Let $\{u_1,u_2,..., u_n\}$ be an orthonormal basis of $\mathbb {C^n}$ as column vectors. Let $M=(u_1,u_2,...,u_{k})$ and $N=(u_{k+1},u_{k+2},...,u_{n})$ and $P$ be a $k \times k$ diagonal matrix with diagonal entries $\alpha_1,\alpha_2,..., \alpha_k \in \mathbb R$. Then which of the following statements are true?
(A) Rank($MPM^*)=k$ whenever $\alpha_i \neq \alpha_j$, $1\leq i,j \leq k.$
(B)Trace $(MPM^*)=\sum_{i=i}^k\alpha_i$
(C)Rank $(M^*N)=\min(k,n-k)$
(D)Rank$(MM^*+NN^*)<n$
Let $ u_1= \begin{bmatrix} u_{11} \\ u_{21} \\ \vdots \\ u_{n1} \end{bmatrix}$, $ u_2= \begin{bmatrix} u_{12} \\ u_{22} \\ \vdots \\ u_{n2} \end{bmatrix},\dots,$ $ u_n= \begin{bmatrix} u_{1n} \\ u_{2n} \\ \vdots \\ u_{nn} \end{bmatrix}.$ By Matrix multiplication,
I got $MPM^*$ $ = \begin{bmatrix} \alpha_1. |u_{11}|^2&0 &\dots &0 \\ 0 &\alpha_2. |u_{22}|^2&\dots &0\\ \vdots \\ 0&0&\dots &\alpha_k. |u_{kk}|^2 \end{bmatrix}$ If any any of $\alpha_j=0$ then, (A) is false. Trace$(MPM^*)=\sum_{i=1}^k|u_{ii}|^2\alpha_i=\sum_{i=1}^k\alpha_i(\because$ $\langle u_i,u_j\rangle=\delta_{ij})$ . So,(B) is true. Hence (B) is the correct answer. Is there any theoretical approach which reduce the calculations? How do I prove (C) and (D) are false?