Show that $\langle x, Ax \rangle + \langle b, x \rangle = c$ can be transformed to $\langle x', Ax' \rangle = 1$

Question

Let $A$ be a real, regular, symmetric $n \times n$ matrix, $b \in \mathbb{R}^n$ and $c \in \mathbb{R}$

How can I show that $$\langle x, Ax \rangle + \langle b, x \rangle = c$$ can be transformed by using $$x' = \alpha x + \beta$$ where $0 \neq \alpha \in \mathbb{R}$, $\beta \in \mathbb{R}^n$ to $$\langle x', Ax' \rangle = 1$$ if $c + \frac{1}{4}\langle b, A^{-1}b \rangle > 0$

I tried to start with $\langle x, Ax \rangle + \langle b, x \rangle = c$ and $\langle x', Ax' \rangle = 1$ by inserting $x' = \alpha x + \beta$ but I always end up with a large equation.

Does anybody see how I can solve this problem?

Thanks a lot!

Answer 1

You want to use "completing the square" from your algebra-2 class.

Roughly, you've got $$ ax^2 + bx = c, $$ and completing the square would rewrite this as $$ x^2 + a^{-1}bx = a^{-1}c $$ and then $$ x^2 + a^{-1}bx + (a^{-1}b)^2/4 = a^{-1}c + (a^{-1}b)^2/4 $$ after which you'd let $$ x' = x + (a^{-1}b)/2, $$ and get $$ x'^2 = a^{-1}c + (a^{-1}b)^2/4. $$

Can you try to follow that pattern here?

In particular, if you let $$ x' = x + \frac{1}{2}A^{-1}b, $$ then \begin{align} \langle x', Ax' \rangle> &= \langle x + \frac{1}{2}A^{-1}b, A(x + \frac{1}{2}A^{-1}b) \rangle \\ &= \langle x , A(x + \frac{1}{2}A^{-1}b) \rangle + \langle \frac{1}{2}A^{-1}b, A(x + \frac{1}{2}A^{-1}b) \rangle \\ &= \langle x , Ax \rangle + \langle x, A\frac{1}{2}A^{-1}b) \rangle + \langle \frac{1}{2}A^{-1}b, Ax \rangle + \langle \frac{1}{2}A^{-1}b, A\frac{1}{2}A^{-1}b) \rangle \\ &= \langle x , Ax \rangle + \frac{1}{2}\langle x, b) \rangle + \frac{1}{2}\langle A^{-1}b, Ax \rangle + \frac{1}{4}\langle A^{-1}b, b \rangle \\ &= \langle x , Ax \rangle + \frac{1}{2}\langle x, b) \rangle + \frac{1}{2}\langle b, x \rangle + \frac{1}{4}\langle A^{-1}b, b \rangle \\ &= \langle x , Ax \rangle + \langle x, b \rangle + \frac{1}{4}\langle A^{-1}b, b \rangle \\ \end{align}

That seems to me as if it's getting pretty close to what you want, no?

Answer 2

Long story short: we can rewrite the above as $$ x^TAx + b^Tx - c = 0 $$ Now, set $x = x' - \frac 12 A^{-1}b $. Substituting this into the above gives you $$ 0 = (x' - \frac 12 A^{-1}b)^TA(x' - \frac 12 A^{-1}b) + b^T(x' - \frac 12 A^{-1}b) - c = \\ \left[(x')^T A x' - b^T x' + \frac 12 b^TA^{-1}b\right] + \left[ b^T x' - \frac 12 b^T A^{-1}b\right] - c =\\ (x')^T A x' - c $$ So, setting $x'$ to $x + \frac 12 A^{-1}b$ gives us the equation $$ (x')^T A x' = c \implies \left(\frac 1{\sqrt c} x'\right)^T A \left(\frac 1{\sqrt c} x'\right) = 1 \implies\\ \left\langle \left(\frac 1{\sqrt c} x'\right), A \left(\frac 1{\sqrt c} x'\right) \right \rangle = 1 $$ So, setting $\tilde x = \frac 1{\sqrt{c}} x' = \frac 1 {\sqrt{c}}(x + \frac 12 A^{-1})$ means that $$ \langle x,Ax \rangle + \langle x,b \rangle = c \iff \langle \tilde x, A \tilde x \rangle = 1 $$