Eigenvalues & Eigenvectors

When a matrix multiplies most vectors, the result points in a completely new direction. But some special vectors only get scaled — their direction stays the same (or flips). These are eigenvectors, and the scale factors are eigenvalues. Together, they reveal the intrinsic behavior of a linear transformation.

Definition

Given a square matrix $A$ , a nonzero vector $\vec{v}$ is an eigenvector of $A$ if:

A\vec{v} = \lambda \vec{v}

where $\lambda$ is a scalar called the eigenvalue. The matrix $A$ acts on $\vec{v}$ by simply stretching (or compressing, or flipping) it — no rotation, no shearing, just scaling by $\lambda$ .

Geometric Intuition

The eigen-decomposition reveals the “natural axes” of a transformation — the eigenvectors are the axes, and the eigenvalues are the scale factors along each:

$\lambda > 1$ : the eigenvector direction is stretched
$0 < \lambda < 1$ : the eigenvector direction is compressed
$\lambda < 0$ : the eigenvector direction is flipped and scaled
$\lambda = 0$ : the eigenvector direction is collapsed (the matrix is singular along this direction)

Computing Eigenvalues

Rearrange $A\vec{v} = \lambda\vec{v}$ :

(A - \lambda I)\vec{v} = \vec{0}

For a nonzero $\vec{v}$ to exist, the matrix $(A - \lambda I)$ must be singular — its determinant must be zero:

\det(A - \lambda I) = 0

This is the characteristic equation. For an $n \times n$ matrix, it yields a polynomial of degree $n$ in $\lambda$ .

2×2 Example

Find the eigenvalues of:

A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}

The characteristic equation:

\det\begin{bmatrix} 3 - \lambda & 1 \\ 0 & 2 - \lambda \end{bmatrix} = (3 - \lambda)(2 - \lambda) - 0 = 0

\lambda^2 - 5\lambda + 6 = 0 \implies (\lambda - 3)(\lambda - 2) = 0

So $\lambda_1 = 3$ and $\lambda_2 = 2$ .

Computing Eigenvectors

For each eigenvalue $\lambda_i$ , solve $(A - \lambda_i I)\vec{v} = \vec{0}$ for $\vec{v}$ .

For $\lambda_1 = 3$ :

(A - 3I)\vec{v} = \begin{bmatrix} 0 & 1 \\ 0 & -1 \end{bmatrix}\vec{v} = \vec{0} \implies v_2 = 0

Eigenvector: $\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ (or any scalar multiple)

For $\lambda_2 = 2$ :

(A - 2I)\vec{v} = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}\vec{v} = \vec{0} \implies v_1 + v_2 = 0

Eigenvector: $\vec{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$ (or any scalar multiple)

Eigenvectors Are Not Unique

If $\vec{v}$ is an eigenvector, so is $c\vec{v}$ for any nonzero scalar $c$ . We typically normalize eigenvectors to unit length, but the choice of sign/scale is arbitrary. What matters is the direction (or subspace).

Symmetric Matrices — A Special Case

Symmetric matrices ( $A = A^T$ ) arise frequently in robotics — inertia tensors, covariance matrices, stiffness matrices. They have two powerful guarantees:

All eigenvalues are real (no complex numbers)
Eigenvectors are orthogonal (perpendicular to each other)

This means a symmetric matrix can always be decomposed as:

A = Q \Lambda Q^T

where $Q$ is an orthogonal matrix of eigenvectors and $\Lambda$ is a diagonal matrix of eigenvalues. This is the spectral decomposition — it decomposes the transformation into independent scalings along perpendicular axes.

Inertia Tensor Robotics Application

The inertia tensor of a rigid body describes how mass is distributed around each axis:

I = \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{xy} & I_{yy} & -I_{yz} \\ -I_{xz} & -I_{yz} & I_{zz} \end{bmatrix}

This matrix is symmetric ( $I = I^T$ ). Its eigenvectors are the principal axes of inertia — the directions around which the body rotates most naturally (without wobbling). Its eigenvalues are the principal moments of inertia.

When you design a robot link, aligning the coordinate frame with the principal axes simplifies the dynamics equations because the inertia tensor becomes diagonal:

I_{\text{principal}} = \begin{bmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{bmatrix}

Eigenvalues of Rotation Matrices

Rotation matrices have a distinctive eigenvalue signature. For a 2D rotation by angle $\theta$ (where $\theta \neq 0, \pi$ ):

R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

The characteristic equation gives:

\lambda^2 - 2\cos\theta \cdot \lambda + 1 = 0 \implies \lambda = \cos\theta \pm i\sin\theta = e^{\pm i\theta}

The eigenvalues are complex — which makes sense geometrically: a nontrivial rotation in 2D doesn’t preserve any real direction.

For a 3D rotation by angle $\theta$ around axis $\hat{k}$ , the eigenvalues are:

\lambda_1 = 1, \quad \lambda_{2,3} = e^{\pm i\theta}

The eigenvector for $\lambda = 1$ is the rotation axis $\hat{k}$ itself — the one direction that the rotation leaves unchanged. This fact is the foundation of the axis-angle representation (covered in Lesson 3).

