Eigenvalues and Eigenvectors Revision Notes for AQA A-Level Further Maths

Eigenvalues and Eigenvectors

Introduction

When we apply a matrix transformation to a vector, the vector usually changes both its direction and magnitude. However, some special vectors only change in magnitude while maintaining their original direction. These vectors are called eigenvectors, and the scaling factor is called the eigenvalue. Understanding eigenvalues and eigenvectors is essential for analysing transformations and solving problems involving powers of matrices.

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Geometric Interpretation:

Eigenvectors represent special directions in space that remain unchanged under a transformation. While other vectors may rotate or change direction, eigenvectors only stretch or compress along their original line.

What are eigenvectors and eigenvalues?

An eigenvector is a non-zero vector that maintains its direction when a transformation is applied to it. The vector may be stretched or compressed, but it continues to point along the same line.

When we apply a matrix transformation A to an eigenvector x, the result is simply a scalar multiple of the original vector. This relationship is expressed as:

$\mathbf{Ax} = \lambda\mathbf{x}$

where:

A is a square matrix
x is the eigenvector (a non-zero vector)
λ (lambda) is the eigenvalue (the scaling factor)

The eigenvalue tells us how much the eigenvector is scaled. If λ > 1, the vector is stretched; if 0 < λ < 1, it is compressed; if λ < 0, it is reversed in direction as well.

chatImportant

Eigenvectors are never the zero vector. If we allowed x = 0, then every λ would satisfy the equation, making the concept meaningless.

Finding eigenvalues: the characteristic equation

To find the eigenvalues of a matrix A, we need to rearrange the eigenvector equation. Starting with:

$\mathbf{Ax} = \lambda\mathbf{x}$

We can write λx as λIx (where I is the identity matrix with the same dimensions as A):

$\mathbf{Ax} = \lambda\mathbf{Ix}$

Rearranging gives:

$\mathbf{Ax} - \lambda\mathbf{Ix} = \mathbf{0}$

Factorising (since matrix multiplication is distributive):

$(\mathbf{A} - \lambda\mathbf{I})\mathbf{x} = \mathbf{0}$

Since x is a non-zero vector (by definition of eigenvectors), the only way this equation can be satisfied is if the matrix (A - λI) has no inverse. In other words, (A - λI) must be singular.

A matrix is singular when its determinant equals zero, so:

$\det(\mathbf{A} - \lambda\mathbf{I}) = 0$

This equation is called the characteristic equation of the matrix A. Solving it gives us the eigenvalues.

chatImportant

Remember the Identity Matrix:

You must write λ as λI (not just λ) when substituting into matrix expressions, because I is an identity matrix with the same dimensions as A. You cannot subtract a scalar from a matrix directly.

Finding eigenvectors

Once you have found the eigenvalues, you can find the corresponding eigenvectors by substituting each eigenvalue back into the equation:

$\mathbf{Ax} = \lambda\mathbf{x}$

This will give you a system of simultaneous equations. Solving these equations will give you the eigenvector(s) corresponding to that eigenvalue.

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Important properties of eigenvectors:

Any non-zero scalar multiple of an eigenvector is also an eigenvector (since the direction remains the same)
This means there are infinitely many eigenvectors for each eigenvalue
We typically express eigenvectors in their simplest form

Worked example 1: finding eigenvalues and eigenvectors of a 2×2 matrix

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Worked Example: Finding Eigenvalues and Eigenvectors

Let's find the eigenvalues and corresponding eigenvectors of the matrix:

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

Step 1: Find the eigenvalues

We need to solve det(A - λI) = 0.

First, calculate A - λI:

$\mathbf{A} - \lambda\mathbf{I} = \begin{pmatrix} 3 & -1 \\ 4 & -2 \end{pmatrix} - \lambda\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3-\lambda & -1 \\ 4 & -2-\lambda \end{pmatrix}$

Now find the determinant:

$\det(\mathbf{A} - \lambda\mathbf{I}) = (3-\lambda)(-2-\lambda) - (-1)(4)$

$= -6 - 3\lambda + 2\lambda + \lambda^2 + 4$

$= \lambda^2 - \lambda - 2$

Setting this equal to zero:

$\lambda^2 - \lambda - 2 = 0$

Factorising:

$(\lambda - 2)(\lambda + 1) = 0$

Therefore, the eigenvalues are λ = 2 and λ = -1.

