Using Principal Minors in Three-Variable Tests

Why Principal Minors Are Useful

A three-variable Hessian contains six distinct entries when it is symmetric. Its eigenvalues can be useful for classifying the matrix, but computing them may require solving a cubic equation. When the entries are exact and simple, leading principal minors often give a faster route to determining whether the Hessian is positive definite or negative definite.

The sign tests are powerful, but they must be used carefully. The correct sign pattern proves definiteness. A failed pattern, by itself, does not prove indefiniteness: a matrix may instead be semidefinite. This tutorial develops the tests, explains their basis, and shows how to distinguish a decisive result from an inconclusive one.

Leading Principal Minors of a Three-by-Three Matrix

Let \(\mathbf{A}\) be a real symmetric matrix. Its leading principal submatrices are formed by keeping the first \(k\) rows and the first \(k\) columns, for \(k=1,2,3\). The determinant of each such submatrix is the \(k\)th leading principal minor, denoted by \(\Delta_k\).

\mathbf{A}= \begin{pmatrix} a&b&d\\ b&c&e\\ d&e&f \end{pmatrix}, \qquad \Delta_1=a,\qquad \Delta_2= \begin{vmatrix} a&b\\ b&c \end{vmatrix} =ac-b^2,\qquad \Delta_3=\det(\mathbf{A}).

The word leading matters: \(\Delta_2\) uses rows and columns 1 and 2, not any chosen pair. A principal minor more generally is the determinant obtained by selecting the same set of row and column indices. For example, the minor using indices 1 and 3 is \(af-d^2\). The strict definiteness tests below use the leading minors in the stated order.

Definition: A symmetric matrix is positive definite if \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}>0\) for every nonzero vector \(\mathbf{x}\). It is negative definite if \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}<0\) for every nonzero \(\mathbf{x}\). It is indefinite if its quadratic form takes both positive and negative values.

A Hessian of a twice continuously differentiable function is symmetric. At a critical point, these matrix properties determine whether the quadratic part of the local Taylor expansion is positive in every direction, negative in every direction, or has both signs. Here the task is to classify the Hessian itself using its minors.

The Sign Patterns

For a real symmetric three-by-three matrix, Sylvester’s criterion gives the following tests. Each condition is necessary and sufficient for the stated strict definiteness.

Theorem: The matrix \(\mathbf{A}\) is positive definite exactly when all three leading principal minors are positive. It is negative definite exactly when the leading principal minors alternate in sign, beginning with a negative sign.

\begin{array}{ll} \text{Positive definite:}&\Delta_1>0,\quad\Delta_2>0,\quad\Delta_3>0,\\[2mm] \text{Negative definite:}&\Delta_1<0,\quad\Delta_2>0,\quad\Delta_3<0. \end{array}

Equivalently, the negative-definite conditions are \((-1)^k\Delta_k>0\) for \(k=1,2,3\). These are strict tests: if one of the required minors is zero, that sign pattern does not establish positive or negative definiteness.

Why the Sign Tests Work

A useful justification comes from completing squares in the quadratic form. Suppose first that \(\Delta_1\ne 0\) and \(\Delta_2\ne 0\), so the divisions below are valid. For \(\mathbf{x}=(x,y,z)^{\top}\), expansion and regrouping give:

\begin{aligned} \mathbf{x}^{\top}\mathbf{A}\mathbf{x} ={}&a\left(x+\frac{b}{a}y+\frac{d}{a}z\right)^2\\ &+\frac{\Delta_2}{a} \left(y+\frac{ae-bd}{\Delta_2}z\right)^2 +\frac{\Delta_3}{\Delta_2}z^2. \end{aligned}

To verify the coefficients, the first square contributes \(a x^2+2bxy+2dxz\), as required. After subtracting that square, the coefficient of \(y^2\) is \(c-b^2/a=\Delta_2/a\), and the coefficient of \(yz\) is \(e-bd/a=(ae-bd)/a\). Completing the square in \(y\) therefore uses the factor \((ae-bd)/\Delta_2\). The remaining coefficient of \(z^2\) is \(\Delta_3/\Delta_2\), as can be confirmed by expanding the determinant of \(\mathbf{A}\).

