3.2 Eigenvectors and Eigenvalues as Invariant Axes of Pure Stretch

When a matrix $A$ transforms space, almost every vector $\mathbf{x}$ gets knocked off the line it originally spanned. Yet for most matrices used in science and machine learning, there exist special directions where the transformation produces zero rotation—vectors along those lines simply get stretched, compressed, or flipped along their own span.

Those invariant directions are the eigenvectors (from the German eigen, meaning "own" or "characteristic"), and the scalar factor by which each eigenvector is scaled is its eigenvalue $\lambda$.

In this Article

• 1. The Span-Preservation Condition: $A\mathbf{v} = \lambda\mathbf{v}$
• 2. Why Eigenvectors Form Entire Lines (Eigenspaces)
• 3. Finding Eigenvectors by Pure Geometric Inspection
• 4. Worked Numerical Example: Testing Candidate Vectors
• 5. Active Recall Check and Key Takeaways

1. The Span-Preservation Condition: $A\mathbf{v} = \lambda\mathbf{v}$

For a nonzero vector $\mathbf{v} \in \mathbb{R}^d$, the line passing through the origin in the direction of $\mathbf{v}$ is its span: $\text{span}(\mathbf{v}) = \{ c\mathbf{v} : c \in \mathbb{R} \}$. Asking whether $A$ keeps $\mathbf{v}$ on its original line is identical to asking whether the output $A\mathbf{v}$ is a scalar multiple of the input $\mathbf{v}$:

$$A\mathbf{v} = \lambda \mathbf{v}, \qquad \mathbf{v} \neq \mathbf{0}$$

• If $A\mathbf{v}$ and $\mathbf{v}$ are parallel (their ratio of components is identical in every coordinate), then $\mathbf{v}$ is an eigenvector.
• If the ratio of components changes—even slightly—then $A\mathbf{v}$ points in a new direction and $\mathbf{v}$ is not an eigenvector.

2. Why Eigenvectors Form Entire Lines (Eigenspaces)

Suppose $\mathbf{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$. What happens if we scale $\mathbf{v}$ by any nonzero number $c \neq 0$?

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

3. Finding Eigenvectors by Pure Geometric Inspection

Before touching any characteristic polynomial, you can spot the eigenvectors of many fundamental matrices immediately from their geometry:

   1. HORIZONTAL SHEAR A = [[1, 1], [0, 1]]      2. PURE 90-DEG ROTATION R = [[0, -1], [1, 0]]
            x2                                            x2
            ^     * --> * (Tilted right!)                 ^     * R(x)
            |    /     /                                  |    /
            +---0-----*-----> x1 (v=[1,0] stays           +---0-----* x (Every real vector
         Only horizontal vectors stay on line!             rotates 90 deg! NO real eigenvectors!)

Matrix $A$	Geometric Action	Eigenvectors & Eigenvalues by Inspection
Diagonal $\begin{bmatrix} 5 & 0 \\ 0 & -2 \end{bmatrix}$	Stretches $x_1$ by $5$, flips/stretches $x_2$ by $-2$	$\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} (\lambda_1=5)$, $\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} (\lambda_2=-2)$
Reflection $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$	Mirrors across the $45^\circ$ line $x_2 = x_1$	$\begin{bmatrix} 1 \\ 1 \end{bmatrix} (\lambda_1=1, \text{on mirror})$, $\begin{bmatrix} -1 \\ 1 \end{bmatrix} (\lambda_2=-1, \text{flipped})$
Shear $\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$	Slides horizontal lines; fixes $x_1$-axis	Only $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} (\lambda=1)$; defective (only 1 line!)
Rotation $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$	Rotates every vector by $90^\circ$	No real eigenvectors ($\lambda = \pm i$ over $\mathbb{C}$)

4. Worked Numerical Example: Testing Candidate Vectors

Let $A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$. Let us test three vectors $\mathbf{a} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\mathbf{b} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$, and $\mathbf{c} = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$:

1. Test $\mathbf{a} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$: $A\mathbf{a} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$. Since $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$ is not a multiple of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\mathbf{a}$ is not an eigenvector.
2. Test $\mathbf{b} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$: $A\mathbf{b} = \begin{bmatrix} 3+2 \\ 1+4 \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \end{bmatrix} = 5\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Thus $\mathbf{b}$ is an eigenvector with $\lambda_1 = 5$!
3. Test $\mathbf{c} = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$: $A\mathbf{c} = \begin{bmatrix} 3(-2)+2(1) \\ 1(-2)+4(1) \end{bmatrix} = \begin{bmatrix} -4 \\ 2 \end{bmatrix} = 2\begin{bmatrix} -2 \\ 1 \end{bmatrix}$. Thus $\mathbf{c}$ is an eigenvector with $\lambda_2 = 2$!

5. Active Recall Check and Key Takeaways

Quick Check: Test Your Understanding

Question: Let $P = \mathbf{u}\mathbf{u}^T$ be the orthogonal projection matrix onto a unit vector $\mathbf{u} \in \mathbb{R}^2$. Without doing any determinant algebra, what are the two eigenvectors and eigenvalues of $P$?

Answer: First, any vector already along $\mathbf{u}$ projects onto itself unchanged: $P\mathbf{u} = 1\mathbf{u}$, so $\mathbf{u}$ is an eigenvector with $\lambda_1 = 1$. Second, any vector $\mathbf{u}_\perp$ perpendicular to $\mathbf{u}$ projects to the zero shadow at the origin: $P\mathbf{u}_\perp = \mathbf{0} = 0\mathbf{u}_\perp$, so $\mathbf{u}_\perp$ is an eigenvector with $\lambda_2 = 0$!

• Pure Stretch Condition: $A\mathbf{v} = \lambda\mathbf{v}$ means $A$ scales $\mathbf{v}$ without rotating it off its span.
• Direction vs. Scale: Any nonzero multiple of an eigenvector is an eigenvector with the same eigenvalue $\lambda$.
• Canonical References: Explore Setosa's Interactive Eigenvectors Playground and Wikipedia Reference on Eigenvalues and Eigenvectors.

Course Navigation & Chapter Links

This article is part of the complete 24-part Synapticz public course: Intuitive Eigenvalues, Eigenvectors, and Principal Component Analysis (Top-Down & Bottom-Up).

• Chapter 3 Overview: Chapter 3: Visualizing Linear Transformations, Eigenvectors, and Eigenvalues Without Memorization
• Microlearning 3.1: 3.1 Matrices as Geometric Grid Transformations: Stretching, Shearing, and Rotating
• Microlearning 3.2: 3.2 Eigenvectors and Eigenvalues as Invariant Axes of Pure Stretch
• Microlearning 3.3: 3.3 Why det(A - λI) = 0 Finds Non-Zero Vectors in the Collapse Nullspace
• Previous Lesson: 3.1 Matrices as Geometric Grid Transformations: Stretching, Shearing, and Rotating
• Next Lesson: 3.3 Why det(A - λI) = 0 Finds Non-Zero Vectors in the Collapse Nullspace