I have an objective function $L:\mathbb{C}^M \to \mathbb{R}$ that can be written:

$$ L(x) = x^H A x - b^H x - x^H b + d $$

where superscript $H$ denotes the conjugate transpose. $A \in \mathbb{C}^M$ is a hermitian matrix of size $M$ (so $A^H = A$), $b$ is a complex vector $b \in \mathbb{C}^M$, and $d$ is a real constant. I believe $L(x)$ is known to be convex, as it can be equivalently written as the squared euclidean norm of an affine function of $x$ (i.e. $L(x) = (Fx - g)^H (Fx - g)$).

I'm looking to solve the following minimization problem, with or without the added constraint that the vector $x$ is real. Formally:

$$ \text{minimize} \; L(x) $$ $$ \text{subject to} \; \text{Im}(x) = 0 $$

In other words, I want to find the real vector $x \in \mathbb{R}^M$ that minimizes the convex loss function, which is formally a function of complex input vector $x$. I have a feeling this should be straightforward but I'm not sure exactly how to go about doing so.

In the case that the "reality" constraint is quite non-trivial, it can be discarded. Discarding the constraint, I believe Wirtinger calculus can be used to solve this. Unfortunately, I'm not 100% clear on how to apply Wirtinger derivatives to actually obtain first-order conditions for a minima.

For example, to solve the unconstrained problem, is it valid to write: $$ L(x) = x^H A x - 2 \text{Re}(b^H x) + d $$ $$ \nabla{L}(\hat{x}) = A\hat{x} - 2\text{Re}(b) = 0$$ $$ A \hat{x} = 2\text{Re}(b) $$

Thus obtaining the solution $\hat{x} = A^{-1} 2\text{Re}(b)$ ?


1 Answer 1


In fact, it is possible to rewrite this as a problem on real matrices without a great deal of effort.

First, write $A = H + i K$, where both $H,K$ are matrices with real entries. The fact that $A$ is Hermitian means that $$ A = A^H \implies H + iK = H^T - i K^T \implies \begin{cases} H = H^T,\\ K = -K^T. \end{cases} $$ From the fact that $K^T = -K$, we find that for any real vector $x$, $$ x^TKx = (x^TKx)^T = x^TK^Tx = -x^TKx \implies x^TKx = 0. $$ Thus, for any such $x$, $$ x^TAx = x^T(H + iK)x = x^THx + ix^TKx = x^THx. $$ Similarly, we note that for any real $x$, we have $$ b^Hx + x^Hb = b^Hx + \overline{b^Hx} = b^Hx + \overline{b^H} x = (b + \bar b)^T x, $$ where $\bar x$ denotes the complex-conjugate of $x$. Note that $v := b + \bar b = 2 \operatorname{Re}(b)$.

Putting everything together, we see that it suffices to minimize the unconstrained objective function $f:\Bbb R^M \to \Bbb R$ given by $$ f(x) = x^THx - v^Tx + d, $$ where $H$ is a real, symmetric, positive definite matrix and $v \in \Bbb R^n$.

  • $\begingroup$ Thank you so much, this makes a lot of sense! :) I had previously been solving the problem by simply "discarding" the complex components of $A$ and $b$, but was not sure at all if that was correct — but now it is clear that this does in fact solve the same optimization problem. Thanks again! $\endgroup$ Jul 20, 2021 at 15:17
  • $\begingroup$ @user3204752350982 Glad to help. It's nice when your intuitive guess happens to be right $\endgroup$ Jul 20, 2021 at 15:20

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