# How to bring $5x_1^2 - 26x_1x_2 + 5x_2^2 + 10x_1 - 26x_2 = 31$ to the form $\langle x',Ax' \rangle = 1$

How can I bring $$5x_1^2 - 26x_1x_2 + 5x_2^2 + 10x_1 - 26x_2 = 31$$ to the form $$\langle x',Ax' \rangle = 1$$ where $x' = \alpha x + \beta$ where $\alpha \in \mathbb{R}^+$ and $\beta \in \mathbb{R}^n$ in order to diagonalize $A$.

I tried to rewrite it to a vector and a matrix. But when I multiply it out I don't get the original equation.

Does anybody can help me?

Thank you a lot!

$$5x_1^2−26x_1x_2+5x_2^2+10x_1−26x_2=5(x_1+1)^2-5-26(x_1+1)x_2+5x_2^2=31$$ So $x^\prime=[x_1+1,x_2]^T$, $A=\frac{1}{36}\Big(\begin{matrix} 5 &-13\\-13&5\end{matrix}\Big)$

• It's not clear to me that we're allowed to change the entries of the matrix $A$; I think we're supposed to scale the $x$ parameters instead to normalize.
– blue
Jul 3, 2014 at 5:33
• Matrix $A$ is defined in the way you want, in the original formula there is no any matrix yet. You can normalize $x^\prime$ instead dividing by $6$. Jul 3, 2014 at 5:36
• There is implicitly already a matrix in the original equation; the matrix defining the quadratic form on the left-hand side. I suspect the problem wouldn't have mentioned a $\alpha\in\Bbb R^+$ parameter if it were fine with changing the matrix instead.
– blue
Jul 3, 2014 at 5:39
• Again you can divide $x^\prime$ by $6$ instead. Jul 3, 2014 at 5:53
• (I knew, and did not fail to read, that.)
– blue
Jul 3, 2014 at 5:58

Writing

$${\bf x}^TA{\bf x}+{\bf b}^T{\bf x}=({\bf x}+{\bf u})^TA({\bf x}+{\bf u})+c$$

$$={\bf x}^TA{\bf x}+{\bf u}^T(A+A^T){\bf x}+({\bf u}^TA{\bf u}+c)$$

tells us we can merely solve for $\bf u$ via $(A+A^T){\bf u}={\bf b}$ and then $c=-{\bf u}^TA{\bf u}$.

Use this to rewrite ${\bf x}^TA{\bf x}+{\bf b}^T{\bf x}=\rm blah$ and then normalize to make it $=1$ on the right side.