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Positive Definite Matrix — Definition, Formula & Examples

A positive definite matrix is a symmetric matrix where every eigenvalue is positive, or equivalently, where xTAx>0\mathbf{x}^T A \mathbf{x} > 0 for every nonzero vector x\mathbf{x}. This property guarantees the matrix behaves like a "positive" quantity in higher dimensions.

A real n×nn \times n symmetric matrix AA is positive definite if and only if the quadratic form xTAx>0\mathbf{x}^T A \mathbf{x} > 0 for all x∈Rn\mathbf{x} \in \mathbb{R}^n with x≠0\mathbf{x} \neq \mathbf{0}. Equivalent conditions include: (1) all eigenvalues of AA are strictly positive, (2) all leading principal minors (upper-left determinants) of AA are strictly positive, and (3) AA can be factored as A=RTRA = R^T R for some invertible matrix RR.

Key Formula

xTAx>0for all x≠0\mathbf{x}^T A \mathbf{x} > 0 \quad \text{for all } \mathbf{x} \neq \mathbf{0}
Where:
  • AA = A real $n \times n$ symmetric matrix
  • x\mathbf{x} = Any nonzero column vector in $\mathbb{R}^n$

How It Works

To test whether a symmetric matrix is positive definite, you can use any of three practical methods. The **eigenvalue test** checks that every eigenvalue is positive. The **leading principal minor test** (Sylvester's criterion) checks that the determinants of the 1×11 \times 1, 2×22 \times 2, ..., n×nn \times n upper-left submatrices are all positive. For small matrices, you can also directly verify xTAx>0\mathbf{x}^T A \mathbf{x} > 0 by expanding the quadratic form and confirming it is always positive for nonzero x\mathbf{x}. The leading principal minor test is often the fastest approach for 2×22 \times 2 and 3×33 \times 3 matrices.

Worked Example

Problem: Determine whether the matrix A=(4223)A = \begin{pmatrix} 4 & 2 \\ 2 & 3 \end{pmatrix} is positive definite using the leading principal minor test.
Step 1: Confirm the matrix is symmetric. Since AT=AA^T = A, it is symmetric and the test applies.
Step 2: Compute the first leading principal minor, which is just the top-left entry.
Δ1=4>0✓\Delta_1 = 4 > 0 \quad \checkmark
Step 3: Compute the second leading principal minor, which is the determinant of the full 2×22 \times 2 matrix.
Δ2=det⁡(A)=(4)(3)−(2)(2)=12−4=8>0✓\Delta_2 = \det(A) = (4)(3) - (2)(2) = 12 - 4 = 8 > 0 \quad \checkmark
Step 4: Since both leading principal minors are strictly positive, the matrix is positive definite.
Answer: The matrix AA is positive definite because Δ1=4>0\Delta_1 = 4 > 0 and Δ2=8>0\Delta_2 = 8 > 0.

Another Example

Problem: Show that the matrix B=(1332)B = \begin{pmatrix} 1 & 3 \\ 3 & 2 \end{pmatrix} is NOT positive definite.
Step 1: Check symmetry: BT=BB^T = B, so the matrix is symmetric.
Step 2: Compute the first leading principal minor.
Δ1=1>0✓\Delta_1 = 1 > 0 \quad \checkmark
Step 3: Compute the second leading principal minor.
Δ2=(1)(2)−(3)(3)=2−9=−7<0×\Delta_2 = (1)(2) - (3)(3) = 2 - 9 = -7 < 0 \quad \times
Step 4: Since Δ2\Delta_2 is negative, the matrix fails the test and is not positive definite.
Answer: BB is not positive definite because det⁡(B)=−7<0\det(B) = -7 < 0.

Why It Matters

Positive definite matrices appear throughout college-level linear algebra and multivariable calculus. In optimization, the Hessian matrix being positive definite confirms that a critical point is a local minimum — a key result in Calculus III and machine learning. In statistics, every valid covariance matrix must be positive semi-definite, so understanding this property is essential in courses on probability, data science, and econometrics.

Common Mistakes

Mistake: Checking that all entries of the matrix are positive and concluding it is positive definite.
Correction: Positive definiteness is about eigenvalues or the quadratic form, not individual entries. A matrix with all positive entries can fail to be positive definite, and a positive definite matrix can have negative off-diagonal entries.
Mistake: Applying the leading principal minor test to a non-symmetric matrix.
Correction: Sylvester's criterion only works for symmetric matrices. If A≠ATA \neq A^T, first symmetrize it as 12(A+AT)\frac{1}{2}(A + A^T) or use eigenvalue methods.

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