Eigenvalues & eigenvectors
The idea
Most vectors get knocked off their line when a matrix acts on them — but a few special directions survive, merely stretched or compressed. An eigenvector v of A satisfies Av = λv for a scalar λ, the eigenvalue: the matrix acts on that direction like simple multiplication. These invariant directions are the skeleton of the transformation, and they power everything from stability analysis of differential equations to vibration modes to ranking algorithms.
The computation runs in two acts. Eigenvalues first: Av = λv rearranges to (A − λI)v = 0, which has a nonzero solution exactly when det(A − λI) = 0 — the characteristic equation, a polynomial in λ. Then, for each root λ, solve (A − λI)v = 0 for the eigenvectors; the matrix A − λI is built to be singular, so the system is guaranteed to have a free variable.
Two corrections worth making early. The zero vector is never an eigenvector — every λ would trivially work, carrying no information — but the number 0 can perfectly well be an eigenvalue, precisely when A is singular. And eigenvectors are only determined up to scale: any nonzero multiple of an eigenvector is an equally valid representative of the same invariant direction.
Worked example
Find the eigenvalues and corresponding eigenvectors of the 2 × 2 matrix A with rows (4, 1) and (2, 3).
- Set up the characteristic equation: det(A − λI) = (4 − λ)(3 − λ) − (1)(2) = λ² − 7λ + 12 − 2 = λ² − 7λ + 10 = 0.
- Factor: λ² − 7λ + 10 = (λ − 5)(λ − 2), so the eigenvalues are λ₁ = 5 and λ₂ = 2. Quick check: their sum 7 equals the trace 4 + 3 and their product 10 equals det(A) = 12 − 2 — both identities hold.
- Eigenvector for λ₁ = 5: A − 5I has rows (−1, 1) and (2, −2), and both rows demand y = x, so v₁ = (1, 1) (or any nonzero multiple).
- Eigenvector for λ₂ = 2: A − 2I has rows (2, 1) and (2, 1), demanding 2x + y = 0, so v₂ = (1, −2).
- Verify by direct action: A(1, 1) = (4 + 1, 2 + 3) = (5, 5) = 5(1, 1), and A(1, −2) = (4 − 2, 2 − 6) = (2, −4) = 2(1, −2) — both directions survive, stretched by exactly their eigenvalues.
Answer. The eigenvalues are 5 and 2, with eigenvectors (1, 1) and (1, −2) respectively (up to nonzero scaling).
Check your understanding
- Why does demanding a nonzero solution of (A − λI)v = 0 force the determinant of A − λI to vanish?
- What do the trace and determinant of a matrix tell you about its eigenvalues before you solve anything?
- How would you interpret a matrix having eigenvalue 0, both algebraically and geometrically?
- Why is any scalar multiple of an eigenvector still an eigenvector, and what does that say about what an eigenvector really represents?
Build the foundations first
Eigenvalues & eigenvectors builds on these concepts. If any feel shaky, start there.