Minimization of a determinant Let 
$
\mathbf{x}_m = \mathbf{z}_m - \alpha\mathbf{v} \in \mathbb{C}^N$, with $m=1,\dots,M$, be $M$ complex vectors such that


*

*$\mathbf{z}_m, m=1,\dots,M$ and $\mathbf{v}$ complex $N$-dimensional vectors,

*$\alpha$ a complex scalar.


Define the matrix $\mathbf{X}$ of dimension $N\times M$ as:
$$
\mathbf{X} = [\mathbf{x}_1|\mathbf{x}_2|\cdots|\mathbf{x}_M].
$$
How can I solve the following minimization problem:
$$
\min_{\alpha}\det(\mathbf{I} + \mathbf{X}^H\mathbf{S}^{-1}\mathbf{X})
$$
where


*

*$\mathbf{I}$ indicates the identity matrix of dimension $M \times M$,

*$\mathbf{S}$ is an invertible, Hermitian and positive definete matrix of dimension $N \times N$.


Thanks!
 A: Define $$\eqalign{
 V & = v1^T \cr
 Z & = [z_1 | z_2| \ldots | z_M] \cr
 X & = \alpha V-Z \cr
 M & = I + X^HS^{-1}X \,\,= M^H \cr
\cr
}$$
Then
$$\eqalign{
 f &= \det M \cr
\cr
 df &= fM^{-T}:dM \cr
    &= fM^{-T}:X^HS^{-1}dX \cr
    &= fM^{-T}:X^HS^{-1}V\,d\alpha \cr
\cr
\frac{df}{d\alpha} &= fM^{-T}:X^HS^{-1}V \cr
 &= f\,1^TM^{-1}X^HS^{-1}v \cr\cr
}$$
Set the derivative to zero and apply the Woodbury formula to the $M$ term
$$\eqalign{
 0 &= 1^T\Big(I-X^H(S+XX^H)^{-1}X\Big)X^HS^{-1}v \cr
 1^TX^HS^{-1}v &= 1^TX^H(S+XX^H)^{-1}XX^HS^{-1}v \cr
\cr
}$$
In order to continue, each of the $X$'s must be expanded into $(\alpha V-Z)$, and then rearranged to solve for $\alpha$.  At this point, an algebraic solution doesn't look very promising.
However, since this is a just a scalar function ($f$) of a scalar variable ($\alpha$), a numerical solution should be easy. And you will have the exact formula for ($df/d\alpha$) at each step in the iteration, which will improve speed and accuracy.
A: In addition to $(V,Z)$ that Hans used, let's also define 
$$\eqalign{
 Q &= S^{-1/2} \cr
 P &= \alpha^2VV^H-\alpha(VZ^H+ZV^H)+ZZ^H+S \cr
}$$
I'm also going to insist that $\alpha$ is real, and all of the "complexity" is baked into the ${\mathbf v}$ vector.
Anyway, Sylvester's Theorem allows us to rewrite the determinant as
$$\eqalign{
 f &= \det(I+X^HQ^HQX) \cr
   &= \det(I+QXX^HQ^H) \cr
   &= \det\Big(Q(S+XX^H)Q\Big) \cr
   &= \det(Q)^2\,\,\det(S+XX^H) \cr\cr
 f\det(S) &= \det(S+XX^H) \cr
   &= \det\Big(S+(\alpha V-Z)(\alpha V-Z)^H\Big) \cr
   &= \det\Big(\alpha^2VV^H-\alpha(VZ^H+ZV^H)+ZZ^H+S\Big) \cr
   &= \det(P) \cr
}$$
The derivative wrt $\alpha$ is
$$\eqalign{
 \frac{d\det(P)}{d\alpha} &= \det(P)P^{-T}:\frac{dP}{d\alpha} \cr
   &= \det(P)P^{-T}:(2\alpha VV^H-VZ^H-ZV^H) \cr
}$$
Setting the derivative to zero yields
$$\eqalign{
 2\alpha(VV^H:P^{-T}) &= (VZ^H+ZV^H):P^{-T} \cr\cr
 \alpha &= \frac{(VZ^H+ZV^H):P^{-T}}{(VV^H+VV^H):P^{-T}} \cr\cr
}$$
This is an iterative equation rather than a solution, since $P$ on the RHS is a function of $\alpha$. I have no idea if the iteration converges or diverges, but you can try it and see.