The three coefficients of the squares are therefore

\frac{\Delta_1}{1},\qquad \frac{\Delta_2}{\Delta_1},\qquad \frac{\Delta_3}{\Delta_2}.

The three expressions being squared can serve as new coordinates: the transformation from \((x,y,z)\) is invertible because its diagonal coefficients are 1. Thus every nonzero vector gives a nonzero triple of new coordinates. If all three coefficients are positive, the quadratic form is positive for every nonzero vector; if all are negative, it is negative for every nonzero vector. The stated minor patterns make all three coefficients positive or all three negative, respectively.

This square completion directly explains the tests when the relevant minors are nonzero. The full criterion, including its necessity, is Sylvester’s criterion for real symmetric matrices. In practice, the sign patterns are applied to the minors as a complete test; they are not replaced by an assumption that a failed pattern must mean indefiniteness.

Worked Examples

Worked Example: Testing for Positive Definiteness

Classify the symmetric matrix

\mathbf{A}= \begin{pmatrix} 2&1&0\\ 1&2&1\\ 0&1&2 \end{pmatrix}.

First, calculate the leading minors in order:

\Delta_1=2.

For the second leading minor,

\Delta_2= \begin{vmatrix} 2&1\\ 1&2 \end{vmatrix} =(2)(2)-(1)(1)=4-1=3.

For the full determinant, expand along the first row:

\begin{aligned} \Delta_3 &=2\begin{vmatrix}2&1\\1&2\end{vmatrix} -1\begin{vmatrix}1&1\\0&2\end{vmatrix} +0\\ &=2(4-1)-(2-0)\\ &=6-2=4. \end{aligned}

All three minors are positive: \(\Delta_1=2>0\), \(\Delta_2=3>0\), and \(\Delta_3=4>0\). The matrix is positive definite by the criterion. As a check on the completed-square interpretation, the square coefficients are \(2\), \(3/2\), and \(4/3\); each is positive, since \(\Delta_2/\Delta_1=3/2\) and \(\Delta_3/\Delta_2=4/3\).

Worked Example: Testing for Negative Definiteness

Classify the matrix

\mathbf{B}= \begin{pmatrix} -2&1&0\\ 1&-2&1\\ 0&1&-2 \end{pmatrix}.

The first leading minor is

\Delta_1=-2.

The second is

\Delta_2= \begin{vmatrix} -2&1\\ 1&-2 \end{vmatrix} =(-2)(-2)-(1)(1)=4-1=3.

Expanding the full determinant along its first row gives

\begin{aligned} \Delta_3 &=(-2)\begin{vmatrix}-2&1\\1&-2\end{vmatrix} -1\begin{vmatrix}1&1\\0&-2\end{vmatrix}\\ &=(-2)(4-1)-(-2-0)\\ &=-6+2=-4. \end{aligned}

The signs are \(\Delta_1<0\), \(\Delta_2>0\), and \(\Delta_3<0\), precisely the negative-definite pattern. Therefore \(\mathbf{B}\) is negative definite. The completed-square coefficients are \(-2\), \(3/(-2)=-3/2\), and \((-4)/3=-4/3\), all negative, which checks the conclusion.

Worked Example: Establishing Indefiniteness with a Principal Minor

Consider

\mathbf{C}= \begin{pmatrix} 1&2&0\\ 2&1&0\\ 0&0&3 \end{pmatrix}.

The leading minors are

\Delta_1=1,\qquad \Delta_2= \begin{vmatrix}1&2\\2&1\end{vmatrix} =(1)(1)-(2)(2)=1-4=-3, \qquad \Delta_3=(-3)(3)=-9.

