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I have a known vector $y\in\mathbb R^n$ and I am looking for the solution $x\in\mathbb R^n$ of the matrix equation: $$(yx^T)(Wx)=(y^Tx)(Wx)$$ where $W\in\mathbb R^{n,n}$ is symmetrical positive definite (so it could be inverted if that's useful).

In this equation, the part on the left side $yx^T$ is a matrix (of rank 1) and part of the right side $y^Tx$ is a scalar. According to the solution of this question, $y^Tx$ is the only non-zero eigenvalue of the matrix $yx^T$ and it will have multiplicity of $1$ (unless $y^Tx=0$ but this case is of no interest). The other eigenvalue will be $0$ with multiciplity $n-1$.

I could postulate that $Wx$ must be the corresponding eigenvector to the eigenvalue $y^Tx$. Is there a way to find the value of $x$ in this situation and prove its unicity, given a choice of $y$ and $W$?

Edit

In the case where $W=I$ (the identity matrix) then we can find that $x=y$ is the trivial solution. In this case, \begin{align} (xx^T)x&=(x^Tx)x\\ x(x^Tx)&=(x^Tx)x\\ x\lVert x\rVert^2&=x\lVert x\rVert^2 \end{align}

PC1
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1 Answers1

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I assume $W$ is (symmetric and) positive definite and that $y^Tx \neq 0$, hence $x,y \neq 0$.

Your equation can be rewritten as $$ (x^TWx) y = (y^Tx)Wx $$ The left is a (non-zero) multiple of $y$, and the right is a multiple of $Wx$. Thus, if this equation has any solution, then there must be some scalar $k$ for which $$ Wx = ky \implies x = kW^{-1}y. $$ Now, substituting this form of $x$ yields the equation $$ k^2 (y^TW^{-1}y)y = k^2 (y^TW^{-1}y)y, $$ Which always holds. Thus, for any $W,y$ satisfying the given assumptions, $x$ is a solution iff $x = kW^{-1}y$ for some scalar $k$.

Ben Grossmann
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