The Characteristic Equation

The algorithm for finding eigenvalues and eigenvectors.

In our last lesson, we developed a deep, intuitive understanding of what eigenvalues and eigenvectors are. We established that they are the key to unlocking the soul of a matrix. We concluded with the master formula, the characteristic equation, which is the key to finding them:

det⁡(A−λI)=0\det(A - \lambda I) = 0

Today, we get our hands dirty. We will use this equation to build a step-by-step algorithm for finding the eigenvalues and eigenvectors of any square matrix.

The process is a two-act play:

  1. Act I: Find the Eigenvalues. We solve the characteristic equation for all possible values of `λ`.
  2. Act II: Find the Eigenvectors. For each eigenvalue `λ` we found, we plug it back into `(A - λI)v = 0` and find the basis for the null space to get the corresponding eigenvector(s) `v`.
The Matrix of the Day
Let's find the eigenvalues and eigenvectors of this matrix `A`:
A=[3−120]A = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}
Act I: Find the Eigenvalues (λ)

Step 1: Set up the matrix `(A - λI)`.

This means we subtract `λ` from every diagonal entry of `A`.

A−λI=[3−120]−λ[1001]=[3−λ−12−λ]A - \lambda I = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3-\lambda & -1 \\ 2 & -\lambda \end{bmatrix}

Step 2: Calculate the determinant of `(A - λI)`.

This resulting polynomial in `λ` is called the **characteristic polynomial**.

det⁡(A−λI)=(3−λ)(−λ)−(−1)(2)=−3λ+λ2+2=λ2−3λ+2\det(A - \lambda I) = (3 - \lambda)(-\lambda) - (-1)(2) = -3\lambda + \lambda^2 + 2 = \lambda^2 - 3\lambda + 2

Step 3: Set the characteristic polynomial to zero and solve for `λ`.

λ2−3λ+2=0\lambda^2 - 3\lambda + 2 = 0

This is a simple quadratic equation. We can factor it:

(λ−1)(λ−2)=0(\lambda - 1)(\lambda - 2) = 0

The solutions are `λ₁ = 1` and `λ₂ = 2`. These are our eigenvalues!

Act II: Find the Eigenvectors (v)
We must do this for each eigenvalue separately.

Case 1: Finding the eigenvector for `λ₁ = 1`

Step 1: Plug `λ = 1` into the matrix `(A - λI)`.

(A−1I)=[3−1−120−1]=[2−12−1](A - 1I) = \begin{bmatrix} 3-1 & -1 \\ 2 & 0-1 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 2 & -1 \end{bmatrix}

Step 2: Solve the system `(A - I)v = 0` by finding the null space.

[2−12−1][xy]=[00]\begin{bmatrix} 2 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

This translates to the equation `2x - y = 0`, which means `y = 2x`.

Step 3: Write the solution in vector form.

v=[x2x]=x[12]v = \begin{bmatrix} x \\ 2x \end{bmatrix} = x \begin{bmatrix} 1 \\ 2 \end{bmatrix}

The basis vector for this null space, `v₁ = [1, 2]`, is our first eigenvector.

Case 2: Finding the eigenvector for `λ₂ = 2`

Step 1: Plug `λ = 2` into `(A - λI)`.

(A−2I)=[3−2−120−2]=[1−12−2](A - 2I) = \begin{bmatrix} 3-2 & -1 \\ 2 & 0-2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 2 & -2 \end{bmatrix}

Step 2: Solve the system `(A - 2I)v = 0`.

[1−12−2][xy]=[00]\begin{bmatrix} 1 & -1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

This gives the equation `x - y = 0`, which means `x = y`.

Step 3: Write the solution in vector form.

v=[xx]=x[11]v = \begin{bmatrix} x \\ x \end{bmatrix} = x \begin{bmatrix} 1 \\ 1 \end{bmatrix}

The basis vector for this null space, `v₂ = [1, 1]`, is our second eigenvector.

Summary: The Algorithm
  1. Find Eigenvalues (`λ`):
    1. Construct the matrix `A - λI`.
    2. Compute its determinant to get the characteristic polynomial: `det(A - λI)`.
    3. Set the polynomial to zero and solve for `λ`. The roots are your eigenvalues.
  2. Find Eigenvectors (`v`):
    1. For **each** eigenvalue `λ`, plug it back into `A - λI`.
    2. Find the basis for the **null space** of this new matrix by solving `(A - λI)v = 0`.
    3. The basis vectors you find are the eigenvectors for that `λ`.

Up Next: Now that we know how to find these fundamental components, we will explore the magical process of **Diagonalization**.