# Eigenvalues and Eigenvectors

> IB Mathematics Applications and Interpretation HL · IB Math AI HL
> Source: https://www.owlsprep.com/study/ib-math-ai-hl-u1-eigenvalues-and-eigenvectors/

This module defines eigenvalues and eigenvectors, derives the characteristic equation, walks through full calculation workflows, and covers AI HL specific applied use cases for linear transformations.

**Prerequisites:** [2x2 and 3x3 matrix arithmetic and determinant calculation](https://www.owlsprep.com/study/ib-math-ai-hl-u1-matrix-operations/); [Solving systems of linear equations via Gaussian elimination](https://www.owlsprep.com/study/ib-math-ai-hl-u1-gaussian-elimination/)

## Learning objectives

- Define eigenvalues and eigenvectors for square 2x2 and 3x3 matrices
- Derive and solve the characteristic equation to find all real eigenvalues
- Calculate corresponding non-zero eigenvectors for each eigenvalue
- Verify eigenvector-eigenvalue pairs and apply properties to transformation problems

## Core Definition of Eigenvalue-Eigenvector Pairs

For any n x n square matrix A, an eigenvector \( \boldsymbol{v} \) is a non-zero vector that, when multiplied by A, returns a scalar multiple of itself. This scalar multiple is the corresponding eigenvalue λ.

**Eigenvalue-Eigenvector Identity** — The core identity that defines the relationship between matrix A, its eigenvalue λ, and non-zero eigenvector v

*Notation:* A\boldsymbol{v} = \lambda \boldsymbol{v}

*Example:* For matrix \( A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \), \( \lambda = 2 \) and \( \boldsymbol{v} = \begin{pmatrix}1 \\ 0\end{pmatrix} \) satisfy the identity

**Worked example:** Verify that \( \boldsymbol{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \) is an eigenvector of \( A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} \) and find its corresponding eigenvalue

1. Calculate the product A v directly
2. $$A\boldsymbol{v} = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ 6 \end{pmatrix}$$
3. Rewrite the resulting vector as a scalar multiple of the original v
4. $$\begin{pmatrix} 8 \\ 6 \end{pmatrix} = 4 \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$
5. Confirm the scalar multiple is the eigenvalue: λ = 4

**Check your understanding**

Test your understanding of the core identity:

1. Which of the following vectors is an eigenvector of \( \begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix} \)?

   - \( \begin{pmatrix} 1 \\ 1 \end{pmatrix} \)
   - \( \begin{pmatrix} 0 \\ 1 \end{pmatrix} \)
   - \( \begin{pmatrix} 2 \\ 2 \end{pmatrix} \)

   *Why:* Multiplying this vector by the matrix returns -2 times the original vector, satisfying the eigenvector identity

## Deriving the Characteristic Equation

Rearranging the core eigenvector identity allows us to derive a polynomial equation whose roots are all eigenvalues of matrix A. This derivation relies on the property that a non-zero vector v can only satisfy \( (A - \lambda I) \boldsymbol{v} = 0 \) if the matrix \( A - \lambda I \) is singular (determinant = 0).

**Derivation:** Derive the characteristic equation for square matrix A

*Starting from:* A\boldsymbol{v} = \lambda \boldsymbol{v}

1. Subtract \( \lambda \boldsymbol{v} \) from both sides: \( A\boldsymbol{v} - \lambda \boldsymbol{v} = 0 \)
2. Factor out the identity matrix I to group terms: \( (A - \lambda I) \boldsymbol{v} = 0 \)
3. For non-zero v to exist, the matrix \( A - \lambda I \) cannot be invertible, so its determinant equals 0
4. This gives the characteristic equation: \( \det(A - \lambda I) = 0 \)

*Conclusion:* The roots of this nth-degree polynomial are all eigenvalues of the n x n matrix A

**Worked example:** Find the characteristic equation for 2x2 matrix \( A = \begin{pmatrix} 5 & 2 \\ 2 & 5 \end{pmatrix} \)

1. Construct the matrix \( A - \lambda I \)
2. $$A - \lambda I = \begin{pmatrix} 5 - \lambda & 2 \\ 2 & 5 - \lambda \end{pmatrix}$$
3. Calculate the determinant of this matrix
4. $$\det(A - \lambda I) = (5 - \lambda)^2 - 4 = \lambda^2 -10 \lambda +21$$
5. Set determinant equal to 0 to get the characteristic equation
6. $$\lambda^2 - 10 \lambda + 21 = 0$$

## Calculating Eigenvectors from Eigenvalues

Once you have solved the characteristic equation for all eigenvalues, substitute each eigenvalue back into the system \( (A - \lambda I) \boldsymbol{v} = 0 \) to solve for the corresponding eigenvector. All non-zero scalar multiples of a valid eigenvector are also valid eigenvectors for that eigenvalue.

