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Assume $u,v\in\mathbb{C}^n$ are complex vectors. I was wondering if there is a closed form expression for the following problem in terms of $u$ and $v$

\begin{equation*} \arg\min_{x\in\mathbb{C}^n} \|uv^*+vu^*-xx^*\|^2_F \end{equation*}

F here denotes Frobenius norm.

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  • $\begingroup$ I should note that on the reals the answer is $\frac{1}{\sqrt{2\|u\|\|v\|}}(u\|v\|+v\|u\|)$ $\endgroup$
    – mohi
    Aug 16, 2013 at 20:57

1 Answer 1

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We assume that the system $u,v$ is $\mathbb{C}$-free. Let $w=uv^*+vu^*$. Note that if $X$ is a solution then $\exp(i\theta)X$ is also a solution. We study the minimum of $trace((w-XX^*)^2)=trace(w^2+XX^*XX^*-wXX^*-XX^*w)=trace(w^2)+trace((X^*X)^2)-2trace(wXX^*)$,

that is we study the minimum of $\phi(X)=||X||^4-2X^*wX$. The hermitian matrix $w$ has rank $2$ and its eigenvalues are $0$ ($n-2$ times), a negative eigenvalue and a positive eigenvalue $\lambda$ (take an orthonormal basis $e_1,e_2$ of the plane $[u,v]$ s.t $e_1$ and $u$ are collinear). Thus $\phi(X)\geq \psi(||X||^2)=||X||^4-2\lambda||X||^2$. Then $\inf(\psi(z))=\psi(\lambda)=-\lambda^2$ and $\phi(X)\geq -\lambda^2$. We can reach this bound: $X$ is an eigenvector of $w$ associated to $\lambda$ and $||X||^2=\lambda$. Note that the previous conditions define $X$ up to a factor $\exp(i\theta)$.

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