Step 2: Find the eigenvectors

For λ = 2, we use Ax = 2x with x = (a, b)ᵀ:

$\begin{pmatrix} 3 & -1 \\ 4 & -2 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = 2\begin{pmatrix} a \\ b \end{pmatrix}$

This gives us:

3a - b = 2a ⟹ a = b
4a - 2b = 2b ⟹ 4a = 4b ⟹ a = b

Both equations give a = b, so a possible eigenvector is:

$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$

For λ = -1, we use Ax = -1x:

$\begin{pmatrix} 3 & -1 \\ 4 & -2 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = -1\begin{pmatrix} a \\ b \end{pmatrix}$

This gives us:

3a - b = -a ⟹ 4a = b
4a - 2b = -b ⟹ 4a = b

So a possible eigenvector is:

$\begin{pmatrix} 1 \\ 4 \end{pmatrix}$

Remember: any non-zero multiple of these vectors is also an eigenvector for the respective eigenvalue.

Worked example 2: finding eigenvalues and eigenvectors of a 3×3 matrix

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Worked Example: 3×3 Matrix Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of:

$\mathbf{T} = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix}$

Step 1: Find the eigenvalues

Calculate det(T - λI) = 0:

$\det\begin{pmatrix} 2-\lambda & 2 & 1 \\ 1 & 1-\lambda & 2 \\ 0 & 0 & 2-\lambda \end{pmatrix} = 0$

Exam Tip:

Use the simplest method to calculate the determinant. In this case, expand using the third row since it has two zeros, which simplifies the calculation significantly.

Expanding along the third row:

$= (2-\lambda)[(2-\lambda)(1-\lambda) - 2]$

$= (2-\lambda)[(2-2\lambda-\lambda+\lambda^2) - 2]$

$= (2-\lambda)(\lambda^2 - 3\lambda)$

$= \lambda(2-\lambda)(\lambda-3)$

Therefore, the eigenvalues are λ = 0, 2, 3.

Step 2: Find the eigenvectors

For λ = 0, use Tx = 0x = 0 with x = (a, b, c)ᵀ:

$\begin{pmatrix} 2 & 2 & 1 \\ 1 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

From the equations:

2c = 0 ⟹ c = 0
2a + 2b + c = 0 with c = 0 ⟹ a + b = 0 ⟹ a = -b

A possible eigenvector is:

$\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$

For λ = 2, use Tx = 2x:

$\begin{pmatrix} 2 & 2 & 1 \\ 1 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix} = 2\begin{pmatrix} a \\ b \\ c \end{pmatrix}$

From the equations:

2c = 2c ⟹ c can be any value
2a + 2b + c = 2a ⟹ 2b + c = 0 ⟹ c = -2b
a + b + 2c = 2b ⟹ a + 2c = b

Substituting c = -2b into the third equation:

a + 2(-2b) = b ⟹ a - 4b = b ⟹ a = 5b

A possible eigenvector is:

$\begin{pmatrix} 5 \\ 1 \\ -2 \end{pmatrix}$

For λ = 3, use Tx = 3x:

Following similar working:

c = 0
2a + 2b = 3a ⟹ a = 2b

A possible eigenvector is:

$\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$

The three eigenvectors are $\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ , $\begin{pmatrix} 5 \\ 1 \\ -2 \end{pmatrix}$ , and $\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ . :::

The Cayley-Hamilton theorem

An important property of matrices and their eigenvalues is stated by the Cayley-Hamilton Theorem:

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The Cayley-Hamilton Theorem:

Every matrix satisfies its own characteristic equation.

This means if you find the characteristic equation of a matrix A (which normally contains λ), you can replace λ with the matrix A itself, and the equation will equal the zero matrix.

For a 2×2 matrix $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the characteristic equation is:

$\lambda^2 - (a+d)\lambda + (ad-bc) = 0$

The Cayley-Hamilton Theorem states that:

$\mathbf{A}^2 - (a+d)\mathbf{A} + (ad-bc)\mathbf{I} = \mathbf{0}$

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Important Detail:

The constant term (ad - bc) must be multiplied by the identity matrix I because you cannot add a scalar to a matrix directly.

Worked example 3: verifying the Cayley-Hamilton theorem

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Worked Example: Verifying Cayley-Hamilton

Verify that the matrix $\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$ satisfies its characteristic equation.

Step 1: Find the characteristic equation

$\det\begin{pmatrix} 1-\lambda & 0 \\ 2 & 3-\lambda \end{pmatrix} = (1-\lambda)(3-\lambda) - 0 = 0$

$\lambda^2 - 4\lambda + 3 = 0$

Step 2: Replace λ with the matrix A

We need to show that A² - 4A + 3I = 0.

