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

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

  1. $\mathbf{I}$ indicates the identity matrix of dimension $M \times M$,
  2. $\mathbf{S}$ is an invertible, Hermitian and positive definete matrix of dimension $N \times N$.


  • $\begingroup$ So H is conjugate transpose or something else? $\endgroup$ – mathreadler Mar 7 '17 at 10:26
  • $\begingroup$ Like most optimization problems: compute the derivative (using Jacobi formula) and set it to zero. $\endgroup$ – Simon Mar 7 '17 at 10:32
  • $\begingroup$ Yes, H is the conjugate transpose. I tried to use the Jacobi formula, but I was not able to obtain a closed form solution... $\endgroup$ – Vuk Mar 7 '17 at 10:59

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.

  • $\begingroup$ Thanks! From other consideration about this problem, I'm quite sure that a close form exists and it is of the form: [ \alpha_0=\frac{1}{M\mathbf{v}^H\mathbf{S}^{-1}\mathbf{v}}\su‌​m\nolimits_{m=1}^{M_‌​1}\mathbf{v}^H\mathb‌​f{S}^{-1}\mathbf{x}_‌​m, ] but I'm not able to prove it... $\endgroup$ – Vuk Mar 28 '17 at 13:20
  • $\begingroup$ Since the $\{x_m\}$ vectors are themselves functions of $\alpha,\,$ I don't see how your assumed expression is a solution. Rather, like my own answer, it appears to be a recursive definition of the form $$\alpha = f(\alpha)$$ $\endgroup$ – greg Mar 28 '17 at 14:17
  • $\begingroup$ Sorry, you are right of course! x_m should be z_m in my previous post $\endgroup$ – Vuk Mar 28 '17 at 16:38

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.


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