LINEAR-ALGEBRA · Interactive Practice | Unit 20 · Video 3

When Eigenvalues Break the Rules: Complex Numbers and Missing Eigenvectors

IKey Formulas & Facts

Concept	Formula / Rule	Notes
Eigenvalue equation	$A\mathbf{x} = \lambda \mathbf{x}$	Vector stays on its own line
Characteristic equation	$\det(A - \lambda I) = 0$	Solve for eigenvalues $\lambda$
Trace & determinant	$\lambda_1 + \lambda_2 = \text{tr}(A),\ \lambda_1 \lambda_2 = \det(A)$	Useful check
Triangular matrix	Eigenvalues $=$ diagonal entries	Upper or lower triangular
Symmetry rule	$A^T = A \Rightarrow$ real $\lambda$ ; $A^T = -A \Rightarrow$ imaginary $\lambda$	Conjugate pairs for real matrices

IIVisualization 1 — Rotation's Missing Real Eigenvector

A rotation by $\theta$ turns every real vector off its own line, so no real $\mathbf{v}$ can be an eigenvector.

Rotation θ 90°

IIIVisualization 2 — A Shear's Single Eigen-Direction

The shear $\begin{bmatrix} 3 & s \\ 0 & 3 \end{bmatrix}$ has one repeated eigenvalue $\lambda = 3$ , yet it fixes only the horizontal line.

Off-diagonal s 1.0

💡 Because the second independent eigenvector is missing, this shear is defective — it cannot be diagonalized.

IVVisualization 3 — Eigenvalues in the Complex Plane

Splitting $A = S + N$ , the symmetric part $S$ pushes the eigenvalues onto the real axis; the anti-symmetric part $N$ pushes them onto the imaginary axis.

Symmetric a 1.5 Anti-symmetric b 0.5

VQuiz Questions

Question 1

Consider the $90°$ rotation matrix $Q = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ .

What are its eigenvalues?

✅ Correct! Anti-symmetric matrices have purely imaginary eigenvalues, always in conjugate pairs.

❌ Not quite. Compute $\det(Q - \lambda I) = \lambda^2 + 1 = 0$ , so $\lambda^2 = -1$ .

Show solution

Solution:

Set up the characteristic equation: $\det(Q - \lambda I) = \det\begin{bmatrix} -\lambda & -1 \\ 1 & -\lambda \end{bmatrix} = \lambda^2 + 1 = 0$

So $\lambda^2 = -1$ , giving $\lambda = \pm i$ .

Check: $\lambda_1 + \lambda_2 = i + (-i) = 0 = \text{tr}(Q)$ , and $\lambda_1 \lambda_2 = i \cdot (-i) = 1 = \det(Q)$ .

This is the signature of an anti-symmetric matrix ( $Q^T = -Q$ ): purely imaginary eigenvalues and no real eigenvectors.

Question 2

True or False: For a real matrix, if $3 + 2i$ is an eigenvalue, then $3 - 2i$ must also be an eigenvalue.

✅ Correct! Real matrices force complex eigenvalues to appear in conjugate pairs.

❌ Not quite. The characteristic polynomial of a real matrix has real coefficients, so complex roots must come in conjugate pairs.

Show solution

Solution:

True. For a matrix with real entries, the characteristic polynomial has real coefficients. Any complex roots of a real polynomial come in complex conjugate pairs.

So if $a + bi$ is an eigenvalue, then $a - bi$ must also be an eigenvalue. This is why the rotation matrix $Q$ has eigenvalues $i$ and $-i$ together — they are conjugates.

Question 3

Consider $A = \begin{bmatrix} 5 & 1 \\ 0 & 5 \end{bmatrix}$ . Find its eigenvalues and determine how many linearly independent eigenvectors it has.

✅ Correct! A repeated eigenvalue with fewer independent eigenvectors than its multiplicity makes the matrix defective.

❌ Not quite. Triangular matrix: eigenvalues are diagonal entries (5, 5). Now solve $(A - 5I)\mathbf{x} = \mathbf{0}$ and count the free parameters.

Show solution

Solution:

Since $A$ is upper triangular, the eigenvalues are the diagonal entries: $\lambda = 5, 5$ (repeated).

Find the eigenvectors: Solve $(A - 5I)\mathbf{x} = \mathbf{0}$ : $A - 5I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

Row 1 gives $x_2 = 0$ . Row 2 gives $0 = 0$ (no info). So $x_1$ is free, $x_2 = 0$ : $\mathbf{x} = t\begin{bmatrix} 1 \\ 0 \end{bmatrix}$

Only 1 independent eigenvector — the x-axis direction. Algebraic multiplicity is 2, but geometric multiplicity is 1. The matrix is defective and cannot be diagonalized.

Question 4

Which statement about a matrix $A$ satisfying $A^T = A$ (symmetric) is TRUE?

✅ Correct! This is the Spectral Theorem — symmetric matrices are the well-behaved ideal case.

❌ Not quite. You're thinking of anti-symmetric or defective matrices. Symmetric matrices ( $A^T = A$ ) are the ideal case.

Show solution

Solution:

Symmetric matrices ( $A^T = A$ ) have two very nice properties (the Spectral Theorem):

All eigenvalues are real — no complex eigenvalues can appear.
They admit a full set of orthogonal eigenvectors — so they are always diagonalizable (never defective).

This is the "well-behaved ideal." The two pathologies we studied happen for other kinds of matrices:

Anti-symmetric ( $A^T = -A$ ) → purely imaginary eigenvalues (e.g., rotations).
Defective matrices (repeated eigenvalues without enough eigenvectors) → not diagonalizable (e.g., shears).

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