Calculate A²:

$\mathbf{A}^2 = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}^2 = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 8 & 9 \end{pmatrix}$

Calculate 4A:

$4\mathbf{A} = 4\begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ 8 & 12 \end{pmatrix}$

Calculate 3I (where I is the 2×2 identity matrix):

$3\mathbf{I} = 3\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$

Now substitute:

$\mathbf{A}^2 - 4\mathbf{A} + 3\mathbf{I} = \begin{pmatrix} 1 & 0 \\ 8 & 9 \end{pmatrix} - \begin{pmatrix} 4 & 0 \\ 8 & 12 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$

$= \begin{pmatrix} 1-4+3 & 0-0+0 \\ 8-8+0 & 9-12+3 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

This confirms that A satisfies its characteristic equation, as required.

Diagonalisation of matrices

A diagonal matrix is a square matrix where all elements are zero except those on the leading diagonal (from top-left to bottom-right). For example:

$\mathbf{D} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{pmatrix}$

Converting a matrix into diagonal form is useful, especially when calculating powers of matrices. This process is called diagonalisation.

A matrix M can be diagonalised by finding matrices P and D such that:

$\mathbf{M} = \mathbf{PDP}^{-1}$

where:

D is a diagonal matrix containing the eigenvalues of M on its leading diagonal
P is a matrix whose columns are the corresponding eigenvectors of M

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Order Matters:

The eigenvectors in the columns of P must appear in the same order as their corresponding eigenvalues in D. If you put the eigenvectors in a different order in P, you must rearrange the eigenvalues in D accordingly.

Worked example 4: diagonalising a 3×3 matrix

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Worked Example: Diagonalisation

Diagonalise the matrix:

$\mathbf{M} = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & -2 & 4 \end{pmatrix}$

by finding matrices P and D such that M = PDP⁻¹.

Step 1: Find the eigenvalues

Calculate det(M - λI) = 0:

$\det\begin{pmatrix} 2-\lambda & 0 & 0 \\ 0 & 1-\lambda & -2 \\ 0 & -2 & 4-\lambda \end{pmatrix} = (2-\lambda)[(1-\lambda)(4-\lambda) - 4] = 0$

$= (2-\lambda)(\lambda^2 - 5\lambda + 4 - 4)$

$= (2-\lambda)(\lambda^2 - 5\lambda)$

$= \lambda(2-\lambda)(\lambda - 5) = 0$

The eigenvalues are λ = 0, 2, 5.

Step 2: Find the eigenvectors

For λ = 0, use Mx = 0x = 0:

$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & -2 & 4 \end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

From the equations:

2a = 0 ⟹ a = 0
b - 2c = 0 ⟹ b = 2c
-2b + 4c = 0 ⟹ b = 2c

A possible eigenvector is:

$\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$

For λ = 2:

2a = 2a (satisfied for any a)
b - 2c = 2b ⟹ -2c = b ⟹ b = -2c
-2b + 4c = 2c ⟹ -2b = -2c ⟹ b = c

From these: b = c = 0, so a possible eigenvector is:

$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

For λ = 5:

2a = 5a ⟹ a = 0
b - 2c = 5b ⟹ -2c = 4b ⟹ c = -2b
-2b + 4c = 5c ⟹ -2b = c ⟹ c = -2b

A possible eigenvector is:

$\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}$

Step 3: Construct D and P

$\mathbf{D} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix}$

$\mathbf{P} = \begin{pmatrix} 0 & 1 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & -2 \end{pmatrix}$

Verification Tip:

Ensure the columns of P are in the correct order matching the eigenvalues in D. You can use your calculator to verify that M = PDP⁻¹.

:::

Calculating powers of matrices using diagonalisation

One of the most useful applications of diagonalisation is calculating powers of matrices. Computing Aⁿ directly by repeated multiplication becomes very laborious, but using diagonalisation makes it much simpler.

