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For a real symmetric matrix $A=\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$, completing the square gives, when $a\ne 0$,Thus $A$ is a positive-definite matrix exactly when $a>0$ and $\det A=ac-b^2>0$. For necessity, test $(1,0)$ first, then $(-b/a,1)$; for sufficiency, the displayed two positive squares give the result. These are the first two leading principal minors in Sylvester's criterion. A positive determinant alone is insufficient: $-I_2$ has determinant $1$ but is a negative-definite matrix.

The eigenvalues provide an equivalent test under the same real-symmetric hypothesis:The spectral theorem for real symmetric matrices supplies coordinates in an orthonormal basis in which $q={\lambda }_{+}{u}^2+{\lambda }_{-}{v}^2$. More generally, a complex Hermitian matrix has an orthonormal eigenbasis and $q(z)={\sum }_j{\lambda }_j|w_j{|}^2$, so it is a Hermitian positive-definite matrix exactly when every eigenvalue is positive. This is a short proof outline: test the eigenvectors for necessity, and sum the positive squared absolute values for sufficiency.^[ref]

For a real positive semidefinite matrix, the condition is instead $q(x,y)\ge 0$ for all vectors, equivalent here to $a\ge 0$, $c\ge 0$, and $ac-b^2\ge 0$. If $a>0$, the same completing the square proves this; if $a=0$, nonnegativity forces $b=0$ and leaves $c{y}^2$ with $c\ge 0$. Merely replacing the strict inequalities on the leading principal minors by nonstrict ones fails: $\mathrm{diag}(0,-1)$ has both leading principal minors zero yet gives $q(0,1)=-1$. An indefinite quadratic form takes both positive and negative values; its real symmetric matrix has eigenvalues of both signs.

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.

For a finite undirected graph without self-loops, give each edge a symmetric nonnegative weight $w_{ij}=w_{ji}$. In the weighted graph, set $(A_w{)}_{ij}=w_{ij}$ for an edge and zero otherwise, and $(D_w{)}_{ii}={\sum }_j{w}_{ij}$ with all other entries zero. The Weighted graph Laplacian isEach edge appears once in the quadratic form. For the path graph $A-B-C$ with weights $w_{AB}=2$ and $w_{BC}=3$,The Weighted graph Laplacian is still a positive semidefinite matrix. Its kernel consists of vectors constant on each connected component of a graph after zero-weight edges are removed: an edge of weight zero imposes no constraint. In an electrical network, taking the weights to be electrical conductances makes $x^T{L}_{w}x$ the total power dissipated by Joule heating when $x$ gives the voltages at the vertices. This physical interpretation explains why stronger connections penalize disagreement more heavily.^[ref]

If $P$ is a real positive semidefinite matrix, then $x^{T}Px=0$ if and only if $Px=0$. Indeed $x^{T}Px=\Vert \sqrt{P}x{\Vert }^2$ by the principal square root of a positive semidefinite matrix, so a zero quadratic form puts $x$ in the kernel of $\sqrt{P}$ and hence of $P$.