Using Leading Principal Minors
For a symmetric matrix, testing positive definiteness can be much simpler than finding its eigenvalues. The entries of the matrix determine a short sequence of determinants, and the signs of those determinants decide the question. The method is particularly useful for small matrices and for families containing a parameter: instead of solving for eigenvalues as functions of that parameter, we can reduce the question to inequalities.
A real symmetric matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is positive definite if \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}>0\) for every nonzero \(\mathbf{x}\in\mathbb{R}^n\). For each \(k\) from \(1\) to \(n\), the leading principal submatrix of order \(k\) is the submatrix in the first \(k\) rows and first \(k\) columns. Its determinant is the \(k\)th leading principal minor, denoted \(\Delta_k\). In particular, \(\Delta_n=\det\mathbf{A}\).
This criterion is the tool for the calculations below. Its hypotheses matter: the matrix must be real and symmetric, and every leading principal minor must be strictly positive. A zero minor does not pass the test. Nor does a single positive determinant suffice: the intermediate leading minors must also be checked.
The Tests in Orders Two and Three
For a symmetric two-by-two matrix, write \(\mathbf{A}=\begin{bmatrix}a&b\\b&c\end{bmatrix}\). Its leading principal minors are \(a\) and \(ac-b^2\). Thus Sylvester’s Criterion becomes two inequalities:
For a symmetric three-by-three matrix, use the notation
The first leading minor is \(a\). The second is the determinant of the upper-left two-by-two block, \(ad-b^2\). Expanding the full determinant gives the third:
Therefore, the matrix is positive definite exactly when all three displayed expressions are positive. In parameter problems, compute each one before solving any inequalities. A parameter value at which one expression becomes zero marks a boundary of the strict test; it is not a positive-definite case.
Why Nonpositive Leading Minors Rule Out Positive Definiteness
The test has a useful necessary direction: a nonpositive leading principal minor rules out positive definiteness. The following argument explains why and also clarifies what a zero or negative value tells us.
Proof. Assume that \(\mathbf{A}\) is positive definite, and fix \(k\) with \(1\le k\le n\). Let \(\mathbf{B}\) be its leading principal submatrix of order \(k\). For any nonzero \(\mathbf{y}\in\mathbb{R}^k\), extend \(\mathbf{y}\) to \(\mathbf{x}\in\mathbb{R}^n\) by appending \(n-k\) zero entries. Then \(\mathbf{x}\ne\mathbf{0}\), and the definition of matrix multiplication gives \(\mathbf{y}^{\top}\mathbf{B}\mathbf{y}=\mathbf{x}^{\top}\mathbf{A}\mathbf{x}>0\). Thus \(\mathbf{B}\) is symmetric and positive definite. By the Spectral Theorem, its eigenvalues are real; positive definiteness makes each eigenvalue positive, since for an eigenvector \(\mathbf{v}\ne\mathbf{0}\) with eigenvalue \(\lambda\), \(\mathbf{v}^{\top}\mathbf{B}\mathbf{v}=\lambda\mathbf{v}^{\top}\mathbf{v}>0\). The determinant of \(\mathbf{B}\) is the product of its positive eigenvalues, so \(\det\mathbf{B}>0\). This holds for each \(k\), and \(\det\mathbf{B}=\Delta_k\). Hence every leading principal minor is positive. \(\square\)
The proposition proves that a zero or negative leading minor is enough to reject positive definiteness. By itself, however, that failure does not classify all values of the quadratic form. For a particular boundary case, examine the quadratic form or use additional information rather than treating “fails the positive-definite test” as a full classification.
Proof. By the Spectral Theorem, the matrix has two real eigenvalues \(\lambda_1,\lambda_2\), and their product is its determinant. If that determinant is negative, one eigenvalue is positive and the other is negative. Let \(\mathbf{v}_1,\mathbf{v}_2\) be corresponding nonzero eigenvectors. For either eigenvector, \(\mathbf{v}_i^{\top}\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i^{\top}\mathbf{v}_i\). Since \(\mathbf{v}_i^{\top}\mathbf{v}_i>0\), these two quadratic-form values have opposite signs. Thus the form takes both signs and cannot be positive definite. \(\square\)
Worked Applications
Worked Example: A Two-by-Two Matrix With a Parameter
For which real values of \(t\) is \(\mathbf{A}(t)=\begin{bmatrix}t&1\\1&2\end{bmatrix}\) positive definite? What happens at the threshold?
