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

IB Mathematics AI SL Eigenvalues and eigenvectors Study Notes

LEARNING OBJECTIVE

  • Eigenvalues and eigenvectors.

Key Concepts: 

  • –

MAI HL and SL Notes – All topics

♦EIGENVALUES AND EIGENVECTORS

Consider the matrix:

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

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.

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 \( \lambda \)).

♦FINDING EIGENVALUES AND EIGENVECTORS

Notice that:

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

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

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

▶️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 $t \neq 0$

NOTICE
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:

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

▶️Answer/Explanation

Solution:

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

No real eigenvalues:

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

▶️Answer/Explanation

Solution:

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

♦ DIAGONALIZATION

For matrices with two distinct eigenvalues, we can write:

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

Example

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

▶️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 $A^n$ for $A = \begin{pmatrix} 1 & 3 \\ 4 & 2 \end{pmatrix}$

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

♦ CAYLEY-HAMILTON THEOREM

Every matrix satisfies its characteristic equation.

Example

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

▶️Answer/Explanation

Solution:

Characteristic equation: $\lambda^2 – 3\lambda – 10 = 0$
Then $A^2 – 3A – 10I = 0$

Verification:
\(A^2 = \begin{pmatrix} 13 & 9 \\ 12 & 16 \end{pmatrix}\)
\(A^2 – 3A – 10I = \begin{pmatrix} 13 & 9 \\ 12 & 16 \end{pmatrix} – \begin{pmatrix} 3 & 9 \\ 12 & 6 \end{pmatrix} – \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)

♦Applications:
1. Compute $A^{-1}$:
\(A^2 – 3A = 10I \Rightarrow A(A – 3I) = 10I \Rightarrow A^{-1} = \frac{A – 3I}{10}\)

2. Compute higher powers recursively:
\(A^2 = 3A + 10I\)
\(A^3 = 3A^2 + 10A = 3(3A + 10I) + 10A = 19A + 30I\)
\(A^4 = 19A^2 + 30A = 19(3A + 10I) + 30A = 87A + 190I\)

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