Interactive Linear Algebra
Dan Margalit & Joseph Rabinoff · Georgia Tech · Open textbook
Read §§5.1 and 5.4 for eigenvectors and diagonalization.
Open the sourceA matrix can reveal directions, not just mix coordinates.
Prerequisites: 08 · A matrix is an action, 10 · Inner products and projection
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]
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.
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.
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.
| Q for diag(1,−1) | Vector | Value |
|---|---|---|
| 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).
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?
For A=[[2,4],[4,2]], what is the eigenvalue of (1,−1)?
For D=diag(2,−2), why can Q(2,2)=0?
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.
Dan Margalit & Joseph Rabinoff · Georgia Tech · Open textbook
Read §§5.1 and 5.4 for eigenvectors and diagonalization.
Open the sourceDeisenroth, 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 sourceGilbert 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 source3Blue1Brown · Video · 17 min
Watch the 17-minute creator explanation of eigenvectors and eigenvalues.
Open the source