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Complex Eigenvalues
& Real Solutions

Discover how complex conjugate eigenvalues produce real-valued solutions involving rotation and scaling in systems of differential equations.

Linear Algebra

What Are Eigenvalues?

Given a square matrix \( A \), an eigenvalue \( \lambda \) and its corresponding eigenvector \( \mathbf{v} \) satisfy the equation:

The matrix \( A \) stretches, flips, or rotates the vector \( \mathbf{v} \), and the scalar \( \lambda \) captures how much. When \( \lambda \) is real, the vector simply scales. When \( \lambda \) is complex, something more interesting happens: rotation.

When Do Complex Eigenvalues Appear?

We find eigenvalues by solving the characteristic equation:

For a \( 2 \times 2 \) matrix, this gives a quadratic in \( \lambda \). Complex eigenvalues appear whenever the discriminant is negative:

Conjugate Pairs Always Travel Together

A key fact: for a real matrix \( A \), complex eigenvalues always come in conjugate pairs.

\( \lambda_1 = \alpha + \beta i \)
With eigenvector \( \mathbf{v}_1 = \mathbf{a} + \mathbf{b}i \)
\( \lambda_2 = \alpha - \beta i \)
With eigenvector \( \mathbf{v}_2 = \mathbf{a} - \mathbf{b}i \)

Eigenvalues on the Complex Plane

Re Im \( \alpha + \beta i \) \( \alpha - \beta i \) \( r \) \( \alpha \) \( \beta \) \( \theta \)

The Core Theorem

For \( \mathbf{x}' = A\mathbf{x} \) where \( A \) has complex eigenvalues \( \lambda = \alpha \pm \beta i \) with eigenvector \( \mathbf{v} = \mathbf{a} + \mathbf{b}i \), two linearly independent real solutions are:

The general solution is then \( \mathbf{x}(t) = c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t) \).

Anatomy of the Solution

\( e^{\alpha t} \)
Controls amplitude: growth (\( \alpha > 0 \)), decay (\( \alpha < 0 \)), or steady (\( \alpha = 0 \))
\( \cos\beta t,\;\sin\beta t \)
Controls rotation: the solution spirals with angular frequency \( \beta \)
\( \mathbf{a},\;\mathbf{b} \)
Real and imaginary parts of the eigenvector set the axes of the ellipse

Three Phase Portrait Cases

\( \alpha > 0 \): Spiral Source
Solutions spiral outward: unstable. Amplitude grows without bound.
\( \alpha < 0 \): Spiral Sink
Solutions spiral inward: stable. All trajectories decay to the origin.
\( \alpha = 0 \): Center
Solutions form closed ellipses: neutrally stable. No growth, no decay.
Spiral Source Spiral Sink Center

How to Find Real Solutions

1
Compute eigenvalues. Solve \( \det(A - \lambda I) = 0 \). Get \( \lambda = \alpha \pm \beta i \).
2
Find one eigenvector. Solve \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) for \( \lambda = \alpha + \beta i \). Write \( \mathbf{v} = \mathbf{a} + \mathbf{b}i \).
3
Separate real and imaginary parts. Identify vectors \( \mathbf{a} \) (real part) and \( \mathbf{b} \) (imaginary part).
4
Write two real solutions. Use \( \mathbf{x}_1 = e^{\alpha t}(\mathbf{a}\cos\beta t - \mathbf{b}\sin\beta t) \) and \( \mathbf{x}_2 = e^{\alpha t}(\mathbf{a}\sin\beta t + \mathbf{b}\cos\beta t) \).
5
Form the general solution. \( \mathbf{x}(t) = c_1\mathbf{x}_1 + c_2\mathbf{x}_2 \) with arbitrary constants.

Example 1: Finding Eigenvalues

Consider the system \( \mathbf{x}' = A\mathbf{x} \) with:

Step 1: Characteristic equation: \( \det(A - \lambda I) = (1 - \lambda)^2 + 2 = 0 \)

Example 1: Eigenvector

Step 2: For \( \lambda = 1 + \sqrt{2}\,i \), solve \( (A - \lambda I)\mathbf{v} = \mathbf{0} \):

From the second row: \( v_1 - \sqrt{2}\,i\,v_2 = 0 \implies v_1 = \sqrt{2}\,i\cdot v_2 \). Taking \( v_2 = 1 \):

Example 1: General Solution

Steps 4–5: With \( \alpha = 1 \), \( \beta = \sqrt{2} \), \( \mathbf{a} = \begin{pmatrix}0\\1\end{pmatrix} \), \( \mathbf{b} = \begin{pmatrix}\sqrt{2}\\0\end{pmatrix} \):

Example 2: A Spiral Sink

Now let \( A = \begin{pmatrix} -1 & -4 \\ 1 & -1 \end{pmatrix} \). The characteristic equation gives:

Here \( \alpha = -1 \) and \( \beta = 2 \). The eigenvector for \( \lambda = -1 + 2i \) works out to \( \mathbf{v} = \begin{pmatrix}2i\\1\end{pmatrix} \), so \( \mathbf{a} = \begin{pmatrix}0\\1\end{pmatrix} \), \( \mathbf{b} = \begin{pmatrix}2\\0\end{pmatrix} \).

Example 3: A Pure Center

Consider \( A = \begin{pmatrix} 0 & -3 \\ 3 & 0 \end{pmatrix} \). The characteristic equation:

Here \( \alpha = 0 \) and \( \beta = 3 \). No exponential factor means no growth or decay.

Common Mistakes

✕ Leaving \( i \) in the Answer
Complex eigenvectors must be separated into real and imaginary parts. Your final solution must be entirely real.
✕ Swapping \( \mathbf{a} \) and \( \mathbf{b} \)
If \( \mathbf{v} = \mathbf{a} + \mathbf{b}i \), mixing up which is \( \mathbf{a} \) (real) vs \( \mathbf{b} \) (imaginary) flips the sign in the formulas.
✕ Forgetting \( e^{\alpha t} \)
When \( \alpha \neq 0 \), the exponential factor is essential. Without it, you describe a center instead of a spiral.

Summary Table

Spiral Source Spiral Sink Center
Eigenvalue \( \alpha \pm \beta i \) \( \alpha \pm \beta i \) \( \pm \beta i \)
Sign of \( \alpha \) \( \alpha > 0 \) \( \alpha < 0 \) \( \alpha = 0 \)
Behavior Spirals outward Spirals inward Closed ellipses
Stability Unstable Stable Neutrally stable
Exp. Factor \( e^{\alpha t} \to \infty \) \( e^{\alpha t} \to 0 \) \( e^{0} = 1 \)

Question 1 of 5

If a \( 2 \times 2 \) real matrix has eigenvalues \( \lambda = 3 \pm 4i \), what type of phase portrait does the system exhibit?

Question 2 of 5

For \( \lambda = -2 \pm 5i \), what are \( \alpha \) and \( \beta \)?

Question 3 of 5

If \( A = \begin{pmatrix} 0 & -5 \\ 5 & 0 \end{pmatrix} \), the eigenvalues are \( \lambda = \pm 5i \). The origin is a:

Question 4 of 5

Given eigenvector \( \mathbf{v} = \begin{pmatrix} 1 - 3i \\ 2 \end{pmatrix} \), what are the real part \( \mathbf{a} \) and imaginary part \( \mathbf{b} \)?

Question 5 of 5

Why do complex eigenvalues of a real matrix always come in conjugate pairs?