Is there a solution to $x \circ a + x \circ (Bx) + c = 0$ for $x$, where $B$ is an $N \times N$ matrix, $x$, $a$ and $c$ are $N \times 1$ column vectors, and $\circ$ is the Hadamard product (element-wise multiplication)? $a$, $B$, and $c$ are given, real, and non-zero. $B$ is also invertible.

  • $\begingroup$ I presume $a$, $B$ and $c$ are given. Obviously there is no solution if $a = 0$ and $B = 0$ but $c \ne 0$. So maybe the question should be, under what conditions can we guarantee the existence of a solution? $\endgroup$ – Robert Israel May 16 '19 at 20:19
  • $\begingroup$ Are you looking for real solutions (where $a, B, c$ are real, or are complex solutions allowed? $\endgroup$ – Robert Israel May 16 '19 at 20:25
  • $\begingroup$ Oh good point Robert Israel, I edited the question. Complex solutions are totally fine. $\endgroup$ – Carter D. May 18 '19 at 12:16

For example, let's take $N=2$. The equation can be written as

$$ \eqalign{ b_{{1,1}}{x_{{1}}}^{2}+b_{{1,2}}x_{{1}}x_{{2}}+a_1 x_{{1}} +c_{{1}} &= 0\cr b_{{2,1}}x_{{1}} x_2 +b_{{2,2}}{x_{{2}}}^{2}+a_2 x_{{2}} +c_{{2}} &= 0\cr} $$ The resultant with respect to $x_2$ is an irreducible quartic in $x_1$:

$$\left( {b_{{1,1}}}^{2}b_{{2,2}}-b_{{1,1}}b_{{1,2}}b_{{2,1}} \right) { x_{{1}}}^{4}+ \left( 2\,a_{{1}}b_{{1,1}}b_{{2,2}}-a_{{1}}b_{{1,2}}b_{{ 2,1}}-a_{{2}}b_{{1,1}}b_{{1,2}} \right) {x_{{1}}}^{3}+ \left( {a_{{1}} }^{2}b_{{2,2}}-a_{{1}}a_{{2}}b_{{1,2}}+2\,b_{{1,1}}b_{{2,2}}c_{{1}}+{b _{{1,2}}}^{2}c_{{2}}-b_{{1,2}}b_{{2,1}}c_{{1}} \right) {x_{{1}}}^{2}+ \left( 2\,a_{{1}}b_{{2,2}}c_{{1}}-a_{{2}}b_{{1,2}}c_{{1}} \right) x_{ {1}}+b_{{2,2}}{c_{{1}}}^{2} $$

Given a root $x_1$ of this, you get $x_2$ by solving the first equation:

$$ x_{{2}}=-{\frac {b_{{1,1}}{x_{{1}}}^{2}+a_1 x_{{1}}+c_{{1}}}{b_{{1, 2}}x_{{1}}}} $$ In general there will be four complex solutions.

For larger $N$, things will get horribly complicated if you try to do everything with symbolic coefficients, but I suspect there will be $2^N$ complex solutions in general.

  • $\begingroup$ Ahh, thanks. There's no simple way to do this I see. Thanks! $\endgroup$ – Carter D. May 20 '19 at 14:46

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