Key Insight

Every 3D rotation matrix has eigenvalue $\lambda = 1$ . The corresponding eigenvector is the rotation axis. This is an elegant way to extract the axis from a rotation matrix.

Diagonalization

A matrix $A$ is diagonalizable if it has $n$ linearly independent eigenvectors. In that case:

A = P \Lambda P^{-1}

where $P = [\vec{v}_1 \mid \vec{v}_2 \mid \ldots \mid \vec{v}_n]$ has eigenvectors as columns, and $\Lambda = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)$ .

This decomposition is powerful because:

Matrix powers: $A^k = P \Lambda^k P^{-1}$ (just raise each eigenvalue to the $k$ -th power)
Matrix exponential: $e^{At} = P \, e^{\Lambda t} \, P^{-1}$ (critical for solving differential equations)

Robotics Applications

1. Stability of Linear Systems

A linear dynamical system $\dot{\mathbf{x}} = A\mathbf{x}$ has behavior determined entirely by the eigenvalues of $A$ :

Eigenvalue property	System behavior
All Re( $\lambda_i$ ) < 0	Stable — all states decay to zero
Any Re( $\lambda_i$ ) > 0	Unstable — at least one state grows unbounded
Re( $\lambda_i$ ) = 0 (others < 0)	Marginally stable — oscillates without growing or decaying

Joint Controller Stability Robotics Application

A PD controller for a single robot joint gives the closed-loop dynamics:

\begin{bmatrix} \dot{e} \\ \ddot{e} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -k_p/m & -k_d/m \end{bmatrix} \begin{bmatrix} e \\ \dot{e} \end{bmatrix}

where $e$ is the position error, $k_p$ is the proportional gain, and $k_d$ is the derivative gain.

The characteristic equation is $\lambda^2 + (k_d/m)\lambda + k_p/m = 0$ . For stability, both eigenvalues must have negative real parts. By the quadratic formula, this requires:

$k_p > 0$ (positive stiffness)
$k_d > 0$ (positive damping)

The eigenvalues also reveal the response character: real eigenvalues give overdamped response (sluggish), complex eigenvalues give underdamped response (oscillatory), and the boundary is critical damping at $k_d^2 = 4mk_p$ .

2. Principal Component Analysis (PCA)

Given a point cloud from a LiDAR sensor, the covariance matrix captures how the data is spread:

\Sigma = \frac{1}{n}\sum_{i=1}^{n} (\mathbf{p}_i - \bar{\mathbf{p}})(\mathbf{p}_i - \bar{\mathbf{p}})^T

The eigenvectors of $\Sigma$ point along the directions of maximum variance (the principal axes of the data cloud), and the eigenvalues measure how much variance there is along each direction.

Surface Normal Estimation Robotics Application

A mobile robot’s LiDAR returns a cluster of 3D points near a surface. To estimate the surface normal:

Compute the covariance matrix $\Sigma$ of the local neighborhood
Find the eigenvalues $\lambda_1 \geq \lambda_2 \geq \lambda_3$
The eigenvector corresponding to $\lambda_3$ (the smallest eigenvalue) is the surface normal

Why? The smallest eigenvalue direction has the least spread — that’s the direction perpendicular to the surface where points vary the least.

If $\lambda_3 \approx 0$ and $\lambda_1, \lambda_2 \gg 0$ : clearly a flat surface. If $\lambda_2 \approx \lambda_3 \approx 0$ : the points lie along a line (edge). If $\lambda_1 \approx \lambda_2 \approx \lambda_3$ : the points form a blob (corner or noise).

3. Vibration Analysis

The generalized eigenvalue problem for a mechanical system:

K\vec{v} = \omega^2 M\vec{v}

where $K$ is the stiffness matrix and $M$ is the mass matrix. The eigenvalues $\omega^2$ give the natural frequencies and the eigenvectors give the mode shapes — the patterns of vibration the system naturally exhibits.

Robot Arm Resonance Robotics Application

A 2-DOF robot arm has mass matrix $M$ and stiffness matrix $K$ . Solving the generalized eigenvalue problem yields two natural frequencies $\omega_1, \omega_2$ and their mode shapes.

If the controller commands motion at a frequency near $\omega_1$ or $\omega_2$ , the arm resonates — vibrations amplify and can damage the system. Knowledge of these eigenvalues lets you:

Design controllers that avoid exciting resonant modes
Add damping targeted at problematic frequencies
Set trajectory acceleration limits to stay below resonance

4. Manipulability Ellipsoid

The matrix $J J^T$ (where $J$ is the robot Jacobian) encodes the end-effector’s ability to move and exert forces in different directions. Its eigenvectors define the axes of the manipulability ellipsoid, and the square roots of its eigenvalues are the semi-axis lengths.

