An eigenvector of A is a direction that the matrix does not turn: A·v is a multiple λ·v of v. The multipliers λ are the roots of λ² − τ·λ + Δ = 0, where τ is the trace and Δ the determinant. For each such pair, e^(λt)·v is a solution that travels along a straight line through the origin.
Core ideas
Three ideas. The first names what to look for, the second says how to find the multipliers, and the third says why the effort pays: each pair gives a solution you can write down.
A non-zero vector v is an eigenvector of A when A·v = λ·v for some number λ, called its eigenvalue. Multiplying by A may stretch, shrink or flip v, but it leaves it on the same line through the origin. Every non-zero multiple of v is an eigenvector for the same λ, so (1, 1) and (2, 2) are one answer written twice.
A·v = λ·v says (A − λ·I)·v = 0, and a non-zero v can satisfy that only when the determinant of A − λ·I is zero. Expanding it gives λ² − τ·λ + Δ = 0 with τ = a + d and Δ = a·d − b·c. Two exact numbers decide the multipliers, and the quadratic formula finishes the job.
If A·v = λ·v then x(t) = e^(λt)·v solves x′ = A·x. The point never leaves the line through v: it slides along it, away from the origin if λ is positive and toward it if λ is negative.
The material
The definition, the quadratic, a full example in fractions, the solution it produces, and what happens when the roots are not rational.
For a matrix A, a number λ is an eigenvalue and a non-zero vector v an eigenvector when A·v = λ·v.
Every non-zero multiple of an eigenvector is an eigenvector with the same eigenvalue, so what is determined is a direction, not a point. The lab prints the smallest whole-number vector with its first non-zero entry positive, so a diagonal matrix gets (0, 1) and (1, 0).
The eigenvalues of A, with rows a b and c d, are the roots of λ² − τ·λ + Δ = 0, where τ = a + d is the trace and Δ = a·d − b·c the determinant.
A·v = λ·v is (A − λ·I)·v = 0, which has a non-zero solution v exactly when A − λ·I has determinant zero, the fact about two-by-two systems from Algebra's Systems and Matrices.
That determinant is (a − λ)·(d − λ) − b·c. Multiplying out gives λ² − (a + d)·λ + (a·d − b·c), which is λ² − τ·λ + Δ. The two roots add to τ and multiply to Δ.
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Take A with rows 2 1 and 1 2. Then τ = 4 and Δ = 3, so λ² − 4λ + 3 = 0, which factors as (λ − 1)·(λ − 3). The eigenvalues are 1 and 3.
For λ = 1, (A − I)·v = 0 is the pair v₁ + v₂ = 0 and v₁ + v₂ = 0, one equation twice, so v = (1, −1). For λ = 3 it is −v₁ + v₂ = 0, so v = (1, 1).
The two equations are always the same equation, up to a multiple, when λ is an eigenvalue. That is the quadratic's doing: the determinant being zero means the two rows of A − λ·I are parallel. So solving for v takes one equation, and any non-zero solution of it will do.
If A·v = λ·v, then x(t) = e^(λt)·v solves x′ = A·x.
Differentiate: x′ = λ·e^(λt)·v, because v is constant.
The other side is A·x = e^(λt)·A·v = e^(λt)·λ·v. They are equal at every t. Each coordinate of x is a multiple of the same exponential, so the point stays on the line through v.
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For the symmetric matrix the straight-line solutions are e^(t)·(1, −1) and e^(3t)·(1, 1). Check the second directly: both coordinates are e^(3t), so x′ = 3e^(3t), and the first equation says x′ = 2x + y = 3e^(3t). They agree. Both solutions move away from the origin, the second faster, because both eigenvalues are positive.
The quadratic does not always factor. For the matrix with rows 1 1 and 1 0, τ = 1 and Δ = −1, so λ² − λ − 1 = 0 and λ = (1 ± √5)/2. The eigenvalues are exact surds, and the lab prints them as such. Their eigenvectors exist, but have irrational entries, and the lab prints a dash for the vectors rather than a rounded guess. That is a limit of the lab, not of the mathematics: the straight-line solutions are there to be found.
A student finds (2, 2) for the first matrix and the lab prints (1, 1). Test (2, 2): A·(2, 2) = (6, 6) = 3·(2, 2). It is an eigenvector with eigenvalue 3.
The two answers describe one line, the diagonal. The lab reports the primitive vector because a direction has no preferred length.
Interactive
Trace, determinant, eigenvalues and exact Euler steps
Method
Trace and determinant first, then the roots, then one equation per root.
τ = a + d and Δ = a·d − b·c. For rows 1 2 and 3 0 they are τ = 1 and Δ = 0 − 6 = −6.
Here λ² − λ − 6 = (λ − 3)·(λ + 2), so λ = 3 and λ = −2. If it does not factor, use the formula and keep the surd.
For λ = 3 the first row is −2·v₁ + 2·v₂ = 0, so v₁ = v₂ and v = (1, 1). For λ = −2 it is 3·v₁ + 2·v₂ = 0, so v = (2, −3). One row is enough.
Test A·v = λ·v on the vector you found: A·(2, −3) = (1·2 + 2·(−3), 3·2 + 0) = (−4, 6) = −2·(2, −3). It is an eigenvector for −2.
e^(3t)·(1, 1) and e^(−2t)·(2, −3). The first runs out along the diagonal and the second runs in along a steeper line. Any non-zero multiple is as good.
The first preset. Find the eigenvalues, the directions and the solution along the diagonal.
The lab prints the eigenvalues 1, 3 and the vectors (1, −1), (1, 1), which is the same list written in the same order. Only the direction of a vector matters. Any other multiple, such as (−1, 1) or (5, −5), is the same eigenvector for the purpose of the solution.
The two eigenlines cut the plane into four regions, and a start in any region stays in it, because a solution cannot cross an eigenline: on the line it never leaves, so off it, it cannot arrive. The next lesson adds the two solutions together and gets all the rest.
You should be able to compute the trace and determinant, solve the quadratic exactly, find one eigenvector per eigenvalue from a single equation, check it by multiplying, and write the straight-line solution e^(λt)·v.
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