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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

Tout était clair ?

Comment pouvons-nous l'améliorer ?

Merci pour vos commentaires !

Section 4. Chapitre 11

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bookIntroduction to Eigenvectors & Eigenvalues

Glissez pour afficher le menu

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

Tout était clair ?

Comment pouvons-nous l'améliorer ?

Merci pour vos commentaires !

Section 4. Chapitre 11
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