Two-by-two positive-definiteness test ($a>0,ac-b^2>0$)

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.