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IB Mathematics AI AHL Eigenvalues and eigenvectors Study Notes - New Syllabus

IB Mathematics AI AHL Eigenvalues and eigenvectors Study Notes

LEARNING OBJECTIVE

  • Eigenvalues and eigenvectors.

Key Concepts: 

  • Characteristic polynomial
  • Diagonalization
  • Applications 

MAI HL and SL Notes – All topics

EIGENVALUES AND EIGENVECTORS

Let \( A \) be a square \( n \times n \) matrix. The value \( \lambda \in \mathbb{R} \) is said to be an eigenvalue of \( A \) if there exists a non-zero vector \( u \) such that:

\( Au = \lambda u \)

Such a vector is called an eigenvector (corresponding to the eigenvalue \( \lambda \)).

Eigenvalues and eigenvectors reveal fundamental properties of linear transformations represented by matrices, offering deep insights into their structure and behavior.

Let

$A = \begin{pmatrix} 1 & 3 \\ 4 & 2 \end{pmatrix}$

Observe what happens when we multiply $A$ by the vector

$u = \begin{pmatrix} 3 \\ 4 \end{pmatrix}:$

$
Au = \begin{pmatrix} 1 & 3 \\ 4 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 15 \\ 20 \end{pmatrix} = 5 \begin{pmatrix} 3 \\ 4 \end{pmatrix} = 5u
$

We obtain a multiple of $u$. In such a case, we say that 5 is an eigenvalue of the matrix $A$, and $u$ is a corresponding eigenvector.

Our task here is to find all such characteristic vectors.

CHARACTERSTICS POLYNOMIAL

♦Characteristic Polynomial

\( \lambda \text{ is an eigenvalue of } A \iff Au = \lambda u \text{ for some } u \neq 0 \)
\( \iff Au – \lambda u = 0 \text{ for some } u \neq 0 \)
\( \iff (A – \lambda I)u = 0 \text{ for some } u \neq 0 \)
\( \iff \text{the system } (A – \lambda I)X = 0 \text{ has a nonzero solution} \)
\( \iff A – \lambda I \text{ has no inverse} \)
\( \iff \det(A – \lambda I) = 0 \)

For a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), the expression \( \det(A – \lambda I) \) is:

\( \begin{vmatrix} a – \lambda & b \\ c & d – \lambda \end{vmatrix} = (a – \lambda)(d – \lambda) – bc = \lambda^2 – (a + d)\lambda + ad – bc \)

This is known as the characteristic polynomial of matrix \( A \). Notice that the constant term is \( \det A \).

In practice:

We find the eigenvalues by solving $\det(A – \lambda I) = 0$
For each eigenvalue $\lambda$, we solve the corresponding system $(A – \lambda I)X = 0$ to find the eigenvectors

The following example will clarify the process:

Example

$A = \begin{pmatrix} 1 & 3 \\ 4 & 2 \end{pmatrix}$

Find

Eigenvalues:

Eigenvectors:

▶️Answer/Explanation

Solution:

\(\det(A – \lambda I) = \begin{vmatrix} 1-\lambda & 3 \\ 4 & 2-\lambda \end{vmatrix} = \lambda^2 – 3\lambda – 10\)

There are two eigenvalues: $\lambda = 5$, $\lambda = -2$

• For $\lambda = 5$, solve $(A – 5I)X = 0$:
\(\begin{pmatrix} -4 & 3 \\ 4 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

Using first equation: $-4x + 3y = 0 \Rightarrow \frac{x}{y} = \frac{3}{4}$
Eigenvectors: $\begin{pmatrix} 3 \\ 4 \end{pmatrix} t$ for $t \neq 0$

• For $\lambda = -2$, solve $(A + 2I)X = 0$:
\(\begin{pmatrix} 3 & 3 \\ 4 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

Using first equation: $3x + 3y = 0 \Rightarrow x = -y$
Eigenvectors: $\begin{pmatrix} -1 \\ 1 \end{pmatrix} t$ for $t \neq 0$

