LINEAR-ALGEBRA · Interactive Practice | Unit 24 · Video 4

Three Tests, One Truth: Positive Definite Matrices

IKey Tests Reference

A symmetric matrix $A = A^T$ is positive definite if and only if any one of these equivalent tests holds:

Test	Condition	Cost
Eigenvalues	All $\lambda_i > 0$	Expensive (iterative for $n \geq 5$ )
Pivots	All pivots from elimination $> 0$	$O(n^3)$ — fast
Leading minors	$\det(A_k) > 0$ for $k = 1, 2, \ldots, n$	Fastest for small $n$

Pivot Sign Theorem: For symmetric $A$ , the number of positive pivots equals the number of positive eigenvalues. Same signs, different values.

Trap: $\det(A) > 0$ alone is not enough — it only tells you there are an even number of negative eigenvalues. Always check every leading minor.

IIVisualization 1 — The Geometry of Positive Definiteness

The level sets of $\mathbf{x}^{T} A \mathbf{x} = a x^{2} + 2 b x y + c y^{2}$ are ellipses when $A$ is definite, hyperbolas when it is indefinite.

a (top-left) 2 b (off-diagonal) 1 c (bottom-right) 2

💡 Trap: set $a = -1$ , $b = 1$ , $c = -2$ — the determinant is positive, yet the level sets are the red (negative) family, so $A$ is negative definite, not positive.

IIIVisualization 2 — Three Tests, One Region

Pivots, eigenvalues, and leading minors all turn positive on exactly the same interval of $t$ .

Matrix family parameter t 1.0

IVPractice Quiz

Question 1

Consider the symmetric matrix $A = \begin{pmatrix} 4 & 2 \\ 2 & 3 \end{pmatrix}.$

Using the leading principal minors test, is $A$ positive definite?

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

❌ Not quite. Recompute $d_2 = (4)(3) - (2)(2)$ and check both leading minors.

Show solution

Solution:

Compute the two leading minors:

$d_1 = 4 > 0$ ✓
$d_2 = \det(A) = (4)(3) - (2)(2) = 12 - 4 = 8 > 0$ ✓

Both leading minors are positive, so $A$ is positive definite.

(The third option — " $\det(A) > 0$ alone is sufficient" — is a tempting trap. For 2×2 it happens to give the right answer here only because we also checked $d_1 > 0$ . In general, $\det > 0$ alone is not enough; the trap matrix $\begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}$ has $\det = 1 > 0$ but is not positive definite.)

Question 2

True or False: If a symmetric matrix $A$ has $\det(A) > 0$ , then $A$ is positive definite.

✅ Correct! A positive determinant only guarantees an even count of negative eigenvalues — possibly zero, possibly two.

❌ Not quite. Consider the matrix from the video: $\begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}$ has $\det = 1 > 0$ but both eigenvalues are negative.

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Solution:

The determinant equals the product of the eigenvalues. So $\det(A) > 0$ only tells us:

Either all eigenvalues are positive (positive definite), or
An even number of eigenvalues are negative (e.g., two negative eigenvalues in a 2×2).

Counterexample: $B = \begin{pmatrix} -1 & 1 \\ 1 & -2 \end{pmatrix}, \quad \det(B) = (-1)(-2) - (1)(1) = 1 > 0.$

But $d_1 = -1 < 0$ , so $B$ fails the very first leading minor. Both eigenvalues are negative.

Lesson: Always check every leading minor — not just $\det(A)$ .

Question 3

Apply Gaussian elimination to $A = \begin{pmatrix} 5 & 2 \\ 2 & 3 \end{pmatrix}.$

What are the pivots, and what do they tell you?

✅ Correct! $p_1 = 5$ , $p_2 = 11/5$ , and their product equals $\det(A) = 11$ .

❌ Not quite. After eliminating below the first pivot, the (2,2) entry becomes $3 - \frac{2}{5}(2) = \frac{11}{5}$ . Pivots are not eigenvalues.

Show solution

Solution:

Eliminate below the first pivot. The first pivot is $p_1 = 5$ . Subtract $\frac{2}{5}$ times row 1 from row 2:

$\begin{pmatrix} 5 & 2 \\ 2 & 3 \end{pmatrix} \to \begin{pmatrix} 5 & 2 \\ 0 & 3 - \tfrac{2}{5}(2) \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 0 & \tfrac{11}{5} \end{pmatrix}.$

So $p_2 = \frac{11}{5}$ . Both pivots are positive, so $A$ is positive definite.

Sanity check: The product of pivots equals the determinant: $p_1 \cdot p_2 = 5 \cdot \tfrac{11}{5} = 11 = \det(A).$

The eigenvalues of $A$ are $4 \pm \sqrt{5} \approx 6.24$ and $1.76$ — not the pivots. But by the pivot sign theorem, both pivots and both eigenvalues share the same signs.

Question 4

For a $100 \times 100$ symmetric matrix, which test for positive definiteness is generally the most efficient in practice?

✅ Correct! A single $O(n^3)$ elimination pass gives all 100 pivots — no iteration, no characteristic polynomial.

❌ Not quite. Eigenvalue computation is iterative for large $n$ , and computing 100 separate determinants is wasteful. One elimination gives you all the pivots at once.

Show solution

Solution:

For large $n$ :

Eigenvalues require iterative methods (no closed form for $n \geq 5$ ). Convergence can be slow, and you only need signs, not exact values — so this is overkill.
Pivots come from a single Gaussian elimination, costing $O(n^3)$ . After elimination you simply read the diagonal. This is direct, non-iterative, and the standard choice for large $n$ .
All leading minors would require computing $n$ separate determinants — far more expensive than one elimination pass.
Just $\det(A) > 0$ is insufficient (as shown in Question 2).

Rule of thumb:

$n = 2, 3$ : leading sub-determinants are fastest.
$n$ large: one Gaussian elimination + check pivot signs.
Eigenvalues: only when you actually need their values.

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