Extension of Sylvester's Theorem $\mathbf{h}_i\in\mathbb{C}^{M}$ are column vectors $\forall i=\{1, 2, \cdots, K\}$.
$q_i\in\mathbb{R}_+$ are scalars $\forall i=\{1, 2, \cdots, K\}$
$\lvert\bullet\rvert$ denotes determinant of a square matrix or Euclidean norm of a vector according to the context. 
From Sylvester's theorem, it's trivial to show that $\lvert\mathbf{I}_M+\mathbf{h}_1q_1\mathbf{h}_1^{\text{H}}\rvert=1+q_i\left|\mathbf{h}_1\right|^2$. Is it possible to extend the theorem to simplify $\lvert\mathbf{I}_M+\sum_{i=1}^K\mathbf{h}_iq_i\mathbf{h}_i^{\text{H}}\rvert$?
P. S. It's part of a problem involving multiple access channels and I am not sure if a definite solution exists. 
 A: You can use  the matrix determinant lemma (see http://en.wikipedia.org/wiki/Matrix_determinant_lemma#Statement) and the induction by $K$:
$$
\lvert\mathbf{I}_M+\sum_{i=1}^K\mathbf{h}_iq_i\mathbf{h}_i^{\text{H}}\rvert = 
\lvert(\mathbf{I}_M+\sum_{i=1}^{K-1}\mathbf{h}_iq_i\mathbf{h}_i^{\text{H}}) + \mathbf{h}_Kq_K\mathbf{h}_K^{\text{H}}\rvert = 
(1+\sum_{i=1}^{K-1}\mathbf{h}_i^{\text{H}}q_i(\mathbf{I}_M+\sum_{i=1}^{K-1}\mathbf{h}_iq_i\mathbf{h}_i^ {\text{H}})^{-1}\mathbf{h}_i)\cdot\lvert \mathbf{I}_M+\sum_{i=1}^{K-1}\mathbf{h}_iq_i\mathbf{h}_i^ {\text{H}}\rvert,
$$
but I am not sure that it will help you ...
A: Yes, the matrix determinant lemma is the key. Define $H= [h_1,\dots,h_K]$, and $Q = \begin{bmatrix}q_1 \\ & q_2 \\ & & \ddots \\& & & q_K  \end{bmatrix}$ then applying the lemma,
$det(A+UV^T) = det(I + V^TA^{-1}U) det(A)$
Now, apply with $A = I_M$, $U=HQ$ and $V = H^T$
$det(I_M + HQH^T) = det (I_K + H^THQ)$
This expression can be simplified further if the vectors $h_i$ are orthonormal.
EDIT: Corrected an error and showed necessary steps.