If A = PDP⁻¹, then:

$\mathbf{A}^n = (\mathbf{PDP}^{-1})^n$

Since matrix multiplication is associative:

$= \mathbf{PDP}^{-1} \cdot \mathbf{PDP}^{-1} \cdot \mathbf{PDP}^{-1} \cdots \mathbf{PDP}^{-1}$

Notice that P⁻¹P = I (the identity matrix), so many terms cancel:

$= \mathbf{PD}(\mathbf{P}^{-1}\mathbf{P})\mathbf{D}(\mathbf{P}^{-1}\mathbf{P})\mathbf{D}\cdots\mathbf{DP}^{-1}$

$= \mathbf{PD} \cdot \mathbf{I} \cdot \mathbf{D} \cdot \mathbf{I} \cdot \mathbf{D} \cdots \mathbf{DP}^{-1}$

$= \mathbf{PD}^n\mathbf{P}^{-1}$

Therefore:

$\text{If } \mathbf{A} = \mathbf{PDP}^{-1} \text{ then } \mathbf{A}^n = \mathbf{PD}^n\mathbf{P}^{-1}$

For a diagonal matrix D, calculating Dⁿ is straightforward because we simply raise each diagonal element to the power n:

$\text{If } \mathbf{D} = \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix} \text{ then } \mathbf{D}^n = \begin{pmatrix} a^n & 0 & 0 \\ 0 & b^n & 0 \\ 0 & 0 & c^n \end{pmatrix}$

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Strategy for calculating powers of a square matrix A:

Diagonalise the matrix using A = PDP⁻¹
Use Aⁿ = PDⁿP⁻¹ to find powers of A
Calculate Dⁿ by raising each diagonal element to the power n
Multiply the matrices P, Dⁿ, and P⁻¹ together

Worked example 5: using diagonalisation to find powers

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Worked Example: Matrix Powers Using Diagonalisation

Given that $\mathbf{A} = \begin{pmatrix} 3 & -1 \\ 0 & 5 \end{pmatrix}$ , show that:

$\mathbf{A}^n = \begin{pmatrix} 3^n & \frac{3^n - 5^n}{2} \\ 0 & 5^n \end{pmatrix}$

Step 1: Find the eigenvalues

$\det\begin{pmatrix} 3-\lambda & -1 \\ 0 & 5-\lambda \end{pmatrix} = (3-\lambda)(5-\lambda) = 0$

The eigenvalues are λ = 3 and λ = 5.

Step 2: Find the eigenvectors

For λ = 3:

$\begin{pmatrix} 3 & -1 \\ 0 & 5 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = 3\begin{pmatrix} a \\ b \end{pmatrix}$

From the equations:

3a - b = 3a ⟹ b = 0
5b = 3b ⟹ b = 0

A possible eigenvector is:

$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$

For λ = 5:

$\begin{pmatrix} 3 & -1 \\ 0 & 5 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} = 5\begin{pmatrix} a \\ b \end{pmatrix}$

From the equations:

3a - b = 5a ⟹ -b = 2a ⟹ b = -2a

A possible eigenvector is:

$\begin{pmatrix} 1 \\ -2 \end{pmatrix}$

Step 3: Form P, D, and find P⁻¹

$\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 0 & -2 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 5 \end{pmatrix}$

To find P⁻¹:

$\mathbf{P}^{-1} = \frac{1}{(1)(-2) - (1)(0)}\begin{pmatrix} -2 & -1 \\ 0 & 1 \end{pmatrix} = \frac{1}{-2}\begin{pmatrix} -2 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{2} \\ 0 & -\frac{1}{2} \end{pmatrix}$

Step 4: Calculate Aⁿ = PDⁿP⁻¹

$\mathbf{D}^n = \begin{pmatrix} 3^n & 0 \\ 0 & 5^n \end{pmatrix}$

$\mathbf{A}^n = \begin{pmatrix} 1 & 1 \\ 0 & -2 \end{pmatrix}\begin{pmatrix} 3^n & 0 \\ 0 & 5^n \end{pmatrix}\begin{pmatrix} 1 & \frac{1}{2} \\ 0 & -\frac{1}{2} \end{pmatrix}$

First multiply Dⁿ and P⁻¹:

$\begin{pmatrix} 3^n & 0 \\ 0 & 5^n \end{pmatrix}\begin{pmatrix} 1 & \frac{1}{2} \\ 0 & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} 3^n & \frac{3^n}{2} \\ 0 & -\frac{5^n}{2} \end{pmatrix}$

Now multiply by P:

$\begin{pmatrix} 1 & 1 \\ 0 & -2 \end{pmatrix}\begin{pmatrix} 3^n & \frac{3^n}{2} \\ 0 & -\frac{5^n}{2} \end{pmatrix} = \begin{pmatrix} 3^n & \frac{3^n}{2} - \frac{5^n}{2} \\ 0 & 5^n \end{pmatrix}$

$= \begin{pmatrix} 3^n & \frac{3^n - 5^n}{2} \\ 0 & 5^n \end{pmatrix}$

as required.

