Unique minimizer of a positive-definite quadratic function ($f(x)=\frac{1}{2}{x}^{T}Ax-b^{T}x+c,x_{\ast }=A^{-1}b$)

Let $A$ be a real positive-definite matrix and minimize the displayed quadratic function over all of ${\mathbb{R}}^n$. Its gradient is $Ax-b$, so its only stationary point solves the linear system $A{x}_{\ast }=b$. The matrix is an invertible matrix, making this solution exist and be unique. Completing the square around that solution proves the stronger global statement:The quadratic function is strictly convex, and its level sets above the minimum are translated ellipsoids. Positivity therefore turns solving a linear system into locating the bottom of a bowl.^[ref]

For the opening matrix $A=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$, take $b=(1,1{)}^T$ and $c=0$. Then $A^{-1}=\frac{1}{8}\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$, $x_{\ast }=(1/4,1/4{)}^T$, and $f(x_{\ast })=-1/4$. Every displacement $h$ raises the value by $\frac{1}{2}{h}^{T}Ah$; along the unit eigenvectors $e_{+}$ and $e_{-}$ the increases are respectively $2s^2$ and $s^2$ for displacement $s{e}_{+}$ or $s{e}_{-}$.

For a singular positive semidefinite matrix, a solution of $Ax=b$ is still a global minimizer, but adding any kernel vector gives another. If $b$ is outside the image of a linear map defined by $A$, some kernel vector $v$ has $b^{T}v\ne 0$, and $f(sv)=c-s{b}^{T}v$ is unbounded below. For an indefinite quadratic form, a negative eigenvalue gives a direction along which the quadratic term tends to $-\infty$ regardless of $b$. Thus a stationary point alone does not establish a minimum. The same energy-difference identity extends to the quadratic variational principle for a symmetric positive operator, with additional existence hypotheses in infinite dimension.