---
title: "Eigenvalues defined via linear transformation"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:eigenvalues-defined-via-linear-transformation"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-defined-via-linear-transformation
site: BCT Engineering
---

# Eigenvalues defined via linear transformation

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2074 Asoj, 2+3 marks
- Video: [Eigenvectors and eigenvalues | Chapter 14, Essence of linear algebra](https://www.youtube.com/watch?v=PFDu9oVAE-g) — 3Blue1Brown
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-defined-via-linear-transformation
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Define eigenvalues and eigenvectors in terms of a linear transformation with matrices as operator. Find the eigenvalues of the matrix [−2 2 −3; 2 1 −6; −1 −2 0].

## Short answer

A non-zero vector X with A X = λ X is an eigenvector and λ is the eigenvalue. For the given matrix, characteristic equation λ³+λ²−21λ−45=0 gives eigenvalues λ = 5, −3, −3.

## Answer

#### Definition through a linear transformation
A square matrix A of order n acts as an operator (a linear transformation) T on vectors: Y = AX takes the vector X to a new vector Y, usually with a different direction and length.

A non-zero vector X is called an **eigenvector** (characteristic vector) of A if the transformation only rescales it, i.e. keeps its line unchanged:

```text
AX = λX ,   X ≠ 0
```

The scalar λ is the corresponding **eigenvalue** (characteristic root). Equivalently (A − λI)X = 0 has a non-trivial solution, which happens only if

```text
|A − λI| = 0      (characteristic equation)
```

**Example:** for A = [2 0; 0 3], the vector (1, 0)ᵀ is mapped to (2, 0)ᵀ = 2(1, 0)ᵀ, so λ = 2 with eigenvector (1, 0)ᵀ: the map stretches that direction without turning it.

#### Eigenvalues of A = [−2 2 −3; 2 1 −6; −1 −2 0]
- Trace = −2 + 1 + 0 = −1
- Sum of principal minors = (0 − 12) + (0 − 3) + (−2 − 4) = −21
- |A| = −2(0 − 12) − 2(0 − 6) − 3(−4 + 1) = 24 + 12 + 9 = 45

```text
λ³ + λ² − 21λ − 45 = 0
(λ − 5)(λ + 3)² = 0
```

**Eigenvalues: λ = 5, −3, −3.**

(Check: sum = 5 − 3 − 3 = −1 = trace ✓; product = 5 · 9 = 45 = |A| ✓.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In terms of the transformation Y = AX, an eigenvector X is one that:
   A. Is mapped to the zero vector
   B. Is only rescaled, keeping its line unchanged
   C. Is rotated by 90 degrees
   D. Is mapped to its transpose
   Answer: B. AX = λX with X not zero: the direction is kept and length scaled by λ. (Key point: Eigenvector meaning)

2. Why must |A - λI| = 0?
   A. So that A is symmetric
   B. So that λ is real
   C. So that (A - λI)X = 0 has a non-trivial solution
   D. So that X is the zero vector
   Answer: C. A non-zero solution exists only if A - λI is singular. (Key point: Characteristic equation)

3. The eigenvalues of the given matrix are:
   A. 5, -3, -3
   B. 5, 3, 3
   C. -5, 3, 3
   D. 1, 5, -3
   Answer: A. λ^3 + λ^2 - 21λ - 45 = (λ-5)(λ+3)^2. (Key point: Eigenvalues)

4. Which check confirms 5, -3, -3?
   A. Sum = 0 and product = 0
   B. Sum = -1 = trace and product = 45 = |A|
   C. Sum = 45 and product = -1
   D. Sum = 5 and product = -9
   Answer: B. 5 - 3 - 3 = -1 equals the trace, 5 x 9 = 45 equals |A|. (Key point: Check)

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