How do you simplify the mutlidimensional quadratic formula?

Question

This is taken from one of the lecture slides, where a simplification of the quadratic formula is attempted: $$ (x-x_{A})^T A (x-x_{A}) + a^T (x-x_{a}) +b $$ and from there I expand the brackets, which yields: $$ (x-x_{A})^T A (x-x_{A}) + a^T(x-x_{a}) +b =x^T A x - x_{A}^T A x -x_A^T A x_A +a^T x - a^T x_a +b $$ In the next step the factorization is used to rearrange the formula: $$ (x-x_{A})^T A (x-x_{A}) + a^T (x-x_{a}) +b = x^T A x + (-A^T x_A - x_A ^T A^T + a)^T x + (x_A^T A x_A +b - a^T x_a) $$ I do not understand how the linear term has been transposed from $x^T A x_A$ to $(x_A^T A^T)^T x $. Can someone explain this ?

Answer 1

What about this simplification?

Let $\bar{x}=\begin{bmatrix}x\\1\end{bmatrix}$, then we have that

$$\begin{array}{rcl} x-x_A&=&\begin{bmatrix}1 & -x_A\end{bmatrix}\bar{x},\\ x-x_a&=&\begin{bmatrix}1 & -x_a\end{bmatrix}\bar{x},\\ a &=& \begin{bmatrix}0 &a\end{bmatrix}\bar{x},\ \mathrm{and}\\ b &=& \bar{x}^T\begin{bmatrix}0 & 0\\0 & b\end{bmatrix}\bar{x}. \end{array}$$

$$\bar{x}^T\left(\begin{bmatrix}1 \\ -x_A^T\end{bmatrix}A\begin{bmatrix}1 & -x_A\end{bmatrix}+\dfrac{1}{2}\begin{bmatrix}0 \\ a^T\end{bmatrix}\begin{bmatrix}1 & -x_a\end{bmatrix}+\dfrac{1}{2}\begin{bmatrix}1 \\ -x_a^T\end{bmatrix}\begin{bmatrix}0 &a\end{bmatrix}+\begin{bmatrix}0 & 0\\0 & b\end{bmatrix}\right)\bar{x}.$$

If we expand that we get

$$\bar{x}^T\begin{bmatrix}A & -Ax_A+a/2\\\star & x_A^TAx_A-a^Tx_a+b\end{bmatrix}\bar{x}.$$