LINEAR-ALGEBRA · Interactive Practice | Unit 22 · Video 3

Exponentiating a Matrix: Why $e^{At}$ Solves Every Linear System

IKey Formulas

Formula	Name	Description
$e^{At} = I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + \cdots$	Series definition	Always converges for any $A$ , $t$
$e^{At} = S\, e^{\Lambda t}\, S^{-1}$	Diagonalization formula	Requires $A$ to have full eigenbasis
$e^{\Lambda t} = \mathrm{diag}(e^{\lambda_1 t}, \ldots, e^{\lambda_n t})$	Diagonal exponential	Just scalar exponentials of eigenvalues
$u(t) = e^{At} u(0)$	Solution to $\dot u = Au$	Propagator of linear dynamics

IIVisualization 1 — Why the Factorials Save Us

The $n!$ denominators tame each power, so the Taylor sum converges where the geometric one explodes.

Eigenvalue a −1 Terms N 3

💡 Each diagonal entry of a matrix exponential is exactly this scalar series, so convergence of $e^{At}$ reduces to the picture above.

IIIVisualization 2 — Eigenvalues Control Stability

Drag the eigenvalue across the complex plane: trajectories decay in the left half, blow up in the right.

Re(λ) = α −0.5 Im(λ) = β 1.0

💡 On the imaginary axis ( $\alpha = 0$ ) the orbits close — the norm-preserving $e^{-iHt}$ evolution of a quantum system.

IVQuiz Questions

Question 1

Suppose $\Lambda = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix}$ is a diagonal matrix.

What is $e^{\Lambda t}$ ?

✅ Correct! Each eigenvalue $\lambda_i$ becomes $e^{\lambda_i t}$ on the diagonal.

❌ Not quite. Remember: $e^{\Lambda t}$ exponentiates each diagonal entry, treating it as the exponent of $e$ — it does not multiply by $t$ or scale linearly.

Show solution

Solution:

For a diagonal matrix $\Lambda$ , the series $e^{\Lambda t} = \sum_{n=0}^{\infty} \frac{(\Lambda t)^n}{n!}$ becomes diagonal because $\Lambda^n$ is diagonal:

$\Lambda^n = \begin{pmatrix} 3^n & 0 \\ 0 & (-2)^n \end{pmatrix}$

Each diagonal entry is just a scalar Taylor series:

$e^{\Lambda t} = \begin{pmatrix} \sum_n \frac{(3t)^n}{n!} & 0 \\ 0 & \sum_n \frac{(-2t)^n}{n!} \end{pmatrix} = \begin{pmatrix} e^{3t} & 0 \\ 0 & e^{-2t} \end{pmatrix}$

So $e^{\Lambda t} = \mathrm{diag}(e^{3t}, e^{-2t})$ .

Question 2

A matrix $A$ has eigenvalues $\lambda_1 = -1 + 2i$ and $\lambda_2 = -1 - 2i$ .

What is the long-term behavior of solutions $u(t) = e^{At} u(0)$ as $t \to \infty$ ?

✅ Correct! Re $(\lambda) < 0$ gives decay, and Im $(\lambda) \neq 0$ gives oscillation.

❌ Not quite. Recall: the real part of $\lambda$ controls growth/decay, and the imaginary part controls oscillation.

Show solution

Solution:

Stability is determined by the real part of the eigenvalues, while the imaginary part determines oscillation.

$\mathrm{Re}(\lambda) = -1 < 0$ → modes decay
$\mathrm{Im}(\lambda) = \pm 2 \neq 0$ → modes oscillate

Each mode contributes $e^{(-1 \pm 2i) t} = e^{-t}(\cos 2t \pm i \sin 2t)$ . The envelope $e^{-t}$ shrinks while the trigonometric factor oscillates.

The result is a stable spiral: trajectories spiral inward toward the origin.

Question 3

True or False: The series $I + At + (At)^2 + (At)^3 + \cdots$ (with no factorials) is equal to $e^{At}$ for all matrices $A$ .

✅ Correct! That series equals $(I - At)^{-1}$ , not $e^{At}$ . The factorials in the matrix exponential are what make it always converge.

❌ Not quite. Look closely — the matrix exponential has factorial denominators $\frac{1}{n!}$ on each term. Without those, the series is something else entirely.

Show solution

Solution:

This is False.

The series without factorials is the geometric series:

$I + At + (At)^2 + (At)^3 + \cdots = (I - At)^{-1}$

It only converges when every eigenvalue of $At$ has magnitude less than 1.

The matrix exponential has factorials in the denominator:

$e^{At} = I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + \cdots$

Those factorials are essential — they make the series converge for every matrix $A$ and every $t$ . Without them, terms blow up.

The two series do agree to first order ( $I + At + \cdots$ ), so $(I - At)^{-1}$ is a useful approximation to $e^{At}$ for small $t$ , but they are not the same object.

Question 4

Suppose $A = S \Lambda S^{-1}$ is diagonalizable. Why does the formula $e^{At} = S\, e^{\Lambda t}\, S^{-1}$ work?

✅ Correct! The telescoping $S^{-1}S = I$ in every power is the engine that makes diagonalization work.

❌ Not quite. The formula works because of a specific cancellation that happens when we compute powers $A^n = (S\Lambda S^{-1})^n$ .

Show solution

Solution:

The key telescoping argument: when we compute powers of $A$ ,

$A^2 = (S\Lambda S^{-1})(S\Lambda S^{-1}) = S\Lambda \underbrace{(S^{-1}S)}_{= I}\Lambda S^{-1} = S\Lambda^2 S^{-1}$

The middle $S^{-1}S$ collapses to the identity. The same cancellation telescopes through every power:

$A^n = S\Lambda^n S^{-1}$

Now plug into the series:

$e^{At} = \sum_{n=0}^{\infty} \frac{(At)^n}{n!} = \sum_{n=0}^{\infty} \frac{S \Lambda^n S^{-1} t^n}{n!} = S \left(\sum_{n=0}^{\infty} \frac{\Lambda^n t^n}{n!}\right) S^{-1} = S\, e^{\Lambda t}\, S^{-1}$

We factor $S$ on the left and $S^{-1}$ on the right out of every term, leaving the diagonal series in the middle.

The other options are wrong:

$S$ does not commute with $A$ in general.
$A$ and $e^{At}$ have related but different eigenvalues ( $\lambda_i$ vs $e^{\lambda_i t}$ ).
Not every matrix series equals its diagonalization — this only works because of the specific telescoping structure.

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Exponentiating a Matrix: Why eAte^{At}eAt Solves Every Linear System

IKey Formulas

IIVisualization 1 — Why the Factorials Save Us

IIIVisualization 2 — Eigenvalues Control Stability

IVQuiz Questions

Exponentiating a Matrix: Why $e^{At}$ Solves Every Linear System