> **tip**
>
> IB exam markers accept any non-zero scalar multiple of the eigenvector you calculate, so you can simplify to integer values for full marks

**Worked example:** Find all eigenvectors for matrix \( A = \begin{pmatrix} 5 & 2 \\ 2 & 5 \end{pmatrix} \) whose characteristic equation is \( \lambda^2 - 10 \lambda + 21 = 0 \)

1. Solve the characteristic equation to get eigenvalues λ = 3 and λ =7
2. Substitute λ=3 into \( (A - \lambda I) \boldsymbol{v} = 0 \)
3. $$\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies 2v_1 + 2v_2 = 0 \implies v_1 = -v_2$$
4. Simplify to get eigenvector for λ=3: \( k \begin{pmatrix} 1 \\ -1 \end{pmatrix} \) where k ≠ 0
5. Substitute λ=7 into the system
6. $$\begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies -2v_1 + 2v_2 = 0 \implies v_1 = v_2$$
7. Simplify to get eigenvector for λ=7: \( m \begin{pmatrix} 1 \\ 1 \end{pmatrix} \) where m ≠ 0

## Exam-Focused Eigenvalue Properties

- The sum of all eigenvalues of matrix A equals the trace of A (sum of diagonal elements)
- The product of all eigenvalues of matrix A equals the determinant of A
- A matrix with a zero eigenvalue is singular (non-invertible)
- Eigenvalues of a diagonal matrix are exactly the elements on its main diagonal

**Exam command terms**

Common IB command terms for this topic:

- **Show that** — You must explicitly substitute the vector and scalar into the A v = λ v identity to prove the result *(Show that \( \lambda = 5 \) is an eigenvalue of matrix A)*

- **Hence find** — You must use the previously calculated eigenvalues to find eigenvectors, no alternative method is accepted

## Common pitfalls

- **Wrong:** Forgetting to subtract λI before calculating the determinant of A
  - Why it fails: Produces a non-constant polynomial that is not a valid characteristic equation
  - Correct: Explicitly write out the full A - λI matrix before expanding the determinant
- **Wrong:** Substituting the eigenvector back into the original matrix A instead of A - λI when solving
  - Why it fails: Leads to an over-constrained system that cannot be solved correctly
  - Correct: Always use the (A - λI) v = 0 system to find eigenvectors
- **Wrong:** Submitting a zero vector as an eigenvector
  - Why it fails: Eigenvectors are defined as non-zero by the IB syllabus, so zero vectors get zero marks
  - Correct: Explicitly state that the scalar multiple constant cannot equal zero
- **Wrong:** Miscalculating the sign of the characteristic polynomial when expanding the determinant
  - Why it fails: Leads to incorrect eigenvalues that do not satisfy the trace and product checks
  - Correct: Verify your eigenvalues by checking their sum equals the trace of A
- **Wrong:** Using decimal approximations for eigenvalues instead of exact radical forms
  - Why it fails: IB exam markers penalize approximate answers for exact eigenvalue calculation questions
  - Correct: Leave eigenvalues in exact surd form unless explicitly told to round

## Cheatsheet

| Task | 2x2 Matrix Workflow | 3x3 Matrix Workflow | Quick Check |
| --- | --- | --- | --- |
| Find characteristic equation | Expand det(A - λI) to get quadratic polynomial | Expand det(A - λI) to get cubic polynomial | Sum of roots equals trace of A |
| Solve for eigenvalues | Factor or use quadratic formula | Use rational root theorem then factor quadratic | Product of roots equals det(A) |
| Find eigenvectors | Solve 1 linear equation for ratio of components | Solve homogeneous 3-variable system | Confirm A v = λ v for your pair |
| Simplify eigenvector | Reduce to smallest integer components | Reduce to smallest integer components | Any non-zero multiple is valid |

## What's next

Mastering eigenvalues and eigenvectors is the critical foundation for advanced AI HL topics that appear frequently in Paper 3 problem sets. You will next apply these skills to diagonalize matrices, which simplifies raising matrices to large integer powers for iterative model calculations. These tools are also core to analyzing Markov chain steady states, a heavily weighted applied topic for population dynamics and predictive modelling assessments. You will also revisit eigenvectors when studying geometric transformations of 2D and 3D shapes, where eigenvectors define lines of invariant direction under transformation.

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