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A Hermitian matrix $A$ is definite when either $x^{†}Ax>0$ for every nonzero vector $x$, or $x^{†}Ax<0$ for every nonzero vector $x$. These are the positive-definite matrix and negative-definite matrix cases. By the spectral theorem, all eigenvalues must have the same strict sign, so every definite matrix is invertible.

A positive semidefinite matrix permits zero eigenvalues and need not be definite. For example, $\mathrm{diag}(1,0)$ is positive semidefinite but not definite, while $\mathrm{diag}(1,-1)$ is an indefinite matrix. Definiteness concerns the sign on every nonzero vector, not merely one diagonal entry or the determinant.

For a real symmetric invertible matrix $H$, the full undamped Newton step for the displayed quadratic function sends every starting point to the unique stationary point:This follows because the gradient is affine and its first-order expansion is exact. If $H$ is a positive-definite matrix the result is the unique global minimum; a negative-definite matrix gives a global maximum, and an indefinite matrix $H$ gives a saddle point. Thus exact arrival does not by itself show that a minimization method has minimized anything. A line search, damping, or a positive-definite matrix replacing the Hessian matrix changes this exact one-step assertion.

One suitable autonomous planar system isIts equilibrium points are exactly the four combinations $(x,y)=(\pm 1,\pm 1)$. Its Jacobian matrix isThus $(-1,-1)$ and $(1,1)$ have eigenvalues $\pm 2$ and are saddle equilibria. At the other two points the eigenvalues are $\pm 2i$. To establish the nonlinear centre equilibrium classifications, use the first integralIts Hessian matrix is $\mathrm{diag}(2x,-2y)$, a negative-definite matrix at $(-1,1)$ and a positive-definite matrix at $(1,-1)$. Thus each is a strict local extremum, and nearby regular level sets are closed curves supporting periodic orbits, by the planar first integral criterion for a centre.

Immediately to the right of $(-1,1)$, the vector field points down because $x^2-1<0$, so that centre is clockwise. Immediately to the right of $(1,-1)$ it points up because $x^2-1>0$, so that centre is counterclockwise. This verifies both the nonlinear classifications and the prescribed orientations, with exactly four equilibrium points.

Put $w_k=C_k^{-1}$, $u_i=H_i^{-1}$, $A={\sum }_k{w}_k$, $B={\sum }_i{u}_i$, and form the weighted meansThe two equations obtained from the score function areConsequently the maximum-likelihood estimators areThe Hessian matrix of the log likelihood isIts first leading principal minor is negative and its determinant is $AB>0$, so it is a negative-definite matrix. Thus the stationary point is the unique global maximum.

Extend a grid vector $v$ by zero to the Dirichlet boundary. Pairing contributions along undirected nearest-neighbour edges gives the discrete energy identitywhere $E$ includes edges from an interior node to a boundary node. This is the three-dimensional version of summation by parts for the seven-point Dirichlet Laplacian.

The right side is nonpositive. If it vanishes, every pair of neighbouring values agrees; connectivity of the grid and the zero boundary values then imply $v=0$. Thus $v^{T}Av<0$ for every nonzero $v$, so $A$ is negative definite. In particular, zero is not an eigenvalue, and therefore

The fixed point branches and their stability are as follows.

which is a negative-definite matrix, so this branch is stable.

Thus the $x$-versus-$\mu$ bifurcation diagram has the stable $x=0$ branch up to $\mu =1$, a stable branch $x=\sqrt{\mu -1}$ joining the bifurcation points $(1,0)$ and $(2,1)$, and the branch $x=\sqrt{\mu /2}$ changing from unstable to stable at $\mu =2$. The origin branch also lies at $x=0$ but remains unstable throughout.

On $X^{+}$, $z^2>x^2+y^2$, so $z\ne 0$ and therefore $a=-2z\ne 0$. Yet $a(I)=-2$ and $a(-I)=2$. The same continuous image of a connected space argument shows that $X^{+}$ is not connected.

Put $r=\sqrt{x^2+y^2}$. The required three disjoint nonempty open sets areTheir union is precisely $X$, the complement of the double cone $z^2=x^2+y^2$. The eigenvalues are $z\pm r$, so $X_1$ consists of positive-definite matrices above the upper sheet, $X_2$ of negative-definite matrices below the lower sheet, and $X_3$ of indefinite matrices outside the double cone. Examples are $I,-I,\mathrm{diag}(1,-1)$ respectively.

The two definite cones are convex sets. The indefinite region is path-connected: shrink $z$ continuously to zero while keeping $r>0$, then change the nonzero vector $(x,y)$ within the punctured plane. Thus these sets are also exactly the three connected components. This is the determinant cone for real symmetric two-by-two matrices.

For each row, a Gershgorin disc has centre $a_{jj}=-2$ and radius $r_j={\sum }_{k\ne j}|a_{jk}|$. Here $r_1=r_3=1$ and $r_2=2$, so the Gershgorin circle theorem givesSince $A$ is a real symmetric matrix, its eigenvalues are real. Intersecting the union of these Gershgorin discs with the real axis yieldsThis bound permits zero but does not assert a zero eigenvalue. The fixed-end Dirichlet boundary conditions exclude it: for any nonzero real vector $v=(v_1,v_2,v_3{)}^T$, extend by $v_0=v_4=0$ and calculateIndeed, equality would force every adjacent value to agree with the zero endpoint. Thus $-A$ is a symmetric positive-definite matrix, $A$ is a negative-definite matrix, and the eigenvalues satisfy $-4\le \lambda <0$. The same argument works for any number of interior grid points, with prefactor $c^2/h^2>0$ in the sum. The fixed string has no zero-frequency rigid translation.

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.