Semidefinite Hessians and Degenerate Three-Variable Points

When Curvature Has One Sign but Does Not Decide

At a critical point, the Hessian records the quadratic part of a function’s local change. If that quadratic part is positive in every nonzero direction, the point is a strict local minimum; if it is negative in every nonzero direction, the point is a strict local maximum. If it takes both signs, the point is a saddle.

A different situation occurs when the quadratic form has one sign but equals zero in at least one nonzero direction. The Hessian is then semidefinite but not definite. In such a direction, the quadratic approximation gives no sign information. The function itself may still have a local minimum, a local maximum, or a saddle, so the second derivative test alone does not settle the classification.

Definitions and the Role of Singularity

Definition: A real symmetric matrix \(\mathbf{H}\) is positive semidefinite if \(\mathbf{v}^{\top}\mathbf{H}\mathbf{v}\geq 0\) for every vector \(\mathbf{v}\). It is negative semidefinite if \(\mathbf{v}^{\top}\mathbf{H}\mathbf{v}\leq 0\) for every \(\mathbf{v}\). It is positive definite if the first inequality is strict for every nonzero \(\mathbf{v}\), and negative definite if the second is strict for every nonzero \(\mathbf{v}\).

These terms overlap: a positive definite matrix is also positive semidefinite, and a negative definite matrix is also negative semidefinite. The inconclusive cases are specifically positive semidefinite but not positive definite, or negative semidefinite but not negative definite. For a symmetric matrix, each such case is singular and has at least one nonzero vector \(\mathbf{v}\) with \(\mathbf{v}^{\top}\mathbf{H}\mathbf{v}=0\).

For a function \(f(x,y,z)\) with continuous second partial derivatives, its Hessian at a point \(\mathbf{p}\) is the symmetric matrix of second partial derivatives. A critical point is a point where all first partial derivatives vanish. Write a nearby point as \(\mathbf{p}+\mathbf{h}\), where \(\mathbf{h}\) is the displacement vector. At a critical point, Taylor’s formula gives

f(\mathbf{p}+\mathbf{h})-f(\mathbf{p}) = \frac{1}{2}\mathbf{h}^{\top}\mathbf{H}(\mathbf{p})\mathbf{h} +o(\|\mathbf{h}\|^2).

The final term divided by \(\|\mathbf{h}\|^2\) tends to zero as \(\mathbf{h}\) tends to zero. Thus, when the quadratic form is strictly positive or strictly negative in every nonzero direction, its sign controls sufficiently small changes. But if the quadratic form vanishes along a nonzero direction, the formula does not tell us the sign of the remaining change in that direction.

Recognizing Semidefinite Hessians

The Hessian at a point is symmetric, so its eigenvalues are real. It is positive semidefinite exactly when all its eigenvalues are nonnegative, and negative semidefinite exactly when all are nonpositive. A zero eigenvalue signals a null direction: a nonzero vector \(\mathbf{v}\) satisfying \(\mathbf{H}\mathbf{v}=\mathbf{0}\). For a symmetric matrix, that direction also has zero quadratic form.

Principal minors provide another test. A principal minor is the determinant of a submatrix formed by selecting the same set of row and column indices. A symmetric three-by-three matrix is positive semidefinite exactly when all its principal minors are nonnegative; it is negative semidefinite exactly when its negative is positive semidefinite. This requires checking all principal minors, not just the leading ones. If the Hessian is semidefinite and at least one eigenvalue is zero, it is not definite and the usual decisive second derivative test does not apply.

Why the Test Is Inconclusive

At a critical point, the Hessian determines the quadratic term in the local expansion, but not what happens when that term vanishes. Along a null direction, the function’s change may be positive, negative, or identically zero. The following examples demonstrate that the same positive semidefinite Hessian at the origin can accompany different types of critical point.

Key takeaway: A singular semidefinite Hessian has a quadratic form of one sign that vanishes in at least one nonzero direction. The second derivative test does not classify the point from that information alone.

Worked Examples

Worked Example: A Positive Semidefinite Hessian at a Strict Minimum

Consider \(f(x,y,z)=x^2+y^2+z^4\). The gradient is \(\nabla f=(2x,2y,4z^3)\), so the origin is a critical point. The Hessian is

\mathbf{H}(x,y,z)= \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&12z^2 \end{pmatrix}, \qquad \mathbf{H}(0,0,0)= \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&0 \end{pmatrix}.

For \(\mathbf{h}=(u,v,w)^{\top}\), the quadratic form at the origin is \(2u^2+2v^2\). It is nonnegative, but it vanishes for every vector of the form \((0,0,w)^{\top}\), so the Hessian is positive semidefinite but not positive definite. The test is inconclusive. Directly, however,

f(x,y,z)-f(0,0,0)=x^2+y^2+z^4\geq 0.

