LINEAR-ALGEBRA

One Matrix, Four Tests: The Hidden Unity of Positive Definiteness

IKey Formulas Reference

For a $2\times 2$ symmetric matrix $A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ , the four equivalent tests for positive definiteness are:

Test	Condition
Eigenvalues	Both $\lambda_1, \lambda_2 > 0$
Leading principal determinants	$a > 0$ and $ac - b^2 > 0$
Pivots	$a > 0$ and $(ac - b^2)/a > 0$
Quadratic form	$x^T A x = a x_1^2 + 2b x_1 x_2 + c x_2^2 > 0$ for all $x \neq 0$

All four tests are logically equivalent. Pass one, pass all. Fail one, fail all.

IIVisualization 1 — The Four Tests, One Verdict

Change any entry of the symmetric matrix and all four tests flip their verdict together.

a (top-left) 2 b (off-diagonal) 6 c (bottom-right) 20

IIIVisualization 2 — The Razor's Edge

With $a=2,\ b=6$ fixed, the sign of $\det = 2c - 36$ is all that separates a saddle from a bowl.

c 18

IVQuiz Questions

Question 1 · Apply the Tests

Consider the symmetric matrix $A = \begin{pmatrix} 3 & 2 \\ 2 & 5 \end{pmatrix}.$

Is $A$ positive definite?

✅ Correct! Both leading principal determinants are positive, so $A$ is positive definite.

❌ Not quite. Remember the formula: $\det = ac - b^2$ , not $b^2 - ac$ . And off-diagonal nonzero entries are perfectly fine — most positive definite matrices have them.

Show solution

Solution:

We apply the leading principal determinants test:

First leading minor: $a = 3 > 0$ ✓
Second leading minor: $\det(A) = ac - b^2 = (3)(5) - (2)^2 = 15 - 4 = 11 > 0$ ✓

Both leading principal determinants are positive, so $A$ is positive definite.

We could also verify with eigenvalues: trace $= 8$ , $\det = 11$ , so $\lambda^2 - 8\lambda + 11 = 0$ giving $\lambda = 4 \pm \sqrt{5}$ . Both positive.

Or the quadratic form: $x^T A x = 3 x_1^2 + 4 x_1 x_2 + 5 x_2^2 = 3(x_1 + \tfrac{2}{3} x_2)^2 + \tfrac{11}{3} x_2^2 > 0$ for $x \neq 0$ .

Question 2 · The Determinant Trap

A student claims: "If $\det(A) > 0$ , then $A$ must be positive definite."

Is this claim True or False?

✅ Correct! The matrix $\text{diag}(-1, -1)$ has determinant $+1$ but is negative definite. You need every leading principal minor positive.

❌ Not quite. Consider $A = \text{diag}(-1, -1)$ : its determinant is $+1$ , yet both eigenvalues are negative. The product of two negative numbers is positive!

Show solution

Solution:

False. A positive determinant alone does NOT prove positive definiteness.

Counterexample: $A = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$

This matrix has $\det(A) = (-1)(-1) - 0 = 1 > 0$ , yet both eigenvalues are $-1$ . So $A$ is negative definite — the exact opposite of positive definite.

For a $2 \times 2$ matrix, you need every leading principal determinant to be positive:

$a > 0$ (the top-left $1\times 1$ block)
$ac - b^2 > 0$ (the full $2\times 2$ determinant)

Both conditions are required, not just the last one. The reason: with two negative eigenvalues, their product (the determinant) is still positive, but the matrix produces negative energy.

Question 3 · Classify the Matrix

Consider $A = \begin{pmatrix} 2 & 6 \\ 6 & 7 \end{pmatrix}.$

Evaluating $x^T A x$ at $x = (1, -1)$ gives $2 - 12 + 7 = -3 < 0$ .

What does this single computation tell us about $A$ ?

✅ Correct! A single negative value of $x^T A x$ is enough to disqualify positive definiteness. (In fact, this matrix is indefinite — one eigenvalue is positive, one negative.)

❌ Not quite. Positive definiteness requires $x^T A x > 0$ for every nonzero $x$ . Finding one $x$ where it's negative is enough to refute it.

Show solution

Solution:

A single negative value of $x^T A x$ instantly rules out positive definiteness. The definition requires $x^T A x > 0$ for every nonzero $x$ . One counterexample is enough to fail the test.

Since we found $x^T A x = -3 < 0$ , the matrix $A$ is not positive definite. It could be indefinite or negative definite (here it turns out to be indefinite, since the trace $= 9 > 0$ means at least one eigenvalue is positive).

Cross-check with the other tests:

$\det(A) = (2)(7) - 36 = -22 < 0$ — negative, fails the leading dets test
Pivots: $2$ , then $-22/2 = -11$ — second pivot negative, fails the pivot test
Eigenvalues: $\lambda^2 - 9\lambda - 22 = 0 \Rightarrow \lambda \approx 11.0$ and $\lambda \approx -2.0$ — one negative

All four tests agree: $A$ is indefinite.

Question 4 · Geometry of the Quadratic Form

For a symmetric matrix $A$ , the graph of $f(x) = x^T A x$ over $\mathbb{R}^2$ is a surface. The level set $\{x : x^T A x = 1\}$ is an ellipse in exactly which case?

✅ Correct! Positive definite ⟺ bowl-shaped graph ⟺ elliptical level sets. The eigenvalues determine the axis lengths, and the eigenvectors determine the axis directions.

❌ Not quite. A saddle gives a hyperbola, a trough gives parallel lines, and an inverted bowl gives no real level set at $+1$ . Only a bowl-shape (positive definite) yields a closed ellipse.

Show solution

Solution:

The level set $\{x : x^T A x = 1\}$ is an ellipse precisely when $A$ is positive definite.

Here's the geometric correspondence:

Type of $A$	Surface shape	Level set $x^T A x = 1$
Positive definite	Bowl (strict min at 0)	Ellipse (closed curve)
Positive semi-definite	Trough (flat valley)	Parallel lines
Indefinite	Saddle	Hyperbola
Negative definite	Inverted bowl	Empty (no real solutions)

When $A$ is positive definite, the eigenvalues $\lambda_1, \lambda_2 > 0$ are the reciprocals of the squared semi-axis lengths of the ellipse. The eigenvectors point along the principal axes. The ellipse becomes a circle only when $\lambda_1 = \lambda_2$ .

This geometric picture is why positive definite matrices appear everywhere: in optimization they guarantee a strict minimum, in statistics they make covariance matrices well-defined, and in physics they yield stable equilibria.

Solved: 0 / 4