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Consider $$\mathbf{x^TDx}+\mathbf{b^Tx}=0 \tag 1.$$

Let $\mathbf{D}$ is full rank diagonal matrix and $\mathbf{b} \neq 0$.

Then solutions $\mathbf{x}=0 \;\text{and}\;\mathbf{x}=-\mathbf{D^{-1}b}$ exist.

Question. Are there conditions that equation $(1)$ has another solutions?

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Consider $$\left\{ \begin{align} \mathbf{y^TD_1y}+\mathbf{b_1^Ty}=0 \\ \mathbf{y^TD_2y}+\mathbf{b_2^Ty}=0 \end{align} \right. \tag 2$$

I believed that the system $(2)$ has a root if there are true conditions that an equation $(1)$ has another solutions, for example, $\mathbf{x}$ has infinite set of values.

Where did I make a mistake?

Updated

Really equation $(1)$ has another solutions, for example in $\mathbb{R^2}$, next root (not only)

$$ \mathbf{x}=\left[\frac{-b_1+\sqrt{-4d_1d_2x^2_2-4b_2d_1x_2+b_1^2}} {2d_1}\;,x_2\right]^\mathbf{T} $$

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    $\begingroup$ Do you consider $x=-D^{-1}b$ to be a "trivial" solution? $\endgroup$ – A.Γ. Mar 19 '17 at 11:35
  • $\begingroup$ Yes. Is it error? $\endgroup$ – Evgeniy Abramov Mar 19 '17 at 11:45
  • $\begingroup$ In the standard terminology, the trivial solution of $Ax=0$ is $x=0$. It may cause a misunderstanding. $\endgroup$ – A.Γ. Mar 19 '17 at 11:50
  • $\begingroup$ Understand. Thank you. $\endgroup$ – Evgeniy Abramov Mar 19 '17 at 11:51
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Pick any $x_0$ with $x_0^TDx_0\ne 0$ and $b^Tx_0\ne 0$ and consider the line $tx_0$. On this line $$ t^2x_0^TDx_0+tb^Tx_0=0\quad\Leftrightarrow\quad t=0\text{ or } t=-\frac{b^Tx_0}{x_0^TDx_0}\ne 0. $$ The second $t$ corresponds to a non-trivial solution $tx_0$, so there is a solution in almost all directions $x_0$.

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  • $\begingroup$ Every quadic can be derived in this diagonal form . Thus, is every system of quadrics an inconsistent if $\mathbf{b}\neq 0$ and $D$ has full rank? $\endgroup$ – Evgeniy Abramov Mar 19 '17 at 12:52
  • $\begingroup$ I do not understand what you mean by inconsistent. Your equation defines a quadric. Full rank, hence, elliptic or hyperbolic. It is huge amount of points in the space. There is no way it is only two points $x=0$ and $x=-D^{-1}b$. $\endgroup$ – A.Γ. Mar 19 '17 at 13:46
  • $\begingroup$ Really equation $(1)$ has another solutions, for example in $\mathbb{R^2}$, next root (not only) $$ \mathbf{x}=\left[\frac{-b_1+\sqrt{-4d_1d_2x^2_2-4b_2d_1x_2+b_1^2}} {2d_1}\;,x_2\right]^\mathbf{T} $$ $\endgroup$ – Evgeniy Abramov Mar 19 '17 at 20:59
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If $D$ is diagonal and full rank, you know for sure that it is invertible, so such solution $\mathbb{x}=-D^{-1}b$ exists.

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  • $\begingroup$ $\mathbf{x}=-\mathbf{D^-b}$ is trivial. Are there anothers? $\endgroup$ – Evgeniy Abramov Mar 19 '17 at 11:31
  • $\begingroup$ @AnnaSdTC Not true. It is possible that $x$ is orthogonal to $Dx+b$. For example, $b=0$, $D=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$, $x=\begin{pmatrix}1\\1\end{pmatrix}$. $\endgroup$ – A.Γ. Mar 19 '17 at 11:38
  • $\begingroup$ @A.G. Yes, you are right. I'm going to delete my comment. $\endgroup$ – Anna SdTC Mar 19 '17 at 11:44

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