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Matrix exponentials and linear systems of ODEs: x(t) = e^Atx(0), diagonalize as Ve^ΛtV⁻¹, and read stability off the eigenvalues

One linear equation x′ = ax has the solution x(t) = e^atx(0). A system x′ = Ax, with a vector x and a square matrix A, has the same solution, x(t) = e^Atx(0), once the exponential of a matrix is defined by the same power series. Computing e^At by hand is an eigenvalue problem, and the eigenvalues also say at a glance whether every solution dies out. The usual mistakes come from treating a matrix like a table of separate numbers.

Why it works

Differentiate the series term by term: the derivative of I + At + A²t²/2 + … is A + A²t + …, which is A times the original series. So e^At satisfies the equation, and at t = 0 it is I, which makes e^Atx(0) start at x(0). The trouble is only computing it.

Along an eigenvector the matrix acts like a number: Av = λv, so x(t) = e^λtv is a solution that stays on the line through v and grows or shrinks at rate λ. When there is a full set of eigenvectors, split x(0) into them, let each piece evolve on its own, and add the pieces back up; written with matrices, that is Ve^ΛtV⁻¹, where V⁻¹ does the splitting and V the adding. The part of each piece that matters for large t is |e^λt| = e^{(Re λ)t}, which is why stability depends only on the real parts.

Example 1: solving x′ = Ax from the eigenvectors

Solve x′ = Ax with A = −211−2 and x(0) = (3, 1).

1. λ² + 4λ + 3 = 0The trace is −4 and the determinant 4 − 1 = 3, so det(A − λI) = λ² + 4λ + 3.
2. λ = −1, λ = −3(λ + 1)(λ + 3) = 0.
3. v₁ = (1, 1), v₂ = (1, −1)A + I has rows (−1, 1), so x = y; A + 3I has rows (1, 1), so y = −x.
4. c₁ = 2, c₂ = 1Solve c₁(1, 1) + c₂(1, −1) = (3, 1): the sum and difference give 2c₁ = 4 and 2c₂ = 2.
5. x = 2e^−tv₁ + e^−3tv₂That is x(t) = (2e^−t + e^−3t, 2e^−t − e^−3t), which is (3, 1) at t = 0. Both eigenvalues are negative, so every solution decays; the e^−3t part dies first, and solutions end up moving in along (1, 1).

Example 2: the whole matrix e^At

For the matrix A of Example 1, compute e^At = Ve^ΛtV⁻¹.

1. V = 111−1The eigenvectors as columns, in the order of the eigenvalues −1, −3.
2. V⁻¹ = ½ 111−1The 2 × 2 inverse: swap the diagonal, negate the corners, divide by det V = −2.
3. e^Λt = e^−t00e^−3tThe exponential of a diagonal matrix is taken entry by entry: this is the one case where that is right.
4. e^At = ½ a + ba − ba − ba + bWith a = e^−t and b = e^−3t. Checks: at t = 0 it is I, and times (3, 1) it gives (2a + b, 2a − b), the answer of Example 1.

Example 3: a repeated eigenvalue with one eigenvector

Solve x′ = Ax with A = 2102 and x(0) = (0, 1).

1. λ = 2, twiceA is triangular, so its eigenvalues are the diagonal entries.
2. A − 2I = 0100 = NIts null space is only the line through (1, 0): one eigenvector, so A cannot be diagonalized.
3. N² = 0So the series for e^Nt stops after two terms: I + Nt.
4. e^At = e^2t(I + Nt)2I commutes with N, and only then may the exponential of the sum be split into a product.
5. x(t) = e^2t(t, 1)I + Nt has rows (1, t) and (0, 1), applied to (0, 1). Check: x′ = e^2t(2t + 1, 2), and Ax = e^2t(2t + 1, 2).

Example 4: complex eigenvalues and a decaying spiral

Compute e^At for A = −14−1−1, and describe the solutions.

1. λ² + 2λ + 5 = 0Trace −2, determinant 1 + 4 = 5.
2. λ = −1 ± 2iThe quadratic formula: (−2 ± √(4 − 20))/2.
3. A = −I + M, M² = −4IM has rows (0, 4) and (−1, 0); squaring it gives −4 on the diagonal and 0 off it.
4. e^Mt = cos 2t I + (sin 2t/2) MWith M² = −4I, the even terms of the series add up to cos 2t and the odd terms to sin 2t over 2, as in e^iθ = cos θ + i sin θ.
5. e^At = e^−tcos 2t2 sin 2t−½ sin 2tcos 2t−I commutes with M, so e^At = e^−te^Mt. The real part −1 makes every solution spiral in to the origin, turning at angular speed 2. Check: det e^At = e^−2t = e^{t · tr A}.

Common mistakes, and the exact line they happen on

1. Exponentiating entry by entry

The matrix of Example 2.

1. At = −2ttt−2tRight so far.
2. e^At = e^−2te^te^te^−2tThis is the mistake. At t = 0 this is the matrix of all ones, not I, so it does not even start at x(0), and its e^t entries grow while every true solution decays. The matrix exponential mixes the entries through the powers A², A³, …; only a diagonal matrix exponentiates entry by entry.

2. Treating a repeated eigenvalue as diagonal

The matrix of Example 3.

1. λ = 2, twiceRight so far.
2. x(t) = e^2t(0, 1)This is the mistake. This treats A as 2I. Then x′ = e^2t(0, 2), but Ax = e^2t(1, 2): the equation fails. With only one eigenvector the solution needs the te^2t term from I + Nt.

3. Reading stability off the diagonal

Is the origin stable for A = −141−1?

1. diagonal −1, −1Right so far as a fact about A, and the trace is −2.
2. so every solution decaysThis is the mistake. The diagonal entries are the eigenvalues only for a triangular matrix. Here det A = 1 − 4 = −3 < 0, a saddle: the eigenvalues are 1 and −3, and the solutions along the eigenvector for 1 grow like e^t. A negative trace is needed for stability, not enough on its own.

Practice

1. Find e^At for A = 100−2.

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   The diagonal matrix with entries e^t and e^−2t.

2. Solve x′ = 0100x with x(0) = (1, 2).

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   The matrix is N with N² = 0, so x(t) = (I + Nt)x(0) = (1 + 2t, 2).

3. Classify the origin for A = 1331.

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   A saddle: det A = −8 < 0, and the eigenvalues are 4 and −2.

4. For the matrix of Example 4, what is det e^At, and is e^At ever singular?

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   e^{t · tr A} = e^−2t, never zero; the inverse of e^At is e^−At.

5. For which k is every solution of x′ = −1k1−1x decaying to 0?

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   k < 1. The trace is −2, so it is stable exactly when det = 1 − k > 0. The eigenvalues are −1 ± √k.