Equality holds only when \(x=y=z=0\), because each term is nonnegative and \(z^4=0\) only when \(z=0\). Therefore the origin is a strict local minimum (in fact, a strict global minimum). Substitution confirms \(f(0,0,0)=0\), as used in the comparison.

Worked Example: The Same Hessian Sign with a Saddle

Now take \(g(x,y,z)=x^2+y^2-z^4\). Its gradient is \(\nabla g=(2x,2y,-4z^3)\), which vanishes at the origin. Its Hessian there is again \(\operatorname{diag}(2,2,0)\), so its quadratic form is \(2u^2+2v^2\), positive semidefinite but not positive definite.

To classify the point, compare values along two lines through the origin. On the \(x\)-axis, \(g(t,0,0)=t^2>0\) for \(t\ne 0\). On the \(z\)-axis, \(g(0,0,t)=-t^4<0\) for \(t\ne 0\). These values of opposite signs occur arbitrarily close to the origin. Since \(g(0,0,0)=0\), every neighbourhood of the origin contains points with function values both above and below the critical value. The origin is a saddle point. This verifies that a positive semidefinite Hessian that is not positive definite does not, by itself, guarantee a minimum.

Worked Example: A Negative Semidefinite Hessian at a Strict Maximum

Consider \(q(x,y,z)=-x^2-y^2-z^4\). Its gradient is \(\nabla q=(-2x,-2y,-4z^3)\), so the origin is critical. Its Hessian at the origin is

\mathbf{H}(0,0,0)= \begin{pmatrix} -2&0&0\\ 0&-2&0\\ 0&0&0 \end{pmatrix}.

For \(\mathbf{h}=(u,v,w)^{\top}\), the quadratic form is \(-2u^2-2v^2\), which is nonpositive and vanishes when \(u=v=0\). Thus the Hessian is negative semidefinite but not negative definite. Direct evaluation gives

q(x,y,z)-q(0,0,0)=-(x^2+y^2+z^4)\leq 0.

Equality holds only at the origin, so the origin is a strict local maximum (and a strict global maximum). Substitution gives \(q(0,0,0)=0\). The function’s values therefore confirm the classification, even though the Hessian test alone does not.

Worked Example: A Negative Semidefinite Hessian at a Saddle

Let \(r(x,y,z)=-x^2-y^2+z^4\). Its gradient is \(\nabla r=(-2x,-2y,4z^3)\), which vanishes at the origin, and its Hessian there is \(\operatorname{diag}(-2,-2,0)\). The quadratic form \(-2u^2-2v^2\) is nonpositive and vanishes for nonzero vectors along the \(z\)-axis. Thus the Hessian is negative semidefinite but not negative definite.

On the \(x\)-axis, \(r(t,0,0)=-t^2<0\) for \(t\ne 0\); on the \(z\)-axis, \(r(0,0,t)=t^4>0\) for \(t\ne 0\). Both types of values occur arbitrarily close to the origin, while \(r(0,0,0)=0\). Hence the origin is a saddle point. This verifies that the negative semidefinite case is inconclusive in the same way.

What the Examples Establish

The first two examples have the same Hessian at the origin, yet one has a strict minimum and the other a saddle. The last two have the same negative semidefinite Hessian, yet one has a strict maximum and the other a saddle. The relevant distinction is not visible in the Hessian’s zero-curvature direction: the function’s actual change along that direction matters.

A semidefinite Hessian is therefore not a classification. It identifies the sign of the quadratic contribution where that contribution is nonzero, but leaves the null directions unresolved. Direct comparisons can sometimes settle a specific example, as above. In other cases, the Hessian alone simply provides no decision.

A Common Misinterpretation

One mistake is to treat “positive semidefinite” as synonymous with “positive definite” and conclude that a local minimum has been proved. For example, at the origin the matrix \(\operatorname{diag}(2,2,0)\) is positive semidefinite, but for \(\mathbf{v}=(0,0,1)^{\top}\),

\mathbf{v}^{\top} \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&0 \end{pmatrix} \mathbf{v} =0,

not a positive number. Thus the matrix is not positive definite. The correction is to check whether the quadratic form is strictly positive for every nonzero vector before applying the positive-definite minimum test. If it is only nonnegative, the Hessian test is inconclusive; the function itself must be examined to classify that particular point.

The corresponding correction applies on the negative side: a nonpositive quadratic form does not establish a maximum unless it is strictly negative for every nonzero direction. A zero eigenvalue must not be ignored.

Check Your Understanding

Use the quadratic form and, where needed, direct function values to answer these questions.

What condition distinguishes a positive semidefinite matrix from a positive definite matrix?
Why does a zero eigenvalue indicate that a semidefinite Hessian is not definite?
At the origin, classify the Hessian \(\operatorname{diag}(3,1,0)\) as positive definite, positive semidefinite but not positive definite, or neither.
For \(s(x,y,z)=x^2+y^2+z^6\), verify that the origin is critical and explain why its Hessian alone does not give the minimum classification.
How can values along two lines through a critical point establish that it is a saddle?