Sylvester’s Criterion for Definite Hessians

Why a Definiteness Test Needs More Than the Determinant

A Hessian at a critical point records the second-order behavior of a function. For a function of three variables, its quadratic form is determined by a symmetric three-by-three matrix. To use the second derivative test, we need to know whether that quadratic form is positive for every nonzero direction or negative for every nonzero direction.

The determinant alone cannot answer that question. Sylvester’s criterion gives an exact test using three determinants: those of the upper-left one-by-one, two-by-two, and three-by-three blocks. The order of these blocks matters; the test uses the leading principal minors, not just any three minors.

Definition: Let \(\mathbf{H}=\begin{pmatrix}a&b&c\\b&d&e\\c&e&f\end{pmatrix}\) be a real symmetric matrix. Its leading principal minors are \[ \Delta_1=a,\qquad \Delta_2=\det\begin{pmatrix}a&b\\b&d\end{pmatrix}=ad-b^2,\qquad \Delta_3=\det(\mathbf{H}). \]

The quadratic form associated with \(\mathbf{H}\) is \(q(\mathbf{x})=\mathbf{x}^{\top}\mathbf{H}\mathbf{x}\), where \(\mathbf{x}=(x,y,z)^{\top}\). The matrix is positive definite if \(q(\mathbf{x})>0\) for every nonzero \(\mathbf{x}\), and negative definite if \(q(\mathbf{x})<0\) for every nonzero \(\mathbf{x}\). These are strict conditions: a zero value in any nonzero direction rules out definiteness of the corresponding kind.

Sylvester’s Criterion in Three Variables

Theorem: For a real symmetric three-by-three matrix \(\mathbf{H}\):

\(\mathbf{H}\) is positive definite if and only if \(\Delta_1>0,\ \Delta_2>0,\ \Delta_3>0\).
\(\mathbf{H}\) is negative definite if and only if \(\Delta_1<0,\ \Delta_2>0,\ \Delta_3<0\).

For positive definiteness, all three leading principal minors must be positive. For negative definiteness, their signs alternate, beginning with a negative sign. In compact form, the negative-definite pattern is \((-1)^k\Delta_k>0\) for \(k=1,2,3\). In particular, the middle minor is positive in both tests.

Why the Positive-Definite Conditions Work

We prove both directions. The proof rests on completing squares, which rewrites the quadratic form using successive pivots. First suppose \(a>0\). Completing the square in \(x\) gives

\begin{aligned} q(x,y,z) &=a\left(x+\frac{b}{a}y+\frac{c}{a}z\right)^2\\ &\quad+\left(d-\frac{b^2}{a}\right)y^2 +2\left(e-\frac{bc}{a}\right)yz +\left(f-\frac{c^2}{a}\right)z^2. \end{aligned}

The remaining quadratic form in \(y,z\) has first diagonal coefficient \(d-b^2/a=\Delta_2/a\). Completing its square in \(y\) then gives the full decomposition

q(x,y,z) = a\left(x+\frac{b}{a}y+\frac{c}{a}z\right)^2 +\frac{\Delta_2}{a} \left(y+\frac{ae-bc}{\Delta_2}z\right)^2 +\frac{\Delta_3}{\Delta_2}z^2.

To see the last coefficient, the determinant of the remaining two-by-two matrix after the first square is removed is \(\Delta_3/a\). Dividing that determinant by its first diagonal coefficient, \(\Delta_2/a\), gives \(\Delta_3/\Delta_2\). Thus the displayed expression is obtained by two ordinary square completions.

If all three minors are positive, each of the three coefficients \(a\), \(\Delta_2/a\), and \(\Delta_3/\Delta_2\) is positive. The successive expressions being squared are obtained from \(x,y,z\) by invertible changes of variables: each step replaces one variable by itself plus a combination of later variables. Consequently, a nonzero \((x,y,z)\) cannot make all three squared expressions zero. The quadratic form is therefore positive for every nonzero vector.

For the converse, suppose \(q\) is positive for every nonzero vector. Taking \((x,y,z)=(1,0,0)\) gives \(a>0\), so \(\Delta_1>0\). Set \(z=0\) and choose \(x=-by/a\). For any nonzero \(y\), completing the square gives

q\left(-\frac{b}{a}y,y,0\right) =\left(d-\frac{b^2}{a}\right)y^2 =\frac{\Delta_2}{a}y^2>0.

Since \(a>0\), this implies \(\Delta_2>0\). Now choose \(x=-(by+cz)/a\), which removes the first square. The remaining quadratic form in \(y,z\) must be positive for every nonzero pair \((y,z)\). Write its coefficients as \(u=d-b^2/a\), \(v=e-bc/a\), and \(w=f-c^2/a\). We already know \(u=\Delta_2/a>0\). Choosing \(y=-v/u\) and \(z=1\) shows that \(w-v^2/u>0\), so \(uw-v^2>0\). The determinant of this remaining two-by-two matrix is \(uw-v^2=\Delta_3/a\). Since \(a>0\), we obtain \(\Delta_3>0\). This proves necessity and completes the positive-definite criterion.