Directions with large eigenvalues: the robot can move/push easily
Directions with small eigenvalues: the robot struggles (near singularity)
The ratio $\sqrt{\lambda_{\max}/\lambda_{\min}}$ is the condition number — it measures how “isotropic” the robot’s capability is

Yoshikawa's Manipulability Measure

A single scalar summarizing overall dexterity is $w = \sqrt{\det(JJ^T)} = \sqrt{\lambda_1 \lambda_2 \cdots \lambda_n}$ . When $w = 0$ , the robot is at a singularity. Maximizing $w$ drives the robot toward configurations with the most uniform capability in all directions.

The Eigenvalue Decomposition Visually

Consider a symmetric 2×2 matrix acting on the unit circle. Because symmetric matrices have orthogonal eigenvectors and real eigenvalues, the eigenvectors define the axes of the resulting ellipse, and the absolute eigenvalues are the semi-axis lengths:

The unit circle maps to an ellipse whose principal axes align with the eigenvectors
The lengths of the semi-axes equal $|\lambda_1|$ and $|\lambda_2|$
If $\lambda_1 = \lambda_2$ , the ellipse is a circle (uniform scaling)
If one eigenvalue is zero, the ellipse collapses to a line (singular matrix)

For non-symmetric matrices, the ellipse axes are instead determined by the singular vectors (from the SVD), and the semi-axis lengths are the singular values — a topic closely related to eigenvalues but beyond this lesson’s scope.

Controls

Symmetric Matrix

[ 2.0, 1.0 ]
[ 1.0, 1.0 ]

a = 2.0

-33

b = 1.0 (off-diagonal)

-33

d = 1.0

-33

Show unit circle

Show eigenvectors

Eigenvalue Analysis

λ₁ = 2.618

λ₂ = 0.382

v₁ = (0.851, 0.526)

v₂ = (0.526, -0.851)

det = 1.000

Positive definite

Try this: Start with "Identity" (circle maps to circle). Increase a to stretch horizontally. Then add off-diagonal b to rotate the eigenvectors. Set "Singular" to see the ellipse collapse to a line (λ₂ = 0).

Practice Problems

Find the eigenvalues and eigenvectors of $A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}$ .
A system matrix is $A = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}$ . Is the system stable?
The covariance matrix of a 2D point cloud is $\Sigma = \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}$ . Find the principal directions and their variances.
A 2D rotation matrix $R(60°)$ has eigenvalues $e^{\pm i\pi/3}$ . Verify this by computing $\det(R - \lambda I) = 0$ .
A robot’s Jacobian at a particular configuration gives $JJ^T = \begin{bmatrix} 9 & 0 \\ 0 & 1 \end{bmatrix}$ . Describe the manipulability ellipsoid. In which direction is the robot most capable?

Answers

Characteristic equation: $(4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda-5)(\lambda-2) = 0$ . So $\lambda_1 = 5$ , $\lambda_2 = 2$ . For $\lambda_1 = 5$ : $(A-5I)\vec{v}=0 \Rightarrow \vec{v}_1 = (2, 1)$ . For $\lambda_2 = 2$ : $(A-2I)\vec{v}=0 \Rightarrow \vec{v}_2 = (-1, 1)$ .
Characteristic equation: $\lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) = 0$ . Eigenvalues: $\lambda_1 = -1$ , $\lambda_2 = -2$ . Both are negative, so the system is stable.
Characteristic equation: $(5-\lambda)(2-\lambda) - 4 = \lambda^2 - 7\lambda + 6 = (\lambda-6)(\lambda-1) = 0$ . So $\lambda_1 = 6$ (direction of max variance), $\lambda_2 = 1$ . For $\lambda_1 = 6$ : $\vec{v}_1 = (2, 1)/\sqrt{5}$ . For $\lambda_2 = 1$ : $\vec{v}_2 = (-1, 2)/\sqrt{5}$ . The data is most spread along the direction $(2,1)$ .
$R(60°) = \begin{bmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{bmatrix}$ . $\det(R - \lambda I) = (1/2 - \lambda)^2 + 3/4 = \lambda^2 - \lambda + 1 = 0$ . $\lambda = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = e^{\pm i\pi/3}$ . Confirmed.
Eigenvalues are 9 and 1. The ellipsoid has semi-axis $\sqrt{9}=3$ along $x$ and $\sqrt{1}=1$ along $y$ . The robot is 3x more capable in the x-direction than the y-direction. Condition number = 3.

Key Takeaways

Eigenvectors are the directions a matrix preserves; eigenvalues are the scale factors
The characteristic equation $\det(A - \lambda I) = 0$ yields eigenvalues
Symmetric matrices have real eigenvalues and orthogonal eigenvectors
Eigenvalues of the system matrix determine stability (negative real parts = stable)
Eigenvalues of covariance matrices reveal principal directions in sensor data
The manipulability ellipsoid uses eigenanalysis to assess robot dexterity

Next Steps

Now that you can analyze the intrinsic properties of matrices, the next lesson covers coordinate frames — how to manage the multiple reference frames that every real robot must juggle.