LINEAR-ALGEBRA

The Vectors That Refuse to Turn: Eigenvalues & Eigenvectors

IKey Formulas Reference

Formula	Name	Description
$A\mathbf{x} = \lambda \mathbf{x}$	Eigenvalue equation	$\mathbf{x} \neq \mathbf{0}$ is an eigenvector; $\lambda$ is its eigenvalue
$\lambda_1 + \lambda_2 + \cdots + \lambda_n = \text{tr}(A)$	Sum of eigenvalues	Equals the trace (sum of diagonal entries)
$\lambda_1 \cdot \lambda_2 \cdots \lambda_n = \det(A)$	Product of eigenvalues	Equals the determinant
$\lambda = 0 \iff A$ is singular	Singularity test	Zero eigenvalue means $\det(A) = 0$

IIVisualization 1 — The Direction That Won't Turn

Which directions does $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ leave on their own line — and by what factor $\lambda$ ?

Direction θ 20°

💡 Because $A$ is symmetric, its two eigen-directions are perpendicular and its eigenvalues are always real.

IIIVisualization 2 — Eigenvalues Stretch Each Axis

Each eigenvalue stretches, flips, or crushes space along one eigen-axis — the other axis is untouched.

λ₁ (along [1, 1]) 2 λ₂ (along [−1, 1]) −1

IVVisualization 3 — Rotation Has No Real Eigenvector

A pure rotation turns every vector by the same angle, so no real direction can return to its own line.

Rotation θ 0°

💡 Every other angle gives complex eigenvalues $e^{\pm i\theta}$ — the reason a real matrix can still force you into complex numbers.

VQuiz Questions

Question 1 · The Eigenvalue Equation

Suppose $A$ is a $2 \times 2$ matrix and $\mathbf{x} \neq \mathbf{0}$ is an eigenvector of $A$ with eigenvalue $\lambda = -3$ .

Which statement is correct?

✅ **Correct!**A negative eigenvalue flips the eigenvector, and the length scales by $|\lambda|$ .

❌ **Not quite.**Remember: $A\mathbf{x} = \lambda \mathbf{x}$ . A negative $\lambda$ means the output is on the same line but flipped, with length $|\lambda|$ times the original.

Show solution

Solution:

The eigenvalue equation says $A\mathbf{x} = \lambda \mathbf{x}$ . With $\lambda = -3$ :

$A\mathbf{x} = -3\,\mathbf{x}$

The magnitude is $|-3| = 3$ times the length of $\mathbf{x}$ .
The sign is negative, so $A\mathbf{x}$ points in the opposite direction.
It is still on the same line as $\mathbf{x}$ (just flipped) — so it's not perpendicular and it's not zero.

Question 2 · Trace and Determinant

A $3 \times 3$ matrix $A$ has eigenvalues $\lambda_1 = 4$ , $\lambda_2 = -1$ , $\lambda_3 = 2$ .

What are the trace and determinant of $A$ ?

✅ **Correct!**Sum of eigenvalues = trace; product of eigenvalues = determinant.

❌ **Not quite.**Remember: $\text{tr}(A) = \sum \lambda_i$ (sum), and $\det(A) = \prod \lambda_i$ (product).

Show solution

Solution:

Trace = sum of eigenvalues: $4 + (-1) + 2 = 5$
Determinant = product of eigenvalues: $4 \cdot (-1) \cdot 2 = -8$

So $\text{tr}(A) = 5$ and $\det(A) = -8$ .

Common mistake: swapping trace and determinant, or using the wrong operation (sum vs. product).

Question 3 · Zero Eigenvalue

True or False: If $\lambda = 0$ is an eigenvalue of matrix $A$ , then $A$ must be singular (i.e., $\det(A) = 0$ ).

✅ Correct! A zero eigenvalue forces $\det(A) = 0$ because the determinant is the product of eigenvalues.

❌ Not quite. Recall that $\det(A) = \prod \lambda_i$ . If any $\lambda_i = 0$ , the product is zero.

Show solution

Solution:

The determinant equals the product of all eigenvalues: $\det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n$

If any eigenvalue is zero, the entire product is zero, so $\det(A) = 0$ and $A$ is singular.

Equivalently: $\lambda = 0$ means there is a nonzero $\mathbf{x}$ with $A\mathbf{x} = \mathbf{0}$ , so $A$ has a nontrivial null space — the hallmark of a singular matrix.

Question 4 · Identifying Eigenvectors

Let $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ (the swap matrix from the video).

Which of the following vectors is an eigenvector of $A$ ?

✅ Correct! $A[-1,1]^T = [1,-1]^T = -1 \cdot [-1,1]^T$ , so $\lambda = -1$ .

❌ **Not quite.**The output $A\mathbf{x}$ must be a scalar multiple of $\mathbf{x}$ — check whether the output is just $\mathbf{x}$ scaled by one number.

Show solution

Solution:

A vector $\mathbf{x}$ is an eigenvector of $A$ when $A\mathbf{x} = \lambda \mathbf{x}$ — the output must be a scalar multiple of the input.

Check each option:

$A[2,3]^T = [3,2]^T$ . Is $[3,2] = \lambda [2,3]$ ? That would need $\lambda = 3/2$ and $\lambda = 2/3$ . ❌ Not an eigenvector.
$A[1,0]^T = [0,1]^T$ . Is $[0,1] = \lambda [1,0]$ ? Need $0 = \lambda \cdot 1$ and $1 = \lambda \cdot 0$ — impossible. ❌
$A[-1,1]^T = [1,-1]^T$ . Is $[1,-1] = \lambda [-1,1]$ ? Yes, with $\lambda = -1$ . ✅ Eigenvector with eigenvalue $-1$ .
$A[1,2]^T = [2,1]^T$ . Is $[2,1] = \lambda[1,2]$ ? Need $\lambda = 2$ and $\lambda = 1/2$ . ❌

So only $[-1, 1]$ qualifies — it's the eigenvector with $\lambda = -1$ .

VISummary

You've explored the core ideas of eigenvalues and eigenvectors:

Eigenvectors are the special directions a matrix can only scale, never rotate off their line.
Eigenvalues tell you how much scaling happens: stretch ( $|\lambda| > 1$ ), compress ( $0 < |\lambda| < 1$ ), flip (negative), or crush to zero ( $\lambda = 0$ ).
$\text{trace}(A) = \sum \lambda_i$ and $\det(A) = \prod \lambda_i$ — two powerful sanity checks.
A zero eigenvalue ⟺ a singular matrix.
Pure rotations have complex eigenvalues — no real direction survives.

These ideas power PageRank, PCA, vibration analysis, and much more.

Solved: 0 / 4