The matrix is symmetric. Its first leading minor and determinant are
Both must be strictly positive. The inequalities \(t>0\) and \(2t-1>0\) combine to give \(t>\tfrac12\). Therefore \(\mathbf{A}(t)\) is positive definite exactly for \(t>\tfrac12\).
At \(t=\tfrac12\), the determinant is zero, so the strict criterion fails. We can see the corresponding zero direction directly:
For \(t<\tfrac12\), the determinant \(2t-1\) is negative, so the second proposition shows that the quadratic form takes both signs. As a direct check, the vector \(\begin{bmatrix}-2\\1\end{bmatrix}\) gives the value \(4t-2<0\), whereas \(\begin{bmatrix}0\\1\end{bmatrix}\) gives \(2>0\). These checks agree with the minor test: the threshold is excluded, values below it are not positive definite, and values above it pass both strict inequalities.
Worked Example: Testing a Three-by-Three Matrix
Decide whether \(\mathbf{B}=\begin{bmatrix}2&1&0\\1&2&1\\0&1&2\end{bmatrix}\) is positive definite.
The matrix is symmetric. Its first leading minor is \(\Delta_1=2\), and its second is
For the full determinant, expand along the first row:
All three leading minors are positive: \(2>0\), \(3>0\), and \(4>0\). Sylvester’s Criterion therefore proves that \(\mathbf{B}\) is positive definite. In particular, checking only its full determinant would not have shown whether the other required inequalities hold; here we have verified each one.
Worked Example: A Three-by-Three Parameter Threshold
Find all real \(t\) for which \(\mathbf{C}(t)=\begin{bmatrix}2&1&0\\1&t&1\\0&1&2\end{bmatrix}\) is positive definite, and interpret \(t=1\).
The matrix is symmetric. Its first two leading minors are
Expanding the full determinant along its first row gives
For positive definiteness, we need \(2>0\), \(2t-1>0\), and \(4t-4>0\). The second inequality requires \(t>\tfrac12\); the third requires \(t>1\). Together these say exactly \(t>1\). At the threshold \(t=1\), the full leading minor is zero, so the matrix is not positive definite. Indeed,
The nonzero vector shown therefore has quadratic-form value zero. For \(t<1\), \(\Delta_3=4t-4<0\), which rules out positive definiteness by the necessary condition proved above. Also, the first coordinate vector gives quadratic-form value \(2>0\); together with the negative determinant, the eigenvalue product is negative, so there is a negative eigenvalue as well. Thus values below the threshold are not positive definite, while values above it satisfy all three strict conditions.
Common Pitfall: Using the Full Determinant Alone
A positive determinant does not by itself show positive definiteness. For example, consider the symmetric matrix \(\mathbf{D}=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\). Its determinant is \((-1)(-1)-0=1>0\). The incorrect conclusion “the determinant is positive, so \(\mathbf{D}\) is positive definite” ignores the first leading minor. In fact, \(\Delta_1=-1<0\), and the vector \(\begin{bmatrix}1\\0\end{bmatrix}\) gives quadratic-form value \(-1\). The correction is to check every leading minor, beginning with \(\Delta_1\), and require each to be strictly positive.
A related caution concerns boundary values. If some \(\Delta_k=0\), the criterion no longer certifies positive definiteness. You may conclude that the matrix is not positive definite, but the zero alone does not tell you the signs of the quadratic form in all other directions. In the examples, we checked a specific vector to understand the boundary case. The same discipline applies to parameterized matrices: solve the strict inequalities first, then inspect any threshold values separately if their behavior matters.
Check Your Understanding
For each matrix, calculate the stated leading minors before drawing a conclusion.
- For \(\begin{bmatrix}3&1\\1&c\end{bmatrix}\), find the values of \(c\) that give positive definiteness.
- For \(\begin{bmatrix}1&1&0\\1&d&1\\0&1&2\end{bmatrix}\), compute \(\Delta_1,\Delta_2,\Delta_3\) and determine the parameter range for positive definiteness.
- For \(\begin{bmatrix}t&2\\2&1\end{bmatrix}\), identify the threshold value and test the quadratic form on a nonzero vector at that value.
- Explain why a zero leading principal minor is enough to rule out positive definiteness, but may not by itself classify all quadratic-form values.
- Give a symmetric two-by-two matrix with positive determinant that is not positive definite, and identify which leading minor fails the test.
