Recover vector $x$ from rank-$1$ matrix $Q=xx^H$

Question

Let the matrix $Q \in\mathbb{C}^{n \times n}$ be known. It is also known that $Q=xx^H$, where $x=[x_1,\ldots,x_n]^T$ and $x^H$ is its conjugate transpose. What is $x$? How to recover it?

Answer 1

If $Q=xx^H$, then $Qx=xx^Hx$. Set $\lambda=x^Hx$; this means $x$ is an eigenvector of $Q$ relative to $\lambda$ (the unique non zero eigenvalue).

On the other hand, if $Q$ is explicitly given, then since it has rank one, you just take a nonzero column $y$ and consider that $x=\mu y$ for some $\mu$. Then do $Q=xx^H=|\mu|^2yy^H$ and you can determine $\mu$ (up to a complex number of modulo $1$). (Assuming $x\ne0$, of course, or the problem is trivial).

Answer 2

Given a rank-$1$ Hermitian matrix $\mathrm Q \in \mathbb C^{n \times n}$, we would like to determine $\mathrm x \in \mathbb C^n$ such that $$\mathrm x \mathrm x^* = \mathrm Q$$

Since $\mathrm Q$ is rank-$1$, only one of its $n$ eigenvalues is nonzero. Let $\lambda$ be this eigenvalue. Hence,

$$\mbox{tr} (\mathrm Q) = \lambda + 0 + \cdots + 0 = \lambda$$

Solving the linear system

$$\left( \mathrm Q - \mbox{tr} (\mathrm Q) \, \mathrm I_n \right) \mathrm v = 0_n$$

we can obtain a normalized eigenvector $\mathrm v \in \mathbb C^n$ corresponding to the eigenvalue $\lambda = \mbox{tr} (\mathrm Q)$. Thus, Hermitian matrix $\mathrm Q$ has the eigendecomposition

$$\mathrm Q = \mbox{tr} (\mathrm Q) \, \mathrm v \mathrm v^* = \left( \sqrt{\mbox{tr} (\mathrm Q)} \, \mathrm v \right) \left( \sqrt{\mbox{tr} (\mathrm Q)} \, \mathrm v \right)^*$$

Hence,

$$\mathrm x := \sqrt{\mbox{tr} (\mathrm Q)} \, \mathrm v$$

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linear-algebra matrices hermitian-matrices rank-1-matrices eigenvalues-eigenvectors