NOTE:

The characteristic polynomial for $2 \times 2$ matrices is quadratic with either:

Two distinct real eigenvalues (our focus)
One repeated real eigenvalue
No real eigenvalues

One eigenvalue:

Find

\(A = \begin{pmatrix} 3 & -1 \\ 1 & 5 \end{pmatrix}\)

Characteristic polynomial:

Eigenvector:

▶️Answer/Explanation

Solution:

Characteristic polynomial: $(\lambda-4)^2 \Rightarrow \lambda = 4$ (repeated)
Eigenvector: $\begin{pmatrix} -1 \\ 1 \end{pmatrix} t$

No real eigenvalues:

Find

\(A = \begin{pmatrix} -1 & -1 \\ 2 & 1 \end{pmatrix}\)

Characteristic polynomial:

▶️Answer/Explanation

Solution:

Characteristic polynomial: $\lambda^2 + 1 = 0 \Rightarrow$ no real solutions

DIAGONALIZATION OF 2x2 MATRICES

Process of Diagonalization:

  1. Determine the eigenvalues of the matrix.
  2. Find the eigenvectors corresponding to each eigenvalue.
  3. Construct a matrix whose columns are the eigenvectors.
  4. Form a diagonal matrix with the eigenvalues placed along the diagonal.

For matrices with two distinct eigenvalues, we can write:

\(A = PDP^{-1} \quad \text{where } D \text{ is diagonal}\)

Important Note: 

 $D$ is the diagonal matrix of eigenvalues.
$P$ is the matrix whose columns are the corresponding eigenvectors.

  • It’s crucial that the order matches:
  • The eigenvalue in position $d_{ii}$ of $D$ must correspond to the eigenvector in the $i$-th column of $P$.
  • If you shuffle eigenvalues without shuffling the eigenvectors in the same way, the diagonalization will be wrong.

Example

Using the concept of diagonalization

 $A = \begin{pmatrix} 1 & 3 \\ 4 & 2 \end{pmatrix}$

Find $A^n$

▶️Answer/Explanation

Solution:

Let $P = \begin{pmatrix} 3 & -1 \\ 4 & 1 \end{pmatrix}$ (columns are eigenvectors)
$D = \begin{pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix}$ (diagonal entries are eigenvalues)

Then $A = PDP^{-1}$, which can be verified by:
\(P^{-1}AP = D = \begin{pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix}\)

Alternative ordering:
\(P = \begin{pmatrix} -1 & 3 \\ 1 & 4 \end{pmatrix} \Rightarrow P^{-1}AP = \begin{pmatrix} -2 & 0 \\ 0 & 5 \end{pmatrix}\)

This diagonalization is useful for computing powers:
\(A^n = PD^nP^{-1} = P\begin{pmatrix} 5^n & 0 \\ 0 & (-2)^n \end{pmatrix}P^{-1}\)

Example

Compute $A^n$ 

$A = \begin{pmatrix} 1 & 3 \\ 4 & 2 \end{pmatrix}$

Using $A = PDP^{-1}$:

▶️Answer/Explanation

Solution:

\(A^n = \begin{pmatrix} 3 & -1 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 5^n & 0 \\ 0 & (-2)^n \end{pmatrix} \begin{pmatrix} \frac{1}{7} & \frac{1}{7} \\ -\frac{4}{7} & \frac{3}{7} \end{pmatrix}\)
\(= \frac{1}{7} \begin{pmatrix} 3 \cdot 5^n & -(-2)^n \\ 4 \cdot 5^n & (-2)^n \end{pmatrix} \begin{pmatrix} 1 & 1 \\ -4 & 3 \end{pmatrix}\)
\(= \frac{1}{7} \begin{pmatrix} 3 \cdot 5^n + 4 \cdot (-2)^n & 3 \cdot 5^n – 3 \cdot (-2)^n \\ 4 \cdot 5^n – 4 \cdot (-2)^n & 4 \cdot 5^n + 3 \cdot (-2)^n \end{pmatrix}\)

For example:
\(A^4 = \begin{pmatrix} 277 & 261 \\ 348 & 364 \end{pmatrix}\)

APPLICATIONS TO POWERS OF 2 × 2 MATRCIES

Diagonalization is a powerful tool for computing high powers of square matrices efficiently.

