Leading Principal Minors as a Positivity Test
A real symmetric matrix is positive definite when its quadratic form is strictly positive at every nonzero vector. That definition asks us to check infinitely many vectors. Sylvester’s criterion replaces that task with a finite list of determinant inequalities. The essential point is that the inequalities must come from the leading principal submatrices: those formed using the first \(k\) rows and first \(k\) columns, for each \(k\). The result is an exact test for positive definiteness, not merely a sufficient condition.
Throughout, matrices and vectors are real. For a symmetric matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\), the associated quadratic form is \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\). Positive definiteness is a strict condition: the value must be positive for every \(\mathbf{x}\ne\mathbf{0}\), not merely nonnegative.
The first leading principal minor is the top-left entry; the last is the determinant of the full matrix. For a \(3\times3\) matrix, the sequence consists of the top-left \(1\times1\), \(2\times2\), and \(3\times3\) determinants. These are not, in general, all the principal minors: a principal submatrix may use a different set of matching row and column indices. Sylvester’s criterion concerns this particular nested sequence.
The Criterion and the Schur Complement
The proof uses induction on the matrix size. Its central step isolates the last row and column. Write a symmetric matrix in block form as
where \(\mathbf{B}\) is an \((n-1)\times(n-1)\) symmetric matrix, \(\mathbf{u}\in\mathbb{R}^{n-1}\), and \(a\in\mathbb{R}\). When \(\mathbf{B}\) is invertible, the scalar \(s=a-\mathbf{u}^{\top}\mathbf{B}^{-1}\mathbf{u}\) is the Schur complement of \(\mathbf{B}\) in \(\mathbf{A}\). Two identities explain its role. First, the determinant factors as \(\det\mathbf{A}=(\det\mathbf{B})s\). To obtain this, subtract \(\mathbf{u}^{\top}\mathbf{B}^{-1}\) times the top block of rows from the last row. This row operation leaves the determinant unchanged and transforms the last row into \(\begin{bmatrix}\mathbf{0}^{\top}&s\end{bmatrix}\); the resulting block upper-triangular matrix has determinant \((\det\mathbf{B})s\).
Second, for \(\mathbf{x}\in\mathbb{R}^{n-1}\) and \(t\in\mathbb{R}\), symmetry of \(\mathbf{B}\) gives the quadratic-form identity
Indeed, expanding the first term on the right gives \(\mathbf{x}^{\top}\mathbf{B}\mathbf{x}+2t\mathbf{u}^{\top}\mathbf{x}+t^2\mathbf{u}^{\top}\mathbf{B}^{-1}\mathbf{u}\). Adding \(s t^2\) gives the block quadratic form because \(s+\mathbf{u}^{\top}\mathbf{B}^{-1}\mathbf{u}=a\). Thus, when \(\mathbf{B}\) is positive definite, the sign of \(s\) determines whether the full form is positive in directions with a nonzero final coordinate.
Proof. We prove the theorem by induction on \(n\). For \(n=1\), write \(\mathbf{A}=[a]\). Its quadratic form is \(a x^2\), which is positive for every \(x\ne0\) exactly when \(a>0\). Since \(\Delta_1=a\), the claim holds.
Suppose now that \(n\ge2\) and that the result holds for symmetric matrices of size \(n-1\). Partition \(\mathbf{A}\) as above, so that \(\mathbf{B}=\mathbf{A}_{n-1}\). First assume \(\mathbf{A}\) is positive definite. For every nonzero \(\mathbf{y}\in\mathbb{R}^{n-1}\), the vector \(\begin{bmatrix}\mathbf{y}\\0\end{bmatrix}\) is nonzero and satisfies \(\begin{bmatrix}\mathbf{y}^{\top}&0\end{bmatrix}\mathbf{A}\begin{bmatrix}\mathbf{y}\\0\end{bmatrix}=\mathbf{y}^{\top}\mathbf{B}\mathbf{y}>0\). Thus \(\mathbf{B}\) is positive definite. By the induction hypothesis, its leading principal minors \(\Delta_1,\dots,\Delta_{n-1}\) are positive. Also, \(\mathbf{B}\) is invertible: if \(\mathbf{B}\mathbf{z}=\mathbf{0}\) for a nonzero \(\mathbf{z}\), then \(\mathbf{z}^{\top}\mathbf{B}\mathbf{z}=0\), contradicting positive definiteness.
Let \(s=a-\mathbf{u}^{\top}\mathbf{B}^{-1}\mathbf{u}\). The vector \(\begin{bmatrix}-\mathbf{B}^{-1}\mathbf{u}\\1\end{bmatrix}\) is nonzero because its last coordinate is \(1\). Substitution into the quadratic form gives \(s>0\). The determinant identity therefore yields \(\Delta_n=\det\mathbf{A}=(\det\mathbf{B})s=\Delta_{n-1}s>0\). This proves that every leading principal minor is positive.
