LINEAR-ALGEBRA · Interactive Practice | Unit 24 · Video 3

The Eigenvalue Shortcut: Why Counting Pivots Beats Solving Polynomials

IKey Formulas

Formula	Name	Meaning
$A = LDL^\top$	Symmetric factorization	The diagonal of $D$ holds the pivots
$\#(\text{pos pivots}) = \#(\text{pos eigenvalues})$	Sign Theorem (Sylvester's Law)	Pivot signs match eigenvalue signs (symmetric $A$ )
$\det(A) = \prod d_i = \prod \lambda_i$	Product Identity	Pivots and eigenvalues share the same product
$\lambda_i(A - cI) = \lambda_i(A) - c$	Shifting Trick	Subtracting $cI$ shifts every eigenvalue by $-c$

IIVisualization 1 — Pivots vs. eigenvalues

Pivots and eigenvalues rarely match value-for-value — yet their sign split and their product always agree.

a₁₁ 2 a₂₂ 3 a₃₃ 1 a₁₂ 1 a₁₃ 0 a₂₃ 1

💡 Challenge: make the matrix indefinite (set $a_{11} = -3$ ) — a red pivot appears exactly when a red eigenvalue does.

IIIVisualization 2 — The shifting trick

How many eigenvalues sit above a threshold $c$ ? Count the positive pivots of $A - cI$ .

Shift c 2.00

💡 Probe at several shifts and subtract the counts to trap each eigenvalue in an interval — the bisection that solvers like LAPACK run on cheap elimination alone.

IVQuiz Questions

Question 1

Suppose you run elimination on a symmetric $4 \times 4$ matrix $A$ and find the pivots $d_1 = 3, \quad d_2 = -2, \quad d_3 = 1, \quad d_4 = -5.$

How many positive eigenvalues does $A$ have?

✅ Correct! The number of positive pivots equals the number of positive eigenvalues.

❌ Not quite. Count the pivots that are positive — those are $3$ and $1$ .

Show solution

Solution:

By the Sign Theorem (Sylvester's Law of Inertia), for a symmetric matrix: $\#(\text{positive pivots}) = \#(\text{positive eigenvalues}).$

The pivots are $\{3, -2, 1, -5\}$ . There are 2 positive pivots ( $3$ and $1$ ), so $A$ has 2 positive eigenvalues (and 2 negative ones).

Note: The actual eigenvalue values are not equal to the pivots — only the sign counts match.

Question 2

A student claims: "For any square matrix $A$ , the number of positive pivots equals the number of positive eigenvalues."

Is this claim true or false?

✅ Correct! Symmetry is essential — the Sign Theorem fails for general matrices.

❌ Not quite. The Sign Theorem (Sylvester's Law) requires the matrix to be symmetric.

Show solution

Solution:

The claim is false. The Sign Theorem requires $A$ to be symmetric.

For non-symmetric matrices:

Eigenvalues can be complex (so "positive" doesn't always make sense).
Even when eigenvalues are real, the sign counts of pivots and eigenvalues can disagree.

Symmetry guarantees that $A = LDL^\top$ (a congruence transformation), and Sylvester's Law of Inertia says congruence preserves the signature. Drop symmetry and the protection is gone.

Question 3

For a symmetric $3 \times 3$ matrix $A$ , you compute the pivots and find $d_1 = 2, \quad d_2 = \tfrac{5}{2}, \quad d_3 = \tfrac{3}{5}.$

Without computing eigenvalues, what is $\det(A)$ ?

✅ Correct! $\det(A) = 2 \cdot \tfrac{5}{2} \cdot \tfrac{3}{5} = 3$ .

❌ Not quite. The determinant equals the product of the pivots.

Show solution

Solution:

The determinant equals the product of the pivots (and equivalently, the product of the eigenvalues): $\det(A) = d_1 \cdot d_2 \cdot d_3 = 2 \cdot \tfrac{5}{2} \cdot \tfrac{3}{5} = 3.$

This is one of the two invariants that pivots and eigenvalues share — products always agree, even though individual values do not.

Question 4

You run elimination on a symmetric $5 \times 5$ matrix $A$ at three different shifts and count positive pivots:

Pivots of $A - 0I$ : 4 positive
Pivots of $A - 5I$ : 2 positive
Pivots of $A - 10I$ : 0 positive

How many eigenvalues of $A$ lie in the interval $(0, 5]$ ?

✅ Correct! Subtract: 4 eigenvalues above 0, minus 2 above 5, leaves 2 in $(0, 5]$ .

❌ Not quite. Use subtraction: (eigenvalues above 0) minus (eigenvalues above 5).

Show solution

Solution:

The number of positive pivots of $A - cI$ equals the number of eigenvalues of $A$ greater than $c$ .

Eigenvalues $> 0$ : 4
Eigenvalues $> 5$ : 2
Eigenvalues $> 10$ : 0

So the number of eigenvalues in the interval $(0, 5]$ is: $\#\{\lambda > 0\} - \#\{\lambda > 5\} = 4 - 2 = 2.$

This is the bisection idea: by combining counts at different shifts, we can locate every eigenvalue in an interval — using only elimination, no polynomial root-finding.

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