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bookIntroduction to Eigenvectors & Eigenvalues

Eigenvalues and eigenvectors are foundational concepts in linear algebra, especially relevant in areas like machine learning, data compression, and system dynamics. They allow you to understand how a matrix stretches or rotates vectors.

What Are Eigenvectors and Eigenvalues?

An eigenvector is a non-zero vector that only gets scaled (not rotated) when a matrix is applied to it.
The scalar is called the eigenvalue.

Av⃗=λv⃗A\vec{v} = \lambda\vec{v}

Where:

  • AA is a square matrix;
  • λ\lambda is the eigenvalue (a scalar);
  • v⃗\vec{v} is the eigenvector (non-zero vector).

Example Matrix and Setup

Suppose:

A=[4123]A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}

We want to find values of λ\lambda and vectors v⃗\vec{v} such that:

Av⃗=λv⃗A \vec{v} = \lambda \vec{v}

Characteristic Equation

To find λ\lambda, solve the characteristic equation:

det⁡(A−λI)=0\det(A - \lambda I) = 0

Substitute:

det⁡[4−λ123−λ]=0\det \begin{bmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{bmatrix} = 0

Compute determinant:

(4−λ)(3−λ)−2=0(4-\lambda)(3-\lambda) - 2 = 0

Solve:

λ2−7λ+10=0λ=5,  λ=2\lambda^2 - 7\lambda + 10 = 0 \\ \lambda = 5, \; \lambda = 2

Find Eigenvectors

Now solve for each λ\lambda.

For λ=5\lambda = 5:

Subtract:

(A−5I)v⃗=0(A - 5I)\vec{v} = 0 [−112−2]v⃗=0\begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \vec{v} = 0

Solve:

v1=v2v_1 = v_2

So:

v⃗=[11]\vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

For λ=2\lambda = 2:

Subtract:

(A−2I)v⃗=0(A - 2I)\vec{v} = 0 [2121]v⃗=0\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \vec{v} = 0

Solve:

v1=−12v2v_1 = -\tfrac{1}{2} v_2

So:

v⃗=[−12]\vec{v} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}

Confirm the Eigenpair

Once you have an eigenvalue λ\lambda and an eigenvector v⃗\vec{v}, verify that:

Av⃗=λv⃗A \vec{v} = \lambda \vec{v}

Example:

A[11]=[55]=5[11]A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 1 \end{bmatrix}
Note
Note

Eigenvectors are not unique.
If v⃗\vec{v} is an eigenvector, then so is any scalar multiple cv⃗c \vec{v} for c≠0c \neq 0.

Example:

[22]\begin{bmatrix} 2 \\ 2 \end{bmatrix}

is also an eigenvector for λ=5\lambda = 5.

Diagonalization (Advanced)

If a matrix AA has nn linearly independent eigenvectors, then it can be diagonalized:

A=PDP−1A = PDP^{-1}

Where:

  • PP is the matrix of eigenvectors as columns;
  • DD is a diagonal matrix of eigenvalues;
  • P−1P^{-1} is the inverse of PP.

You can confirm diagonalization by checking A=PDP−1A = PDP^{-1}.
This is useful for computing powers of AA:

Example

Let:

A=[3102]A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}

Find eigenvalues:

det⁡(A−λI)=0\det(A - \lambda I) = 0

Solve:

λ=3,  λ=2\lambda = 3, \; \lambda = 2

Find eigenvectors:

For λ=3\lambda = 3:

v⃗=[10]\vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

For λ=2\lambda = 2:

v⃗=[−11]\vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}

Construct P,DP, D and P−1P^{-1}:

P=[1−101],D=[3002],P−1=[1101]P = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}, \quad P^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

Compute:

PDP−1=[3102]=APDP^{-1} = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} = A

Confirmed.

Why this matters:

To compute powers of AA, like AkA^k. Since DD is diagonal:

Ak=PDkP−1A^k = P D^k P^{-1}

This makes calculating matrix powers much faster.

Important Notes

  • Eigenvalues and eigenvectors are directions that remain unchanged under transformation;
  • λ\lambda stretches v⃗\vec{v};
  • λ=1\lambda = 1 keeps v⃗\vec{v} unchanged in magnitude.

  6. Which equation finds eigenvalues?
    a) Av⃗=v⃗A \vec{v} = \vec{v}
    b) Av⃗=0A \vec{v} = 0
    c) A=PDP−1A = PDP^{-1}
    d) det⁡(A−λI)=0\det(A - \lambda I) = 0 ✅

  7. In the result v⃗=[−12]\vec{v} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}, the eigenvector is:
    a) [1,1][1, 1]
    b) [2,2][2, 2]
    c) [0,1][0, 1]
    d) [−1,2][-1, 2] ✅

1. What does λ\lambda represent?

2. What is required for v⃗\vec{v} to be an eigenvector?

3. What is the characteristic equation used for?

4. If λ=3\lambda = 3 is an eigenvalue of AA, what do you subtract?

5. Which matrix has eigenvalues 2 and 3?

question mark

What does λ\lambda represent?

Select the correct answer

question mark

What is required for v⃗\vec{v} to be an eigenvector?

Select the correct answer

question mark

What is the characteristic equation used for?

Select the correct answer

question mark

If λ=3\lambda = 3 is an eigenvalue of AA, what do you subtract?

Select the correct answer

question mark

Which matrix has eigenvalues 2 and 3?

