Eigenvalues and Eigenvectors Explained

What are Eigenvalues and Eigenvectors?

In German, 'eigen' means 'characteristic' or 'own'. In linear algebra, eigenvalues and eigenvectors represent the characteristic directions and scaling factors of a square matrix transformation. When a matrix A multiplies a typical vector, the vector changes both its direction and length. However, there are special vectors that do not change direction when multiplied by A—they are simply scaled (stretched, shrunk, or flipped). These vectors are eigenvectors, and the scaling factors are the corresponding eigenvalues.

The relationship is represented by the eigenvector equation:

A × v = λ × v

where v is the eigenvector (must be non-zero) and λ (lambda) is the eigenvalue.

How to Calculate Eigenvalues

To solve the equation Av = λv, we rewrite it as:

Av - λv = 0 ⇒ (A - λI)v = 0

where I is the Identity matrix. For a non-zero vector v to satisfy this equation, the matrix (A - λI) must be singular (have no inverse). A matrix is singular if and only if its determinant is zero. Therefore, we find the eigenvalues by solving the characteristic equation:

det(A - λI) = 0

Solving this determinant yields a polynomial (the characteristic polynomial) whose roots are the eigenvalues of the matrix.

How to Calculate Eigenvectors

Once we have found the eigenvalues, we calculate the eigenvectors for each eigenvalue λ individually:

1. Substitute the value of λ back into the matrix (A - λI).
2. Solve the homogeneous system of linear equations (A - λI)v = 0.
3. The solution set is the null space of (A - λI). The basis vectors that span this null space are the eigenvectors, defining the eigenspace corresponding to λ.

Worked 2x2 Numeric Example

Let's find the eigenvalues and eigenvectors of the following matrix A:

First, we set up (A - λI):

We solve det(A - λI) = 0:
det = (4 - λ)(3 - λ) - (2 × 1) = λ² - 7λ + 12 - 2 = λ² - 7λ + 10 = 0.
Factoring the polynomial: (λ - 5)(λ - 2) = 0.
So, the eigenvalues are λ₁ = 5 and λ₂ = 2.

Finding eigenvectors for λ₁ = 5:
We substitute λ = 5 into (A - λI):

We solve the system (A - 5I)v = 0. Row reducing this matrix yields:

This gives the equation v₁ - v₂ = 0, or v₁ = v₂. Setting the free variable v₂ = 1, we obtain the eigenvector: v₁ = [1, 1]^T.

Finding eigenvectors for λ₂ = 2:
We substitute λ = 2 into (A - λI):

Solving (A - 2I)v = 0. Row reducing yields:

This gives the equation v₁ + 1/2 v₂ = 0, or v₁ = -1/2 v₂. Setting the free variable v₂ = 2, we obtain the eigenvector: v₂ = [-1, 2]^T.