If $\exists x_i,y_i \in C^n $ such that $A=\sum_{i=1}^{k}x_i y_i^*$ where $^*$ is conjugate transpose. Then what can be said about rank of $A$?
I was able to show that if rank of $A$, $r(A)=k$ then $\exists x_i,y_i \in C^n $ such that $A=\sum_{i=1}^{k}x_i y_i^*$ as follow:
$r(A)=k \rightarrow \; col(A)=span\{v_1, v_2\ldots v_k\}$
Then $A=AI_n= A[e_1\; e_2 \; \ldots e_n]= [Ae_i \; Ae_2\ldots \;Ae_n] = \big[\sum_{j=1}^{k}m_1^jv_j\; \; \sum_{j=1}^{k}m_2^jv_j\; \ldots\;\sum_{j=1}^{k}m_n^jv_j\big]= \sum_{j=1}^{k}\big[m_1^jv_j\; m_2^jv_j\ldots\; m_n^jv_j \big]= \sum_{j=1}^{k} v_jm_j^*$ where
$m_j=\big[\overline{m_1^j} \; \overline{m_2^j}\ldots\; \overline{m_n^j}\big]^T$
What can I say about converse ?
All we can say is that the rank of $A$ is at most $k$.
One argument is as follows: let $X = \pmatrix{x_1 & \cdots & x_k}$ and $Y = \pmatrix{y_1 & \cdots & y_k}$. Then by block-matrix multiplication, we have $A = XY^*$ . Now it suffices to note that $r(XY^*) \leq \min\{r(X),r(Y^*)\} \leq k$. The rank is at most $k$ because both $X$ and $Y$ have $k$ columns.