Positive Definite Matrix: Which Test to Use, and When It Fails

A positive definite matrix is a real symmetric matrix A for which xᵀAx is greater than zero for every nonzero vector x. The expression xᵀAx is the quadratic form of A: for A = [[a, b], [b, c]] it equals ax² + 2bxy + cy². Symmetric means A equals its transpose Aᵀ. Positive semidefinite means xᵀAx ≥ 0 for every x; indefinite means xᵀAx takes both signs.

Symmetry is part of the definition in most exam settings because the quadratic form cannot see the asymmetric part of a matrix: xᵀAx = xᵀ((A + Aᵀ)/2)x. Some textbooks therefore define definiteness by xᵀAx > 0 alone. Then a non-symmetric matrix is positive definite exactly when its symmetric part (A + Aᵀ)/2 is (Wikipedia, Definite matrix). Read the question's definition before you apply a test.

For a symmetric matrix, three statements are equivalent: xᵀAx > 0 for all nonzero x; every eigenvalue is positive; every leading principal minor is positive. The eigenvalue link has a one-line reason. If Av = λv with v ≠ 0, then vᵀAv = λ(vᵀv), and vᵀv > 0, so λ must be positive.

Positive eigenvalues do not rescue a non-symmetric matrix. Take A = [[1, 3], [0, 1]]. Both eigenvalues equal 1, yet for x = (1, −1) we get xᵀAx = −1. Its symmetric part [[1, 1.5], [1.5, 1]] has eigenvalues 2.5 and −0.5.

A matrix is positive definite exactly when it is symmetric and every leading principal minor D₁, D₂, … Dₙ is positive, so the tests run from the cheapest rejection to that decisive one. The characteristic-polynomial and quadratic-form routes below are alternatives for when they are quicker.

A leading principal minor Dₖ is the determinant of the top-left k×k block of the matrix.

Is the matrix symmetric?

Pass: every entry aᵢⱼ equals aⱼᵢ. Fail: any pair differs. If the question's definition includes symmetry, a non-symmetric matrix is not positive definite. If it defines the term by xᵀAx > 0 only, run the remaining steps on (A + Aᵀ)/2.

Does a quick check rule it out?

Fail the matrix at once if any diagonal entry is zero or negative, or if any pair gives aᵢᵢaⱼⱼ − aᵢⱼ² ≤ 0. The first fails because x = eᵢ (a vector with a single 1) gives xᵀAx = aᵢᵢ; the second is the same test on a 2×2 block. Passing proves nothing: these are fail checks only.

Are all leading principal minors positive?

Pass: D₁ > 0, D₂ > 0, … Dₙ > 0. This is Sylvester's criterion. Fail: the first Dₖ ≤ 0 ends the test. For a 2×2 matrix [[a, b], [b, c]] it reduces to a > 0 and ac − b² > 0.

Can the characteristic polynomial settle it faster?

For a symmetric matrix every root of det(λI − A) is real, and all roots are positive exactly when the coefficients alternate strictly in sign (+, −, +, −, …). Pass: strict alternation. Fail: a missing term or a repeated sign. This is Descartes' rule of signs applied to a polynomial known to have only real roots.

It pays off when you have the polynomial already, for example from a Cayley Hamilton theorem problem. If you have already diagonalised the matrix, the eigenvalues are the diagonal entries; our diagonalization of a matrix page covers that route.

What if the question gives a polynomial instead of a matrix?

Write the symmetric matrix by halving each cross-term coefficient, then run the steps above. For ax² + bxy + cy², the matrix is [[a, b/2], [b/2, c]]. The form is positive definite exactly when a > 0 and b² < 4ac.

What changes for positive semidefinite?

Leading minors that are merely non-negative do not prove positive semidefinite. A symmetric matrix is positive semidefinite exactly when all principal minors are non-negative, not only the leading ones: 2ⁿ − 1 minors for an n×n matrix (both rules: Definite matrix, Wikipedia). Eigenvalues ≥ 0 are usually quicker.

On a computer, attempting a Cholesky factorisation (writing A = RᵀR with R upper triangular) is the fastest numerical test. The matrix is positive definite if the factorisation completes; if not, it typically breaks down in the early stages (Nick Higham, 2020).