Select the correct answer

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Секція 4. Розділ 11

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bookIntroduction to Eigenvectors & Eigenvalues

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Eigenvalues and eigenvectors are foundational concepts in linear algebra, especially relevant in areas like machine learning, data compression, and system dynamics. They allow you to understand how a matrix stretches or rotates vectors.

What Are Eigenvectors and Eigenvalues?

An eigenvector is a non-zero vector that only gets scaled (not rotated) when a matrix is applied to it.
The scalar is called the eigenvalue.

Av⃗=λv⃗A\vec{v} = \lambda\vec{v}

Where:

  • AA is a square matrix;
  • λ\lambda is the eigenvalue (a scalar);
  • v⃗\vec{v} is the eigenvector (non-zero vector).

Example Matrix and Setup

Suppose:

A=[4123]A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}

We want to find values of λ\lambda and vectors v⃗\vec{v} such that:

Av⃗=λv⃗A \vec{v} = \lambda \vec{v}

Characteristic Equation

To find λ\lambda, solve the characteristic equation:

det⁡(A−λI)=0\det(A - \lambda I) = 0

Substitute:

det⁡[4−λ123−λ]=0\det \begin{bmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{bmatrix} = 0

Compute determinant:

(4−λ)(3−λ)−2=0(4-\lambda)(3-\lambda) - 2 = 0

Solve:

λ2−7λ+10=0λ=5,  λ=2\lambda^2 - 7\lambda + 10 = 0 \\ \lambda = 5, \; \lambda = 2

Find Eigenvectors

Now solve for each λ\lambda.

For λ=5\lambda = 5:

Subtract:

(A−5I)v⃗=0(A - 5I)\vec{v} = 0 [−112−2]v⃗=0\begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \vec{v} = 0

Solve:

v1=v2v_1 = v_2

So:

v⃗=[11]\vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

For λ=2\lambda = 2:

Subtract:

(A−2I)v⃗=0(A - 2I)\vec{v} = 0 [2121]v⃗=0\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \vec{v} = 0

Solve:

v1=−12v2v_1 = -\tfrac{1}{2} v_2

So:

v⃗=[−12]\vec{v} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}

Confirm the Eigenpair

Once you have an eigenvalue λ\lambda and an eigenvector v⃗\vec{v}, verify that:

Av⃗=λv⃗A \vec{v} = \lambda \vec{v}

Example:

A[11]=[55]=5[11]A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 1 \end{bmatrix}
Note
Note

Eigenvectors are not unique.
If v⃗\vec{v} is an eigenvector, then so is any scalar multiple cv⃗c \vec{v} for c≠0c \neq 0.

Example:

[22]\begin{bmatrix} 2 \\ 2 \end{bmatrix}

is also an eigenvector for λ=5\lambda = 5.

Diagonalization (Advanced)

If a matrix AA has nn linearly independent eigenvectors, then it can be diagonalized:

A=PDP−1A = PDP^{-1}

Where:

  • PP is the matrix of eigenvectors as columns;
  • DD is a diagonal matrix of eigenvalues;
  • P−1P^{-1} is the inverse of PP.

You can confirm diagonalization by checking A=PDP−1A = PDP^{-1}.
This is useful for computing powers of AA:

Example

Let:

A=[3102]A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}

Find eigenvalues:

det⁡(A−λI)=0\det(A - \lambda I) = 0

Solve:

λ=3,  λ=2\lambda = 3, \; \lambda = 2

Find eigenvectors:

For λ=3\lambda = 3:

v⃗=[10]\vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}

For λ=2\lambda = 2:

v⃗=[−11]\vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}

Construct P,DP, D and P−1P^{-1}:

P=[1−101],D=[3002],P−1=[1101]P = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}, \quad P^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

Compute:

PDP−1=[3102]=APDP^{-1} = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} = A

Confirmed.

Why this matters:

To compute powers of AA, like AkA^k. Since DD is diagonal:

Ak=PDkP−1A^k = P D^k P^{-1}

This makes calculating matrix powers much faster.

Important Notes

  • Eigenvalues and eigenvectors are directions that remain unchanged under transformation;
  • λ\lambda stretches v⃗\vec{v};
  • λ=1\lambda = 1 keeps v⃗\vec{v} unchanged in magnitude.

  6. Which equation finds eigenvalues?
    a) Av⃗=v⃗A \vec{v} = \vec{v}
    b) Av⃗=0A \vec{v} = 0
    c) A=PDP−1A = PDP^{-1}
    d) det⁡(A−λI)=0\det(A - \lambda I) = 0 ✅

  7. In the result v⃗=[−12]\vec{v} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}, the eigenvector is:
    a) [1,1][1, 1]
    b) [2,2][2, 2]
    c) [0,1][0, 1]
    d) [−1,2][-1, 2] ✅

1. What does λ\lambda represent?

2. What is required for v⃗\vec{v} to be an eigenvector?

3. What is the characteristic equation used for?

4. If λ=3\lambda = 3 is an eigenvalue of AA, what do you subtract?

5. Which matrix has eigenvalues 2 and 3?

question mark

What does λ\lambda represent?

Select the correct answer

question mark

What is required for v⃗\vec{v} to be an eigenvector?

Select the correct answer

question mark

What is the characteristic equation used for?

Select the correct answer

question mark

If λ=3\lambda = 3 is an eigenvalue of AA, what do you subtract?

Select the correct answer

question mark

Which matrix has eigenvalues 2 and 3?

Select the correct answer

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Як ми можемо покращити це?

Дякуємо за ваш відгук!

Секція 4. Розділ 11
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