# Eigenvalues and Eigenvectors

Business I · Mathematics I · https://tryals.app/learn/business-i/eigenvalues-and-eigenvectors

## The Directions a Matrix Cannot Bend

Apply a matrix to most vectors and they rotate and stretch. But special **eigenvectors** only get scaled along their line. Their scaling factor is the **eigenvalue** $\lambda$:

$$Av = \lambda v, \qquad v \ne 0$$

**Finding them**: Rewrite as $(A - \lambda I)v = 0$. A nonzero solution exists only if $A - \lambda I$ is singular, meaning its determinant is zero:

$$\det(A - \lambda I) = 0$$

This **characteristic polynomial** yields the eigenvalues as its roots. For each root, solving $(A-\lambda I)v = 0$ gives its **eigenspace**.

For a triangular matrix, eigenvalues sit right **on the diagonal**. Always use quick checksums to catch errors before moving forward.

## The Eigen-Recipe & Dynamics

| Step / Check | Equation | Yields | Scope |
| --- | --- | --- | --- |
| 1. Polynomial | $\det(A - \lambda I) = 0$ | Eigenvalues $\lambda$ | Roots |
| 2. Nullspace | $(A-\lambda I)v = 0$ | Eigenspace $v$ | Vectors |
| Check: Trace | $\sum \lambda_i = tr(A)$ | Sum test | Errors |
| Check: Det | $\prod \lambda_i = \det A$ | Product test | Errors |

**Why dynamics care**: Repeated multiplication aligns states with the **dominant eigenvector** (largest $|\lambda|$).

$$|\lambda| > 1: \text{growth} \quad |\lambda| < 1: \text{decay} \quad \lambda = 1: \text{steady}$$

Markov matrices have $\lambda = 1$; their eigenvector gives the long-run market share or Google's PageRank.

## Practice questions

5 of this lesson's 13 practice questions, with answers. The full set is in the app.

### 1. Match each object to its role.

**Answer:**

- Eigenvector → Direction the matrix only scales
- Eigenvalue → The scaling factor on that direction
- Characteristic polynomial → $\det(A - \lambda I)$, whose roots are the $\lambda$s
- Eigenspace → All eigenvectors sharing one $\lambda$

**Why:** Direction, factor, detector, and family, the four nouns of eigen-theory.

Page: https://tryals.app/practice/business-i/eigenvalues-and-eigenvectors/match-each-object-to-its-role

### 2. Two phone carriers trade customers monthly by a fixed transition matrix. Whatever the starting split, shares always drift to 60/40 and stay. In eigen-language, $(0.6, 0.4)$ is…

A. An eigenvector of eigenvalue 0
B. Proof that the matrix is singular
C. The eigenvector of eigenvalue 1
D. The characteristic polynomial

**Answer:** C. The eigenvector of eigenvalue 1

**Why:** "Drifts there and stays" is precisely $Mv = v$: eigenvalue 1, eigenvector = the long-run market share. The drift from any start is the dominant eigenvector asserting itself over the fading ones.

Page: https://tryals.app/practice/business-i/eigenvalues-and-eigenvectors/two-phone-carriers-trade-customers-monthly-by-a-fixed-transition

### 3. A market-share simulation is run from three wildly different starting splits and all three converge to the same final split. What guarantees that?

A. That the starting splits were secretly similar once normalised
B. The dominant eigenvector, which suppresses every other component
C. That the simulation was run long enough for random fluctuations to average out
D. Nothing guarantees it; the agreement is a coincidence of the three chosen starts

**Answer:** B. The dominant eigenvector, which suppresses every other component

**Why:** Every component is scaled by its eigenvalue each step, so the largest survives and the rest die away, the destination belongs to the switching behaviour, not the history. It is why churn matrices predict long-run share and why PageRank works on exactly the same mathematics.

Page: https://tryals.app/practice/business-i/eigenvalues-and-eigenvectors/a-market-share-simulation-is-run-from-three-wildly-different-starting

### 4. A subscriber base evolves with dominant eigenvalue $\lambda = 1.2$ per year, starting aligned with the eigenvector at 100k subscribers. Set the slider to the size after 3 years (thousands).

**Answer:** 172.8 (within ±8)

**Why:** $100 \cdot 1.2^3 = 172.8$k. On an eigenvector, matrix dynamics collapse to plain compounding, that reduction of a many-variable system to one growth rate is the entire practical value of eigenvalues.

Page: https://tryals.app/practice/business-i/eigenvalues-and-eigenvectors/a-subscriber-base-evolves-with-dominant-eigenvalue-1-2-per-year

### 5. A triangular matrix wears its eigenvalues on the diagonal, and a zero among them means the matrix is singular. Why do those two facts fit together?

A. Because the determinant is the product of the eigenvalues
B. Because eigenvalues on the diagonal are only approximate for triangular matrices
C. Because the trace of a singular matrix must be zero
D. Because triangular matrices are singular whenever they have repeated eigenvalues

**Answer:** A. Because the determinant is the product of the eigenvalues

**Why:** Both routes give the same number. A zero eigenvalue means some nonzero vector is sent to the origin, the map collapses a direction, which is singularity seen geometrically.

Page: https://tryals.app/practice/business-i/eigenvalues-and-eigenvectors/a-triangular-matrix-wears-its-eigenvalues-on-the-diagonal-and-a-zero