Conversely, assume \(\Delta_k>0\) for all \(k=1,\dots,n\). The leading principal minors of \(\mathbf{B}=\mathbf{A}_{n-1}\) are \(\Delta_1,\dots,\Delta_{n-1}\). By the induction hypothesis, \(\mathbf{B}\) is positive definite. It is therefore invertible by the same null-vector argument. The determinant identity gives \(s=\Delta_n/\det\mathbf{B}=\Delta_n/\Delta_{n-1}>0\). For any \(\begin{bmatrix}\mathbf{x}^{\top}&t\end{bmatrix}^{\top}\ne\mathbf{0}\), use the quadratic-form identity. If \(t\ne0\), its term \(s t^2\) is positive and its \(\mathbf{B}\)-quadratic term is nonnegative (zero only if \(\mathbf{x}+\mathbf{B}^{-1}\mathbf{u}t=\mathbf{0}\)). If \(t=0\), then \(\mathbf{x}\ne\mathbf{0}\), and the form equals \(\mathbf{x}^{\top}\mathbf{B}\mathbf{x}>0\). Hence \(\mathbf{A}\) is positive definite. Both implications hold, completing the induction. \(\square\)
Reading the Test in Small Matrices
Worked Example: A Positive Definite \(2\times2\) Matrix
Determine whether \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\) is positive definite using its leading principal minors.
The first leading principal submatrix is \([2]\), so \(\Delta_1=2\). The second is the full matrix, so \(\Delta_2=2\cdot2-1\cdot1=3\). Both leading principal minors are positive. Since \(\mathbf{A}\) is real and symmetric, Sylvester’s criterion says it is positive definite.
Check: For \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\), its quadratic form is \(2x^2+2xy+2y^2=(x+y)^2+x^2+y^2\). If \((x,y)\ne(0,0)\), at least one of \(x^2,y^2\) is positive, so the form is positive, in agreement with the test.
Worked Example: Checking Three Leading Minors
Test the symmetric matrix \(\mathbf{A}=\begin{bmatrix}2&1&0\\1&2&1\\0&1&2\end{bmatrix}\) for positive definiteness.
The first minor is \(\Delta_1=2\). The second is \(\Delta_2=\det\begin{bmatrix}2&1\\1&2\end{bmatrix}=4-1=3\). Expanding the full determinant along its first row gives \(\Delta_3=2(2\cdot2-1\cdot1)-1(1\cdot2-1\cdot0)+0=2(3)-2=4\). All three are positive, so the criterion gives positive definiteness.
Check: For \(\mathbf{x}=\begin{bmatrix}x\\y\\z\end{bmatrix}\), the quadratic form is \(2x^2+2y^2+2z^2+2xy+2yz=(x+y)^2+(y+z)^2+x^2+z^2\). If this sum is zero, then \(x=z=0\), and the two remaining squares force \(y=0\). Thus only the zero vector gives zero, confirming strict positivity.
Worked Example: Positive Diagonal Entries Are Not Enough
Consider \(\mathbf{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\). Its diagonal entries are both positive, but that fact alone does not decide whether the matrix is positive definite.
The first leading principal minor is \(\Delta_1=1\). The second is \(\Delta_2=1\cdot1-2\cdot2=-3\). Since \(\Delta_2\) is not positive, Sylvester’s criterion says that \(\mathbf{A}\) is not positive definite.
Check: At the nonzero vector \(\begin{bmatrix}1\\-1\end{bmatrix}\), the quadratic form equals \(1-4+1=-2<0\). This directly confirms the failure of positive definiteness.
Common Mistake: Stopping at the First Minor
A common incorrect argument is: “The top-left entry is positive, and the diagonal entries are positive, so the matrix is positive definite.” The preceding example is a concrete correction: the entries on the diagonal of \(\begin{bmatrix}1&2\\2&1\end{bmatrix}\) are positive, but the second leading principal minor is \(-3\), and the quadratic form takes the negative value \(-2\) at \(\begin{bmatrix}1\\-1\end{bmatrix}\). The test requires every leading principal minor to be positive; checking only the first, or only the diagonal entries, is insufficient.
The symmetry hypothesis also matters: Sylvester’s criterion as stated is a test for real symmetric matrices. The proof uses symmetry in the block form and in the quadratic-form identity. Do not apply the criterion to a nonsymmetric matrix as though it characterized positivity of its quadratic expression. Finally, strict inequalities matter. A zero leading principal minor does not satisfy this criterion for positive definiteness, which requires strict positivity in every nonzero direction.
The proof also explains why the determinants are tested in order. At each induction step, the preceding leading block \(\mathbf{B}\) must already be positive definite, and the ratio \(\Delta_n/\Delta_{n-1}\) is the Schur complement \(s\). Positivity of that last scalar extends positivity from the first \(n-1\) coordinates to all \(n\) coordinates. In the other direction, evaluating the full quadratic form in the direction \(\begin{bmatrix}-\mathbf{B}^{-1}\mathbf{u}\\1\end{bmatrix}\) forces that same scalar to be positive. This is why the determinant conditions are both necessary and sufficient.
Check Your Understanding
Use the definition and Sylvester’s criterion to justify each conclusion.
- For \(\mathbf{A}=\begin{bmatrix}3&1\\1&2\end{bmatrix}\), calculate both leading principal minors and decide whether \(\mathbf{A}\) is positive definite.
- For \(\mathbf{B}=\begin{bmatrix}1&1&0\\1&3&1\\0&1&2\end{bmatrix}\), calculate its three leading principal minors.
- Explain why the induction proof establishes that \(\mathbf{B}=\mathbf{A}_{n-1}\) is invertible when \(\mathbf{A}\) is positive definite.
- In the block proof, why does \(\Delta_n/\Delta_{n-1}\) equal the Schur complement \(s\), when the denominator is nonzero?
- Give a symmetric \(2\times2\) matrix with positive diagonal entries that is not positive definite, and verify the failure using a leading principal minor or a nonzero test vector.
