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Linear Algebra in
Differential Equations

Discover how eigenvalues and eigenvectors unlock the solutions to systems of differential equations: the bridge between matrix algebra and dynamic behavior.

Linear Algebra

What Is a System of Differential Equations?

Instead of a single equation with one unknown function, we often need to model multiple quantities that change simultaneously and influence each other. A system of linear DEs looks like:

For example, with two unknowns \(x_1(t)\) and \(x_2(t)\), the system becomes two coupled equations whose behavior depends entirely on the matrix \(A\).

Why Do Eigenvalues Matter Here?

Decouple the System
Eigenvalues let us transform a coupled system into independent equations we can solve one at a time.
Predict Behavior
The sign of each eigenvalue tells us whether solutions grow, decay, or oscillate.
Build the Solution
Each eigenvector provides a direction in which the solution evolves exponentially.

Eigenvalue and Eigenvector Refresher

A scalar \(\lambda\) is an eigenvalue of matrix \(A\) if there exists a nonzero vector \(\mathbf{v}\) such that:

\(\mathbf{v}\)

\(A\mathbf{v} = \lambda\mathbf{v}\)

 Original After multiplication by A →

\(\times A\)

The vector \(\mathbf{v}\) only gets scaled (by \(\lambda\)), never rotated. This special property is what makes eigenvectors perfect for solving DEs.

The Characteristic Equation

To find eigenvalues, we solve:

1
Subtract \(\lambda\) from each diagonal entry of \(A\) to form \(A - \lambda I\).
2
Compute the determinant of \(A - \lambda I\): this yields a polynomial in \(\lambda\).
3
Solve for \(\lambda\): the roots are the eigenvalues.

For a \(2 \times 2\) matrix, this gives a quadratic in \(\lambda\), easy to solve with the quadratic formula.

Core Solution Theorem

Why? Differentiate: \(\mathbf{x}'(t) = \lambda e^{\lambda t}\mathbf{v}\). Compare with \(A\mathbf{x}(t) = e^{\lambda t}A\mathbf{v} = e^{\lambda t}\lambda\mathbf{v}\). They match!

Eigenvalue Sign and Phase Portraits

Stable Node

\(\lambda_1 < 0, \lambda_2 < 0\)

 Decays to origin Unstable Node

\(\lambda_1 > 0, \lambda_2 > 0\)

 Grows from origin Saddle Point

\(\lambda_1 > 0, \lambda_2 < 0\)

 Mixed behavior

Single DE vs. System of DEs

Single DE System of DEs
Form \(x' = ax\) \(\mathbf{x}' = A\mathbf{x}\)
Unknown One function \(x(t)\) Vector \(\mathbf{x}(t)\)
Growth rate Constant \(a\) Eigenvalues \(\lambda_1, \lambda_2, \ldots\)
Solution \(x(t) = ce^{at}\) \(\mathbf{x}(t) = \sum c_i e^{\lambda_i t}\mathbf{v}_i\)
Directions Not applicable Eigenvectors \(\mathbf{v}_i\)

How to Solve \(\mathbf{x}' = A\mathbf{x}\)

1
Write the matrix \(A\) from the system of equations.
2
Find eigenvalues: solve \(\det(A - \lambda I) = 0\) for \(\lambda\).
3
Find eigenvectors: for each \(\lambda_i\), solve \((A - \lambda_i I)\mathbf{v} = \mathbf{0}\).
4
Build the general solution: \(\mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2\).
5
Apply initial conditions \(\mathbf{x}(0)\) to find constants \(c_1, c_2\).

Example 1: Finding Eigenvalues

Solve the system \(\mathbf{x}' = A\mathbf{x}\) where:

Step 1: Characteristic equation.

Example 1: Finding Eigenvectors

Step 2: For \(\lambda_1 = 2\), solve \((A - 2I)\mathbf{v} = \mathbf{0}\):

For \(\lambda_2 = -2\), solve \((A + 2I)\mathbf{v} = \mathbf{0}\):

Example 2: Negative Eigenvalues

Consider the matrix:

Since \(A\) is upper triangular, the eigenvalues are the diagonal entries:

\(\lambda_1 = -3\)
Eigenvector: \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\)
\(\lambda_2 = -2\)
Eigenvector: \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\)

Example 3: Applying an Initial Condition

Using Example 2 with \(\mathbf{x}(0) = \begin{pmatrix} 4 \\ 2 \end{pmatrix}\), find \(c_1\) and \(c_2\).

1
From the second row: \(0 \cdot c_1 + 1 \cdot c_2 = 2\), so \(c_2 = 2\).
2
From the first row: \(c_1 + c_2 = 4\), so \(c_1 = 2\).

Visualizing Exponential Decay

Both components of our solution from Example 3 decay as \(t\) increases:

t x 0 2 4

\(x_1(t)\)

\(x_2(t)\)

Common Mistakes to Avoid

✕ Forgetting the Eigenvector
Writing \(\mathbf{x}(t) = c\,e^{\lambda t}\) without multiplying by \(\mathbf{v}\). The solution is a vector, not a scalar.
✕ Wrong Determinant
Subtracting \(\lambda\) from off-diagonal entries instead of only the diagonal when forming \(A - \lambda I\).
✕ Confusing Eigenvalue Sign
Saying negative eigenvalues cause growth. Negative \(\lambda\) means \(e^{\lambda t} \to 0\): that is decay, not growth.

Summary Table

Eigenvalue Type Solution Form Behavior
Both \(\lambda < 0\) Decaying exponentials Stable node
Both \(\lambda > 0\) Growing exponentials Unstable node
Mixed signs One grows, one decays Saddle point
Complex \(\lambda = \alpha \pm \beta i\) Oscillation with \(e^{\alpha t}\) Spiral (stable if \(\alpha < 0\))
Pure imaginary \(\lambda = \pm \beta i\) Pure oscillation Center (ellipses)

Question 1 of 5

If \(\lambda\) is an eigenvalue of \(A\) with eigenvector \(\mathbf{v}\), what is a solution to \(\mathbf{x}' = A\mathbf{x}\)?

Question 2 of 5

What is the first step in solving \(\mathbf{x}' = A\mathbf{x}\)?

Question 3 of 5

A \(2 \times 2\) system has eigenvalues \(\lambda_1 = -5\) and \(\lambda_2 = -1\). As \(t \to \infty\), what happens to the solutions?

Question 4 of 5

For \(A = \begin{pmatrix} 4 & 0 \\ 0 & -1 \end{pmatrix}\), what are the eigenvalues?

Question 5 of 5

In the general solution \(\mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2\), what determines the constants \(c_1\) and \(c_2\)?