# Solving equation of the form $Axb^Tx = y$

I have a square, invertible $n\times n$ matrix $A$, and column vectors $b$ and $y$. I'd like to find a column vector $x$ such that $Axb^Tx=y$. I suspect there's some way to get it into a QP form, but I haven't been able to wrangle it into that shape. Any suggestions? Am I off base with the QP?

-
If $a_i$ is the $i^{th}$ row of $A$, then you have that $$(a_ix)(b^Tx)=y_i$$ Which has the LHS as the product of two "numbers". It may help to expand this out to write in quadratic form.
Ahhh, that's interesting. From there I can get to $x^T(a_ib)x = y_i$, which starts to look like a QCQP. Thanks! –  Sneftel Oct 24 '12 at 11:09