If a matrix $A$ is diagonalizable, we can write:

$A^n = P D^n P^{-1}$

where:

$P$ is the matrix of eigenvectors,
$D$ is a diagonal matrix of eigenvalues,
$n$ is a positive integer.

This approach is especially helpful because $D^n$ is easy to compute-just raise each diagonal entry (eigenvalue) to the power $n$.

Example

Find $A^{6}$ for $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$

Give final answer in a $2 \times 2$ matrix.

▶️Answer/Explanation

Solution:

We first diagonalize $A$.

Eigenvalues: $\lambda_1 = 2$, $\lambda_2 = 3$
Eigenvectors:
$v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

So,
$P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$

Then,
$A^6 = P D^6 P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2^6 & 0 \\ 0 & 3^6 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$

$\therefore A^6 = \begin{pmatrix} 64 & 665 \\ 0 & 729 \end{pmatrix}$

REAL-WORLD APPLICATIONS OF EIGENVALUES AND EIGENVECTORS

Population Movement 

Eigenvalues and eigenvectors play a critical role in modeling population distribution over time, such as how people or animals move between regions.

Context: Suppose a population moves between two cities or regions over time.
Model: The movement is represented by a transition matrix $A$, where each column represents how individuals in one region are distributed to both regions in the next time step.
Goal: Find the steady-state distribution – a population vector $u$ such that $Au = u$. This is equivalent to finding an eigenvector corresponding to eigenvalue $\lambda = 1$.

In steady state, the population distribution remains unchanged with time.

Predator-Prey Models

Eigenvalues and eigenvectors help describe the stability and dynamics of ecosystems, particularly in predator-prey interactions.

Context: Consider populations of rabbits (prey) and foxes (predators).
Model: Linearize the system near an equilibrium and write the model as a system of linear differential equations, represented in matrix form as $\frac{d\mathbf{x}}{dt} = A\mathbf{x}$.

Interpretation:

If eigenvalues are real and negative, the population returns to equilibrium (stable).
If eigenvalues are complex with positive real parts, populations oscillate and diverge (unstable).
Eigenvectors describe directions of motion in phase space.

Example

Every year, people move between City A and City B according to this rule:

  • $80\%$ of people in City A stay, $20\%$ move to City B.
  • $30\%$ of people in City B move to City A, $70\%$ stay.
▶️Answer/Explanation

Solution:

Transition matrix:

$ A = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix} $

Let $\vec{x}_n = \begin{pmatrix} a_n \\ b_n \end{pmatrix}$ represent the population in year $n$, where $a_n$ and $b_n$ are proportions in Cities A and B, respectively.

$ \vec{x}_{n+1} = A \vec{x}_n $

Steady-state vector $\vec{x}$

$ A\vec{x} = \vec{x} \quad \Rightarrow \quad (A – I)\vec{x} = \vec{0} $

$(A – I)\vec{x} = 0$

$ A – I = \begin{pmatrix} -0.2 & 0.3 \\ 0.2 & -0.3 \end{pmatrix} $

$(A – I)\vec{x} = 0$:

$ -0.2a + 0.3b = 0 \Rightarrow a = \frac{3}{2}b $

One eigenvector is:

$ \vec{x} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \Rightarrow \text{Steady-state ratio is } 3:2 \text{ (City A : City B)} $

$ \vec{x}_{\text{steady}} = \frac{1}{5} \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 0.6 \\ 0.4 \end{pmatrix} $

Over time, regardless of the starting distribution, 60% of the population will be in City A, and 40% in City B. This steady-state distribution corresponds to the eigenvector of $A$ associated with eigenvalue $\lambda = 1$.

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