All lessons

Eigenvectors and the sign of a geometry

A matrix can reveal directions, not just mix coordinates.

Prerequisites: 08 · A matrix is an action, 10 · Inner products and projection

Understand

Objectives

  • Verify two eigenpairs by direct multiplication.
  • Compute a symmetric quadratic form and report its three-entry signature.
  • Distinguish zero vector, a nonzero null vector of Q, and a zero eigenvalue.

Start with something you can see

Apply a matrix to several arrows. Most turn; an eigenvector stays on its own line, possibly reversing or collapsing to zero. Predict the result for e₁=(1,0) and e₂=(0,1) under A=diag(2,−1): one doubles and the other reverses. The MIT eigenvalue passage includes negative and zero multipliers at 1:40. [1]

Give the idea a precise name

For a matrix A:ℝ²→ℝ², a nonzero vector v is an eigenvector if Av=λv for a real number λ, its eigenvalue. Zero can be an eigenvalue even though the zero vector is never an eigenvector. Its quadratic form Q(v)=vᵀAv is a scalar, and its signature (positive, negative, zero) counts eigenvalues with each sign, including multiplicity.

Av=λv; Q(v)=vᵀAv

The transpose vᵀ turns the vector into a row, so vᵀAv is one real number. For an eigenvector, Q(v)=λ‖v‖², making its sign match λ. But Q(v)=0 for a nonzero vector need not make that vector an eigenvector or force λ=0: positive and negative contributions can cancel.

Every real symmetric matrix has an orthogonal basis of eigenvectors. Along each basis direction, the sign of Q matches the sign of that direction’s eigenvalue.

Work one small world

For the symmetric A=[[1,2],[2,1]], A(1,1)=3(1,1) and A(1,−1)=−(1,−1). Thus eigenvalues are 3 and −1 and the signature is (1,1,0). Its quadratic form is Q(x,y)=x²+4xy+y²: Q(1,1)=6 and Q(1,−1)=−2. The two eigenvectors are orthogonal.

The picture uses the simpler form Q(x,y)=x²−y² of D=diag(1,−1), which appears again in the counterexample below.

Indefinite quadratic form x squared minus y squared has positive regions near the horizontal axis, negative regions near the vertical axis, and zero on two diagonalsQ>0Q<0Q=0
Q for diag(1,−1)VectorValue
Positive direction(1,0)1
Negative direction(0,1)−1
Nonzero null vector(1,1)0

Counterexample to “Q(v)=0 means A has a zero eigenvalue”: D=diag(1,−1) has eigenvalues 1 and −1, neither zero, but its nonzero vector (1,1) has Q(1,1)=0. This null vector of the quadratic form is not the zero vector and is not in the kernel of D. By contrast diag(2,0) really has a zero eigenvalue with nonzero eigenvector (0,1).

THE BRIDGEThe later Chow-ring degree pairing has the same matrix [[1,2],[2,1]]; its one positive and one negative direction will become numerically inspectable.
WHERE THIS IDEA STOPSOrthogonal diagonalization holds for real symmetric matrices, not every real matrix. An indefinite quadratic form is not a squared Euclidean norm.

Experiment

Compute A(1,1), A(1,−1) and their Q-values. Then sketch the sign regions of x²−y²; use the table to verify the diagonals carry nonzero null vectors.

What changes when a matrix has one positive direction and all remaining directions negative?

Open the interactive experiment

Check & explain

  1. For A=[[2,4],[4,2]], what is the eigenvalue of (1,−1)?

  2. For D=diag(2,−2), why can Q(2,2)=0?

    1. D must have eigenvalue zero
    2. Positive and negative contributions cancel for a nonzero vector
    3. The zero vector is an eigenvector

Teach back

Find the eigenpairs and signature of [[1,2],[2,1]], then use diag(1,−1) to explain why a nonzero null vector of Q need not signal a zero eigenvalue.

  • Verify each matrix-vector multiplication.
  • Use the (positive,negative,zero) signature order.
  • Keep Q(v)=0 distinct from Av=0 and v=0.

Read deeper

Interactive Linear Algebra

Dan Margalit & Joseph Rabinoff · Georgia Tech · Open textbook

Read §§5.1 and 5.4 for eigenvectors and diagonalization.

Open the source

Mathematics for Machine Learning

Deisenroth, Faisal & Ong · Freely readable textbook

Read §§3.2.3, 4.2 and 4.4, printed pp.105 and 115 for the eigenvalue sections.

Open the source

21. Eigenvalues and Eigenvectors

Gilbert Strang · MIT OpenCourseWare · Lecture · 51 min

Start at 1:40. Watch the short core 1:40–4:40. The full 51-minute lecture is available from the same link.

Open the source

Eigenvectors and eigenvalues | Chapter 14, Essence of linear algebra

3Blue1Brown · Video · 17 min

Watch the 17-minute creator explanation of eigenvectors and eigenvalues.

Open the source

Open the interactive lesson