I got this question and I need help from the experts in it.
The centering matrix C, together with a vector $x$ and it’s transpose $x'$ form a quadratic form $x'Cx$. We discussed the example where
which equaled
This choice of C is not the only one that would produce this equation.
Question:
Construct a symmetric matrix D that when substituted for C in $x'Cx$. also
Any advice will be greatly appreciated.
On
One can write down such a symmetric matrix directly from the quadratic form, as illustrated in another answer, but we can also derive $D$ from $C$.
Observe that only the symmetric part $\frac12(A+A^T)$ of a matrix $A$ contributes to the quadratic form: For any skew-symmetric matrix $S$, we have $(\mathbf x^TS\mathbf x)^T = \mathbf x^TS^T\mathbf x = -\mathbf x^TS\mathbf x$, but this expression is a scalar, so $(\mathbf x^TS\mathbf x)^T=\mathbf x^TS\mathbf x=0$. Hence, \begin{align}\mathbf x^TC\mathbf x &= \mathbf x^T\left(\frac12(C+C^T)+\frac12(C-C^T)\right)\mathbf x \\ &= \mathbf x^T\left(\frac12(C+C^T)\right)\mathbf x + \mathbf x^T\left(\frac12(C-C^T)\right) \\ &= \mathbf x^T\left(\frac12(C+C^T)\right)\mathbf x\end{align} since $C-C^T$ is skew-symmetric.
Therefore, you can take $$D = \frac12(C+C^T) = \begin{bmatrix}1&3&\frac52\\3&7&2\\\frac52&2&5\end{bmatrix}.$$
Write the symmetric matrix $D$ with rows $(d_1,d_2,d_3)$, $(d_2,d_4,d_5)$ and $(d_3,d_5,d_6)$. Denote by $f=f(x_1,x_2,x_3)$ the given polynomial. Then $x^TDx-f=0$ is given by a direct calculation as $$ x_1^2(d_1 - 1) + 2x_1x_2(d_2 - 3) + x_1x_3(2d_3 - 5) + x_2^2(d_4 - 7) + 2x_2x_3(d_5 - 2) + x_3^2(d_6 - 5)=0. $$ Hence we obtain $d_1=1$, $d_2=3$, $d_3=5/2$ , $d_4=7$, $d_5=2$ and $d_6=5$. Done.