Finding the minimum of $z=x^TAx+2b^Tx+1$ How can we find a minimum of $z=x^TAx+2b^Tx+1$?
Where $A$ is a positive definite.
$$
A=\begin{bmatrix}
1 & 1 & 0\\
2 & 3 & 1\\
1 & 1 & 4\\
\end{bmatrix}
$$
$$
x=\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}^T\\
b^T = \begin{bmatrix} 1   & -2 & 1 \end{bmatrix}\\
$$
I know probably this may not comply with the community standard but I have no idea how to start.
 A: Hint:
$\phi(x)=x^\intercal Ax +2b^\intercal x +1$,
From
$$\phi(x+h) = x^\intercal A x + 2b^\intercal x + 1 + 2b^\intercal h +  x^\intercal (A + A^\intercal)h + h^\intercal A h$$
it follows that
$$\phi'(x)=x^\intercal(A+A^\intercal) + 2b^\intercal$$
and  Hessian of $\phi$ is
$$\phi''(x)= A+ A^\intercal $$
Standards methods of Calculus can be use here:

*

*Find $x^*$ such that $\phi'(x^*)=0$

*Check whether $\phi''(x^*)$ is positive definite (or negative definite or neither) the decide whether $x^*$ is a local minimum, (or local max, or neither), etc.

A: This type of question can be solved in the following fashion:
Define a function $$f:R^n\to R$$ as follows
$$f(x)=x^TAx+2b^Tx+1$$
and calculate the gradient of this function and requiring that it will vanish:
$$\nabla f(x) = \nabla x^TAx+\nabla 2b^Tx+\nabla1=(A+A^T)x+2b^T=0$$
which is the same as $(A+A^T)x=-2b^T$. Now, to determine minimality, we need to calculate the Hessian matrix for this function (the differential of the gradient - that happens to be a linear transformation)
$$D\nabla f(x)=2A$$
Minima will correspond to all eigenvalues of the hessian being positive, but that is true from the positively definite.
A: Let $LL^T$ be the Cholesky decomposition of $B=(A+A^T)/2$, which is symmetric positive definite, and $v=L^{-1}b$. Then $x^TAx=x^TBx$ and
$$||L^Tx+v||^2=(L^Tx+v)^T(L^Tx+v)=(x^TL+v^T)(L^Tx+v)\\=x^TLL^Tx+x^TLv+v^TL^Tx+v^Tv=x^TBx+2v^TL^Tx+v^Tv\\=x^TAx+2b^Tx+v^Tv$$
Hence, $x^TAx+2b^Tx$ is minimized for $x=-L^{-T}v=-B^{-1}b$. And the minimum value is $-v^Tv=-b^TL^{-T}L^{-1}b=-b^TB^{-1}b$. Add $1$ to get the minimum of your function $z$.