Invariant lines and planes

Eigenvalues and eigenvectors have important geometric interpretations when considering transformations.

Invariant lines

The eigenvectors of a transformation matrix T determine the direction of invariant lines through the origin. An invariant line is a line whose direction is preserved by the transformation (though points on the line may move along it).

If a transformation represented by matrix A has an eigenvalue λ = 1, then for all corresponding eigenvectors v:

$\mathbf{Av} = 1 \cdot \mathbf{v} = \mathbf{v}$

This means these vectors remain unchanged by the transformation.

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Line of Invariant Points:

If a transformation given by matrix T has an eigenvalue of 1, then the corresponding eigenvectors determine the direction of a line of invariant points through the origin.

These are fixed points - they don't move at all under the transformation.

Invariant planes

When a matrix has a repeated eigenvalue, the situation is different. You can find two non-parallel eigenvectors corresponding to this repeated eigenvalue. Any linear combination of these two eigenvectors is also an eigenvector with the same eigenvalue.

This means the eigenvalue has an entire plane of eigenvectors.

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Invariant Planes:

If a transformation given by matrix T has a repeated eigenvalue, then T has an invariant plane defined by linear combinations of two corresponding non-parallel eigenvectors.

Worked example 6: invariant lines and planes

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Worked Example: Geometric Significance of Eigenvectors

The matrix $\mathbf{A} = \begin{pmatrix} 4 & -3 & 3 \\ 6 & -5 & 3 \\ 0 & 0 & -2 \end{pmatrix}$ represents a transformation.

Given that the eigenvalues of A are 1, -2, and -2:

a) For each eigenvalue, find a full set of eigenvectors.

b) Describe the geometric significance of the eigenvectors of A in relation to the transformation that A represents.

Solution

Part a:

For λ = 1, use Av = v with v = (x, y, z)ᵀ:

$\begin{pmatrix} 4 & -3 & 3 \\ 6 & -5 & 3 \\ 0 & 0 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

From the equations:

-2z = z ⟹ z = 0
4x - 3y + 3z = x ⟹ 3x - 3y = 0 ⟹ x = y

Therefore, the set of eigenvectors corresponding to eigenvalue 1 is all vectors of the form:

$\alpha\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$

where α is any non-zero scalar.

For λ = -2, use Av = -2v:

$\begin{pmatrix} 4 & -3 & 3 \\ 6 & -5 & 3 \\ 0 & 0 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = -2\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

From the equations:

4x - 3y + 3z = -2x ⟹ 6x - 3y + 3z = 0 ⟹ 2x - y + z = 0
6x - 5y + 3z = -2y ⟹ 6x - 3y + 3z = 0 (same as first equation)
-2z = -2z (satisfied for any z)

The equation 2x - y + z = 0 represents a plane. We need to choose any two non-parallel vectors that lie on this plane:

$\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \text{ and } \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$

(You can verify: 2(0) - 1 + 1 = 0 ✓ and 2(1) - 2 + 0 = 0 ✓)

Therefore, the eigenvectors corresponding to eigenvalue -2 are linear combinations:

$\beta\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} + \gamma\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$

where β and γ are scalars (not both zero).

Part b:

The eigenvector $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ (corresponding to eigenvalue λ = 1) gives the direction of a line of invariant points through the origin. Points on this line remain fixed under the transformation.

The set of eigenvectors $\beta\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} + \gamma\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$ (corresponding to the repeated eigenvalue λ = -2) represents an invariant plane. This entire plane is preserved by the transformation, though points on it are scaled by -2 (reversed and doubled in magnitude).

Remember

bookmarkSummary

Key Points to Remember:

Eigenvectors are non-zero vectors that maintain their direction under a transformation, satisfying Ax = λx where λ is the eigenvalue.
To find eigenvalues, solve the characteristic equation det(A - λI) = 0. Always remember to include the identity matrix I when substituting λ.
Once eigenvalues are found, substitute each back into (A - λI)x = 0 to find the corresponding eigenvectors. Any non-zero scalar multiple is also an eigenvector.
Diagonalisation writes a matrix as M = PDP⁻¹ where D contains eigenvalues on the diagonal and P has eigenvectors as columns (in matching order).
To calculate Aⁿ efficiently, use Aⁿ = PDⁿP⁻¹ where Dⁿ simply raises each diagonal element to the power n.
An eigenvalue of λ = 1 indicates an invariant line (fixed points), while a repeated eigenvalue indicates an invariant plane.

Eigenvalues and Eigenvectors (AQA A-Level Further Maths): Revision Notes