Why the Negative-Definite Conditions Work

A symmetric matrix \(\mathbf{H}\) is negative definite exactly when \(-\mathbf{H}\) is positive definite: its quadratic form is multiplied by \(-1\). The leading minors of \(-\mathbf{H}\) are \(-\Delta_1,\Delta_2,-\Delta_3\). Indeed, multiplying the leading \(k\)-by-\(k\) block by \(-1\) multiplies its determinant by \((-1)^k\). Applying the positive-definite criterion to \(-\mathbf{H}\) gives \(-\Delta_1>0,\ \Delta_2>0,\ -\Delta_3>0\), which is precisely the stated negative-definite pattern.

Worked Examples

Worked Example: Testing for Positive Definiteness

Consider \(\mathbf{H}=\begin{pmatrix}2&1&0\\1&2&1\\0&1&2\end{pmatrix}\). Calculate the three leading minors:

\Delta_1=2,\qquad \Delta_2=\det\begin{pmatrix}2&1\\1&2\end{pmatrix}=4-1=3,

\Delta_3 =2(2\cdot2-1\cdot1)-1(1\cdot2-1\cdot0) =2(3)-2=4.

All three are positive, so Sylvester’s criterion says \(\mathbf{H}\) is positive definite. As a check, its quadratic form can be written

\begin{aligned} q(x,y,z) &=2x^2+2xy+2y^2+2yz+2z^2\\ &=2\left(x+\frac{y}{2}\right)^2 +\frac{3}{2}\left(y+\frac{2z}{3}\right)^2 +\frac{4}{3}z^2. \end{aligned}

Expanding the squares recovers the original quadratic form. Each coefficient is positive, and the three squared expressions can vanish together only when \(z=0\), then \(y=0\), then \(x=0\). This confirms positive definiteness.

Worked Example: Testing for Negative Definiteness

Consider \(\mathbf{H}=\begin{pmatrix}-2&1&0\\1&-2&1\\0&1&-2\end{pmatrix}\). Its leading minors are

\Delta_1=-2,\qquad \Delta_2=(-2)(-2)-1^2=3,

\Delta_3=(-2)\bigl((-2)(-2)-1\bigr)-1\bigl(1(-2)-0\bigr) =-6+2=-4.

The signs are negative, positive, negative, so the matrix is negative definite. To check directly, its quadratic form is

q(x,y,z) =-2\left(x-\frac{y}{2}\right)^2 -\frac{3}{2}\left(y-\frac{2z}{3}\right)^2 -\frac{4}{3}z^2.

Expansion gives \(-2x^2+2xy-2y^2+2yz-2z^2\), the quadratic form of the given matrix. It is strictly negative for every nonzero vector, as the squared terms cannot all vanish unless \(x=y=z=0\).

Worked Example: Finding a Parameter Range

For which values of \(t\) is \(\mathbf{H}(t)=\begin{pmatrix}t&1&0\\1&2&1\\0&1&2\end{pmatrix}\) positive definite? The leading minors are

\Delta_1=t,\qquad \Delta_2=2t-1,\qquad \Delta_3=t(4-1)-2=3t-2.

Positive definiteness requires \(t>0\), \(2t-1>0\), and \(3t-2>0\). These are equivalent to \(t>0\), \(t>1/2\), and \(t>2/3\), respectively. All three hold exactly when \(t>2/3\).

At the endpoint \(t=2/3\), the determinant is zero. The nonzero vector \((3,-2,1)^{\top}\) gives zero in the quadratic form: its value is \(6-12+8-4+2=0\). Thus the endpoint is correctly excluded. As a check inside the range, at \(t=1\) the form factors as \((x+y)^2+(y+z)^2+z^2\), which is positive for every nonzero vector.

A Common Sign Error

A frequent mistake is to assume that negative definiteness requires all three leading minors to be negative. The second example exposes the error: its middle minor is \(\Delta_2=3>0\), yet the matrix is negative definite. The correct pattern is \(\Delta_1<0,\ \Delta_2>0,\ \Delta_3<0\), not three negative signs.

Also, the criterion is an if-and-only-if test for definiteness, not a rule that forces a matrix failing the positive or negative pattern into one particular alternative. If the required strict inequalities fail, that definiteness conclusion is unavailable; the matrix might have other behavior. This tutorial’s test alone does not classify every such case.

Using the Criterion with a Hessian

At a point where a twice-differentiable function has a symmetric Hessian, compute the leading minors of that Hessian in the given coordinate order. If the Hessian is positive definite, the second derivative test identifies a strict local minimum at a critical point; if it is negative definite, it identifies a strict local maximum there. The criterion supplies the definiteness check. Its symmetry assumption matters: Hessians are symmetric when the relevant second partial derivatives are continuous near the point.

For a three-variable Hessian, the practical sequence is therefore to calculate \(\Delta_1\), then \(\Delta_2\), then \(\Delta_3\), and compare their signs with the relevant pattern. Keeping the middle condition visible prevents the most common error: for both positive and negative definiteness, \(\Delta_2\) must be positive.

Check Your Understanding

Answer each question using the leading-principal-minor criterion.

State the three leading principal minors of a symmetric three-by-three matrix.
What strict sign pattern characterizes positive definiteness?
What strict sign pattern characterizes negative definiteness, and why is its middle sign positive?
For \(\Delta_1=3,\ \Delta_2=2,\ \Delta_3=5\), what definiteness conclusion follows?
Why does failure of a strict sign condition mean that the corresponding definiteness test does not apply?