Every shortcut except the full leading-minor test passed matrices that are not positive definite. We checked all 15,625 symmetric 3×3 matrices with integer entries from −2 to 2 (5⁶ combinations of the six independent entries), comparing each shortcut with the eigenvalues. Of these, 96 (0.61%) are positive definite and 241 are positive semidefinite.

Every one of the 6,960 matrices with a positive determinant that fail has exactly two negative eigenvalues, so the determinant's sign hides them: two negatives multiply to a positive. Strict Sylvester agreed with the eigenvalue test on all 15,625.

This grid is our model of the shortcuts, not a measure of how often exams use them. The percentages belong to this grid of entries and would change on another. The pattern is general: a positive determinant can hide two negative eigenvalues, and non-negative leading minors do not prove semidefiniteness.

How to redo it (run in November 2026; a matrix counts as positive definite when its smallest eigenvalue exceeds 10⁻⁹, and as positive semidefinite when it exceeds −10⁻⁹):

import itertools
import numpy as np

rows = []
for a, b, c, d, e, f in itertools.product(range(-2, 3), repeat=6):
    A = np.array([[a, b, c], [b, d, e], [c, e, f]], float)
    D = (a, a*d - b*b, round(np.linalg.det(A)))      # D1, D2, D3
    w = np.linalg.eigvalsh(A)
    rows.append(dict(D=D, diag=(a > 0 and d > 0 and f > 0),
                     pd=w.min() > 1e-9, psd=w.min() > -1e-9,
                     two_neg=int((w < -1e-9).sum()) == 2))

def show(label, test):
    hit = [r for r in rows if test(r)]
    print(label, len(hit), "positive definite:", sum(r["pd"] for r in hit),
          "positive semidefinite:", sum(r["psd"] for r in hit))

print("all", len(rows), "PD", sum(r["pd"] for r in rows), "PSD", sum(r["psd"] for r in rows))
show("det > 0", lambda r: r["D"][2] > 0)                    # 7056, 96 PD
print("det > 0, not PD, two negative eigenvalues:",
      sum(r["D"][2] > 0 and not r["pd"] and r["two_neg"] for r in rows))  # 6960
show("diagonal > 0", lambda r: r["diag"])                   # 1000, 96 PD
show("D1, D2 > 0", lambda r: r["D"][0] > 0 and r["D"][1] > 0)  # 1250, 96 PD
show("D1, D2, D3 >= 0", lambda r: all(x >= 0 for x in r["D"]))  # 749, 241 PSD
show("D1, D2, D3 > 0", lambda r: all(x > 0 for x in r["D"]))     # 96, 96 PD

The "passed but wrong" column is the matrices that pass minus those correctly classified: 7,056 − 96, 1,000 − 96, 1,250 − 96 and 749 − 241.

Two of the six are positive definite: the 3×3 matrix in row 1 and the form in row 4. The other four fail for four different reasons: a zero minor, an indefinite form, minors that look fine but are not, and a non-symmetric matrix with positive eigenvalues. Work each one first, then read the answer.

CSIR NET and GATE MA name quadratic forms in their linear algebra syllabus. IIT JAM Mathematics (the MA paper) does not. Only the JAM Mathematical Statistics paper spells out positive definite matrices. Here is what each official document says:

CSIR HRDG syllabus file dated November 2018, checked November 2026. Syllabuses change between cycles, so confirm against the current CSIR NET notice and the current GATE and JAM notices.

Our reading is that "classification of quadratic forms" means deciding whether a form is positive definite, positive semidefinite, negative definite (xᵀAx < 0 for every nonzero x) or indefinite, so the tests above apply to CSIR NET and GATE MA questions even when the words "positive definite" are absent. For the full GATE list, see our GATE MA syllabus page.

Our Linear Algebra - Master Course, taught by Sunil Bansal (SB Sir), covers quadratic forms, eigenvalues and eigenvectors, and diagonalization, in Hindi and English, with quizzes and live tests. As of November 2026 it costs ₹2,499, and the course page lists a start date of 1 September 2026. Details are on the Linear Algebra - Master Course page.

SB Sir is CSIR-NET qualified with an All India Rank of 11 (December 2017) and is GATE MA qualified. If you are still choosing between linear algebra courses for your exam, our linear algebra course online guide sets out what to check.