Finding Eigenvectors

Overview

Once eigenvalues are found, eigenvectors are computed by solving the homogeneous system $(A - \lambda I)\mathbf{v} = \mathbf{0}$. The solution space forms the eigenspace for that eigenvalue.

Method

For each eigenvalue $\lambda$:

1. Form the matrix $A - \lambda I$
2. Solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$
3. The solution set (excluding zero) gives eigenvectors

Step-by-Step Process

Step 1: Substitute $\lambda$

Replace $\lambda$ in $A - \lambda I$ with the known eigenvalue.

Step 2: Row Reduce

Row reduce $A - \lambda I$ to echelon form.

Step 3: Solve the System

Find the null space of $A - \lambda I$.

Step 4: Write Eigenvectors

Express solutions in parametric form.

Example: 2×2 Matrix

$A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \quad \text{Eigenvalues: } \lambda_1 = 5, \lambda_2 = 2$

For $\lambda_1 = 5$:

$A - 5I = \begin{bmatrix} 4-5 & 2 \\ 1 & 3-5 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 1 & -2 \end{bmatrix}$

Row reduce:

$\begin{bmatrix} -1 & 2 \\ 1 & -2 \end{bmatrix} \to \begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix}$

$x_1 - 2x_2 = 0 \Rightarrow x_1 = 2x_2$

Let $x_2 = t$: $\mathbf{v} = t\begin{bmatrix} 2 \\ 1 \end{bmatrix}$

Eigenvector: $\mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ (or any scalar multiple)

For $\lambda_2 = 2$:

$A - 2I = \begin{bmatrix} 4-2 & 2 \\ 1 & 3-2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 1 & 1 \end{bmatrix}$

Row reduce:

$\begin{bmatrix} 2 & 2 \\ 1 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$

$x_1 + x_2 = 0 \Rightarrow x_1 = -x_2$

Let $x_2 = t$: $\mathbf{v} = t\begin{bmatrix} -1 \\ 1 \end{bmatrix}$

Eigenvector: $\mathbf{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$

Eigenspace

The eigenspace $E_\lambda$ is the set of all eigenvectors for $\lambda$ plus the zero vector:

$E_\lambda = \text{Null}(A - \lambda I)$

It's a subspace with dimension = geometric multiplicity of $\lambda$.

Geometric Multiplicity

The dimension of the eigenspace:

$\text{geo.mult}(\lambda) = \dim(E_\lambda) = n - \text{rank}(A - \lambda I)$

Always: $1 \leq$ geometric multiplicity $\leq$ algebraic multiplicity

Example: Repeated Eigenvalue

$A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix} \quad p(\lambda) = (\lambda-2)(\lambda-3)$

Wait, let me reconsider:

$A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \quad p(\lambda) = (\lambda-2)^2 \Rightarrow \lambda = 2 \text{ (mult. 2)}$

For $\lambda = 2$:

$A - 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

$x_2 = 0$, $x_1$ is free

$E_2 = \text{span}\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}$

Geometric multiplicity = 1 < 2 = algebraic multiplicity

Example: Full Eigenspace

$A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \quad \text{Eigenvalue: } \lambda = 2 \text{ (mult. 2)}$

$A - 2I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

All non-zero vectors are eigenvectors!

$E_2 = \mathbb{R}^2 = \text{span}\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}$

Geometric multiplicity = 2 = algebraic multiplicity

Verifying Eigenvectors

Always verify by checking $A\mathbf{v} = \lambda\mathbf{v}$:

$A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \quad \lambda = 5$

$A\mathbf{v} = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4(2)+2(1) \\ 1(2)+3(1) \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} = 5\begin{bmatrix} 2 \\ 1 \end{bmatrix} \checkmark$

Common Mistakes

• Forgetting that eigenvectors cannot be zero
• Not checking all eigenvalues
• Missing linearly independent eigenvectors for repeated eigenvalues
• Computational errors in row reduction