These do not match either strict-definiteness pattern. To prove that the matrix is indefinite, evaluate its quadratic form on two vectors. For \(\mathbf{u}=(1,0,0)^{\top}\),

\mathbf{u}^{\top}\mathbf{C}\mathbf{u}=1>0.

For \(\mathbf{v}=(1,-1,0)^{\top}\), direct multiplication gives

\mathbf{C}\mathbf{v} = \begin{pmatrix}-1\\1\\0\end{pmatrix}, \qquad \mathbf{v}^{\top}\mathbf{C}\mathbf{v} =(1)(-1)+(-1)(1)+0=-2<0.

The quadratic form takes both signs, so \(\mathbf{C}\) is indefinite. In this example the negative second leading minor already signals indefiniteness of the restriction to the first two coordinates; the two vector calculations explicitly verify the opposite signs in the full quadratic form.

What a Failed Test Does Not Tell You

If a matrix fails the positive-definite pattern, it is not automatically indefinite. Consider the diagonal matrix

\mathbf{D}= \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&2 \end{pmatrix}.

Its leading minors are \(\Delta_1=1\), \(\Delta_2=0\), and \(\Delta_3=0\), so it fails the strict positive-definiteness test. But its quadratic form is \(x^2+2z^2\), which is never negative. Thus the failed strict test alone does not prove that the form takes both signs. Matrices on this boundary require separate analysis; the zero-minor case is not settled by the strict criteria.

Important: The sign patterns are conclusive when they hold. If neither pattern holds, do not label the matrix indefinite without further evidence. To establish indefiniteness, show that the quadratic form has a positive value for one vector and a negative value for another, or use another valid criterion.

Using the Test on a Hessian

At a critical point of a twice continuously differentiable function of three variables, form the Hessian and evaluate its entries at that point. Since the Hessian is symmetric, calculate \(\Delta_1\), \(\Delta_2\), and \(\Delta_3\) from that evaluated matrix, then compare their signs with the two patterns. Keep the order of the variables fixed while computing the leading minors.

A positive-definite Hessian gives positive curvature in every nonzero direction, while a negative-definite Hessian gives negative curvature in every nonzero direction. If the signs fail both patterns, the test has not by itself resolved whether the Hessian is indefinite or has zero directions. This distinction is important: the minor test is a classification tool, not permission to infer more than its conditions establish.

Common Pitfall: Treating Any Negative Minor as the Answer

For positive definiteness, all three leading minors must be positive; for negative definiteness, they must follow the alternating pattern. A single negative minor cannot be interpreted without checking which minor it is and what pattern is required. For instance, the negative value \(\Delta_2=-3\) in the third example rules out both definite patterns, but the reason we concluded indefiniteness was the additional verification of positive and negative quadratic-form values.

Another common error is to calculate a non-leading principal minor and substitute it into a test that specifically requires a leading one. The minor \(af-d^2\), for example, uses coordinates 1 and 3; it is a principal minor, but it is not \(\Delta_2\). Compute the prescribed leading minors first, and use other principal minors or vector evaluations only when they help resolve a question the strict tests leave open.

Summary

For a symmetric three-by-three matrix, the leading principal minors are \(\Delta_1\), \(\Delta_2\), and \(\Delta_3\). Their sign patterns classify strict definiteness: all positive means positive definite, while negative, positive, negative means negative definite. Completing the square explains why these patterns control the signs of the quadratic form. If neither pattern holds, further evidence is needed before calling the matrix indefinite.

Check Your Understanding

Use the definitions and tests in this tutorial to answer the following questions.

For a symmetric three-by-three matrix, which rows and columns define the second leading principal minor?
What signs must \(\Delta_1,\Delta_2,\Delta_3\) have for negative definiteness?
Why does failure of the positive-definiteness test not, by itself, prove indefiniteness?
For \(\mathbf{A}=\begin{pmatrix}3&0&0\\0&2&0\\0&0&1\end{pmatrix}\), calculate all three leading minors and classify the matrix.
What two kinds of quadratic-form values are needed to establish that a symmetric matrix is indefinite?