $$ P = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix} \qquad C = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} $$ Then $$ P + (I - P) C = \begin{pmatrix} 1 & 1/2 \\ 0 & 1/2 \end{pmatrix}. $$ Note this is not symmetric and so cannot be positive definite, it does however have positive eigenvalues $\{1, 1/2\}$.