LINEAR-ALGEBRA · Interactive Practice | Unit 22 · Video 1

From Calculus to Linear Algebra: How Eigenvalues Solve Differential Equations

IKey Formulas

Formula	Name	Description
$\dfrac{du}{dt} = Au$	Linear ODE system	Coupled first-order linear differential equations
$u(t) = e^{\lambda t}\, x$	Exponential ansatz	Pure-mode solution from eigenpair $(\lambda, x)$
$A x = \lambda x$	Eigenvalue equation	Plugging the ansatz into the ODE produces this
$u(t) = c_1 e^{\lambda_1 t} x_1 + c_2 e^{\lambda_2 t} x_2$	General solution	Superposition of eigenmodes; $c_i$ from $u(0)$

Eigenvalue signs: $\lambda > 0$ growth, $\lambda = 0$ steady state, $\lambda < 0$ decay.

IIVisualization 1 — The Two-Tank System

Water sloshes between two tanks until the split settles: the decaying mode fades while the $\lambda = 0$ steady state holds.

Time t 0.00

💡 The total $u_1 + u_2$ never changes — that conserved direction is the $\lambda = 0$ eigenvector, and the gap to steady state closes at rate $e^{-3t}$ , the $\lambda = -3$ mode.

IIIVisualization 2 — Eigenvalue Signs Determine Behavior

The sign of $\lambda$ alone fixes a single mode's fate: grow, hold, or decay.

Eigenvalue λ −1.00

💡 A coupled $\tfrac{du}{dt} = Au$ superposes one such mode per eigenvalue: a single $\lambda > 0$ makes the whole system grow, while all $\lambda < 0$ forces every mode to zero.

IVVisualization 3 — Phase Portrait of the Two-Tank System

Every starting point flows along $(1,-1)$ onto the same steady line $\operatorname{span}(2,1)$ .

u₁(0) 1.0 u₂(0) 0.0

💡 The field vanishes exactly on $\operatorname{span}(2,1)$ — that entire line stands still, the $\lambda = 0$ eigenspace of fixed points.

VQuiz Questions

Question 1

We try the ansatz $u(t) = e^{\lambda t}\, x$ in $\dfrac{du}{dt} = A u$ . After differentiating and cancelling $e^{\lambda t}$ , what equation must $\lambda$ and $x$ satisfy?

✅ Correct! The differential equation collapses into the eigenvalue equation.

❌ Not quite. Differentiating $e^{\lambda t} x$ pulls out a factor of $\lambda$ , while $A$ acts on $x$ on the right side.

Show solution

Solution:

Differentiate the ansatz: $\frac{d}{dt}\bigl(e^{\lambda t} x\bigr) = \lambda\, e^{\lambda t}\, x.$

The right-hand side of the ODE is $A\bigl(e^{\lambda t} x\bigr) = e^{\lambda t}\, A x.$

Setting both sides equal and cancelling the scalar $e^{\lambda t}$ : $\lambda x = A x \quad\Longleftrightarrow\quad A x = \lambda x.$

This is the eigenvalue equation. So $\lambda$ must be an eigenvalue of $A$ and $x$ its eigenvector.

Question 2

For the matrix $A = \begin{bmatrix} -1 & 2 \\ 1 & -2 \end{bmatrix}$ , the trace is $-3$ and the determinant is $0$ . Without solving the characteristic polynomial directly, what are the eigenvalues?

✅ Correct! A singular matrix forces one eigenvalue to be 0; the trace then pins down the other.

❌ Not quite. Use both facts together: the eigenvalues sum to the trace and multiply to the determinant. Since $\det = 0$ , one eigenvalue must be 0.

Show solution

Solution:

For a $2 \times 2$ matrix, the eigenvalues $\lambda_1, \lambda_2$ satisfy:

$\lambda_1 + \lambda_2 = \text{trace}(A)$
$\lambda_1 \cdot \lambda_2 = \det(A)$

Since $\det(A) = 0$ , the product $\lambda_1 \lambda_2 = 0$ , so at least one eigenvalue is zero. Take $\lambda_1 = 0$ .

Then $\lambda_2 = \text{trace}(A) - \lambda_1 = -3 - 0 = -3$ .

So the eigenvalues are $\boxed{\lambda_1 = 0,\; \lambda_2 = -3}$ .

We can confirm with the characteristic polynomial: $\lambda^2 - (\text{trace})\lambda + \det = \lambda^2 + 3\lambda = \lambda(\lambda + 3) = 0$ .

Question 3

True or False: If every eigenvalue of $A$ has a negative real part, then every solution of $\dfrac{du}{dt} = A u$ decays to the zero vector as $t \to \infty$ .

✅ Correct! Each eigenmode decays independently, so their superposition does too.

❌ Not quite. Recall that $u(t)$ is a sum of $e^{\lambda_i t} x_i$ terms. If every $\lambda_i < 0$ , every term — and therefore the whole sum — tends to zero.

Show solution

Solution: True.

Every solution can be written as a superposition of eigenmodes: $u(t) = c_1 e^{\lambda_1 t} x_1 + c_2 e^{\lambda_2 t} x_2 + \cdots$

If every $\lambda_i$ has negative real part, then every term $e^{\lambda_i t} \to 0$ as $t \to \infty$ . The whole sum decays to the zero vector regardless of the initial condition — i.e., regardless of the constants $c_i$ .

Contrast this with our two-tank system: it had $\lambda_1 = 0$ , so one mode persisted forever (the steady state). Without that zero eigenvalue, the solution would have decayed all the way to the origin.

Question 4

For the two-tank system with $A = \begin{bmatrix} -1 & 2 \\ 1 & -2 \end{bmatrix}$ and initial condition $u(0) = \begin{bmatrix}1\\0\end{bmatrix}$ , the solution is

$u(t) = \tfrac{1}{3}\begin{bmatrix}2\\1\end{bmatrix} + \tfrac{1}{3} e^{-3t}\begin{bmatrix}1\\-1\end{bmatrix}.$

What is the steady state $\displaystyle\lim_{t \to \infty} u(t)$ ?

✅ Correct! The decaying mode vanishes and only the $\lambda=0$ component survives.

❌ Not quite. $(1,0)$ is the initial condition, not the long-time limit. Take $t \to \infty$ , so $e^{-3t} \to 0$ .

❌ Not quite. Because one eigenvalue is zero (not negative), one mode does not decay. The steady state is nonzero.

❌ Not quite. Set $e^{-3t} = 0$ and read off what's left: $\tfrac{1}{3}(2,1)$ .

Show solution

Solution:

As $t \to \infty$ , the term $e^{-3t} \to 0$ , so the decaying mode vanishes: $\lim_{t\to\infty} u(t) = \tfrac{1}{3}\begin{bmatrix}2\\1\end{bmatrix} + 0 = \begin{bmatrix}2/3\\1/3\end{bmatrix}.$

So the steady state is $(2/3,\, 1/3)$ .

Notice that this point lies on the eigendirection of $\lambda = 0$ (which is $(2,1)$ scaled), and that $2/3 + 1/3 = 1$ matches the conserved total $u_1 + u_2 = 1$ from the initial condition.

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