Eigenvalues and Eigenvectors
Source lecture(s): PHY621 Ch. 2
Intuition
A linear map generally rotates and stretches vectors. Eigenvectors are the directions it does not rotate — it only scales them, by the eigenvalue:
These are the natural axes of the operator. In the eigenbasis the map is diagonal, and every hard question about it becomes easy. Almost all of applied linear algebra is: find the eigenbasis, do the trivial thing there, come back.
Finding them
\(A\mathbf{v} = \lambda\mathbf{v}\) means \((A - \lambda I)\mathbf{v} = 0\) with \(\mathbf{v}\neq0\), so \(A - \lambda I\) must have a non-trivial null space, so it must be singular:
the characteristic equation — a degree-\(n\) polynomial with \(n\) roots counted with multiplicity. For each root, solve \((A-\lambda I)\mathbf{v} = 0\) for the eigenvector.
Two invariants worth knowing because they are quick checks on any hand calculation:
Both are basis-independent, as they must be — they are properties of the operator, not of the matrix you wrote it in.
Algebraic vs geometric multiplicity
A subtlety that decides whether the operator is diagonalisable at all:
- Algebraic multiplicity: how many times \(\lambda\) repeats as a root of the characteristic polynomial.
- Geometric multiplicity: the dimension of its eigenspace, \(\dim\ker(A - \lambda I)\).
Geometric \(\le\) algebraic always. When they are equal for every eigenvalue the matrix is diagonalisable; when geometric is strictly smaller the matrix is defective and the best you can do is Jordan form. The standard example:
has \(\lambda = 1\) with algebraic multiplicity 2 but only one eigenvector \((1,0)^{\mathsf T}\). It cannot be diagonalised.
Hermitian and normal matrices are never defective — which is exactly why physics leans on them so heavily.
Why physicists care
| Setting | Eigenvalue | Eigenvector |
|---|---|---|
| Normal modes | \(\omega^2\) | mode shape |
| Quantum mechanics | measured value | stationary state |
| Rigid-body rotation | principal moment | principal axis |
| Stability analysis | growth rate | unstable mode |
| Data analysis (PCA) | variance | principal component |
| Markov chains | 1 (largest) | steady state |
The pattern is always the same: the eigenvalue is what happens, the eigenvector is to what.
A stability analysis is the cleanest illustration — linearise about equilibrium, find the eigenvalues of the Jacobian, and the sign of the largest real part decides everything. Positive means an instability, and its eigenvector tells you the shape it will take. That is the entire logic of hydrodynamic and MHD stability.
Worked example
Find the eigenvalues and eigenvectors of \(A = \begin{pmatrix}2&1\\1&2\end{pmatrix}\).
\(\det(A - \lambda I) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = 0\), so \(\lambda = 1, 3\).
Check: \(\operatorname{tr}A = 4 = 1+3\) ✓, \(\det A = 3 = 1\times3\) ✓.
For \(\lambda = 3\): \((A-3I)\mathbf{v} = \begin{pmatrix}-1&1\\1&-1\end{pmatrix}\mathbf{v} = 0\) gives \(\mathbf{v} = (1,1)^{\mathsf T}/\sqrt2\). For \(\lambda = 1\): \(\mathbf{v} = (1,-1)^{\mathsf T}/\sqrt2\).
They are orthogonal, as guaranteed for a real symmetric matrix. Physically this is two equal masses coupled by a spring: the symmetric mode (both move together) and the antisymmetric mode (they move oppositely, and it costs more energy, hence the larger eigenvalue).
Common mistakes
- Forgetting eigenvectors are directions. \(\mathbf{v}\) and \(5\mathbf{v}\) are the same eigenvector. Normalise by convention, not necessity.
- Assuming \(n\) distinct eigenvalues. Repeated roots are common, especially when there is a symmetry, and then you must check the geometric multiplicity.
- Assuming diagonalisability. Defective matrices exist. (Not among the normal ones.)
- Expecting real eigenvalues from a real matrix. A rotation matrix is real with complex eigenvalues \(e^{\pm i\theta}\) — it has no real invariant direction in the plane, which is geometrically obvious once stated.
Related concepts
- Characteristic equation (eq.)
- Diagonalization · Hermitian matrices
- Normal modes — the physical payoff
- Rank–nullity — a zero eigenvalue means a null vector
- Normal modes lab — eigenvalues you can watch
Knowledge graph position
Prerequisites: matrices & determinants, vector spaces. Leads to: diagonalization, normal modes, stability analysis, quantum mechanics, PCA.
Quiz
Q1 (computational). A \(3\times3\) matrix has \(\operatorname{tr}A = 6\) and \(\det A = 6\), with two eigenvalues 1 and 2. What is the third?
Answer
From the trace, \(\lambda_3 = 6 - 1 - 2 = 3\). Check with the determinant: \(1\times2\times3 = 6\) ✓.
Q2 (conceptual). Why can \(\begin{pmatrix}1&1\\0&1\end{pmatrix}\) not be diagonalised?
Answer
Its only eigenvalue \(\lambda = 1\) has algebraic multiplicity 2 but geometric multiplicity 1 — solving \((A-I)\mathbf{v} = 0\) gives only \((1,0)^{\mathsf T}\). With just one eigenvector there is no eigenbasis, so no similarity transformation can diagonalise it. It is defective.
Q3 (MCQ). A real rotation matrix in the plane has:
- (a) two real eigenvalues
- (b) complex eigenvalues \(e^{\pm i\theta}\), since no real direction is left invariant
- (c) a zero eigenvalue
- (d) no eigenvalues at all
Answer
(b). A rotation by \(\theta \neq 0,\pi\) moves every real direction, so there is no real eigenvector. Over \(\mathbb{C}\) the eigenvalues are \(e^{\pm i\theta}\) — unit modulus, as required for a length-preserving (unitary) map.