Find the bilinear form associated to a quadratic form.

Question

Let $$Q(x_1,x_2,x_3,x_4)=x_1^2+2x_2^2+3x_3^2+4x_4^2+5x_1x_2+6x_1x_3+7x_1x_4+8x_2x_3+9x_2x_4+10x_3x_4.$$

How can I find $f$ bilinear s.t. $Q(x)=f(x,x)$ ?

It's written in my book that if $Q(x)=\sum_{i,j}\alpha _{ij}x_ix_j$, then $$\frac{\partial q}{\partial x_i}=2\sum_{j}\alpha _{ij}x_j,$$ and thus $Mat(f)=(\alpha _{ij})$, but I don't really understand what it mean. I don't see the correlation between $\frac{\partial q}{\partial x_i}$ and $f$.

Answer 1

For the correlation between $\frac{\partial Q }{\partial x_i}$ and $f$ if $Q(x)=f(x,x)$ then: $$Q(x)=Q\left(\sum_k x_k e_k\right)=f \left(\sum_k x_k e_k, \sum_j x_j e_j \right)=\sum_{k,j} x_k x_j f(e_i,e_j)$$ so for a fixed $i$, differentiating with respect to $x_i$: $$\frac{\partial Q }{\partial x_i}=\sum_j x_j f(e_i,e_j)+\sum_k x_k f(e_k,e_i)$$ and if you are looking for a symmetric $f$ you obtain: $$\frac{\partial Q }{\partial x_i}=2 \sum_j x_j f(e_i,e_j)$$ and by definition: $$Mat(f)_{i,j}=f(e_i,e_j)$$

Answer 2

the Hessian matrix of second partial derivatives is $$ H = \left( \begin{array}{cccc} 2 & 5 & 6 & 7\\ 5 & 4 & 8 & 9 \\ 6 & 8 & 6 & 10 \\ 7 & 9 & 10 & 8 \end{array} \right) $$ Note $H^T = H.$

If I write column vectors $x$ and $y$ with four entries each, then $q(x) = \frac{1}{2} x^THx $ and $f(x,y) = \frac{1}{2} x^THy = \frac{1}{2} y^THx $