(Sorry if the title is not informative)
How can I find the value of matrices $F$ and $d$, in the following equation:

$$y'Ay+b'y+c'c = (y-d)'F(y-d)$$

Given $A:n \times n$, which is positive definite and $b,c:n\times 1$.

My effort:
I used choleskey decomposition for A: $$A=Q'^{-1}Q^{-1}$$

and replaced $z=Q^{-1}y$ and as a result: $$z'z+h'z+c'c=(z-k)'Q'FQ(z-k)$$

In which, $h'=b'Q$ and $k=Q^{-1}d$. I though maybe I should set $F=A$. If so we will have this: $$z'z+h'z+c'c=(z-k)'(z-k)$$

and that's where I do not know how to proceed to find $k$.


This, in general, is not possible. If you want $$ y'Ay+b'y+c'c = (y-d)'F(y-d)=y'Fy-2d'Fy+d'Fd, $$ by comparing coefficients, you get $F=A$ and $d=\frac12A^{-1}b$. But then $d'Fd=\frac14b'A^{-1}b$, which is not necessarily equal to $c'c$. You may, however, express $y'Ay+b'y+c'c$ as $(y+\frac12A^{-1}b)'A(y+\frac12A^{-1}b)+(cc'-\frac14b'A^{-1}b)$.


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