Eigenvalues and Eigenvectors

• A scalar $\lambda$ is an eigenvalue of an $n \times n$ matrix $A$ if there exists a nontrivial solution $\vec{x}$ to the equation $A\vec{x} = \lambda \vec{x}$.
• The vector $\vec{x}$ in the equation $A\vec{x} = \lambda \vec{x}$ is called an eigenvector of $A$ corresponding to the eigenvalue $\lambda$.

Finding Eigenvalues of A

• To find eigenvalues, solve $A\vec{x} = \lambda \vec{x}$, which can be rewritten as $(A - \lambda I)\vec{x} = \vec{0}$.
• For a non-trivial solution to exist, the determinant of $(A - \lambda I)$ must equal zero: $\text{det}(A - \lambda I) = 0$.
• The characteristic polynomial of $A$ is $\text{det}(A - \lambda I)$, which is a polynomial in $\lambda$.
• The characteristic equation of $A$ is $\text{det}(A - \lambda I) = 0$.

Example 1

• Given $A = \begin{bmatrix} 2 & 1 \ 1 & 2 \end{bmatrix}$, subtract $\lambda I$ to get $\begin{bmatrix} 2-\lambda & 1 \ 1 & 2-\lambda \end{bmatrix}$.
• Calculate the determinant: $(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 3)(\lambda - 1)$.
• The eigenvalues are the solutions to $(\lambda - 3)(\lambda - 1) = 0$, which are $\lambda_1 = 3$ and $\lambda_2 = 1$.

Finding Eigenvectors of A

• To find eigenvectors, solve the equation $(A - \lambda I)\vec{x} = \vec{0}$ for each eigenvalue $\lambda$.

Example 2

• For $\lambda_1 = 3$, $A - 3I = \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}$.
• Solve $\begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$, which gives $-x_1 + x_2 = 0$, so $x_1 = x_2$.
• The eigenvector is $\vec{x} = \begin{bmatrix} x_2 \ x_2 \end{bmatrix} = x_2 \begin{bmatrix} 1 \ 1 \end{bmatrix}$.
• For $\lambda_2 = 1$, $A - I = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}$.
• Solve $\begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$, which gives $x_1 + x_2 = 0$, so $x_1 = -x_2$.
• The eigenvector is $\vec{x} = \begin{bmatrix} -x_2 \ x_2 \end{bmatrix} = x_2 \begin{bmatrix} -1 \ 1 \end{bmatrix}$.

Theorem 5.1.2

• If $A$ is an $n \times n$ triangular matrix, its eigenvalues are the entries on its main diagonal.

Example 3

• Given $A = \begin{bmatrix} 3 & 6 & -8 \ 0 & 0 & 6 \ 0 & 0 & 2 \end{bmatrix}$, the eigenvalues are $\lambda_1 = 3, \lambda_2 = 0, \lambda_3 = 2$.

Theorem 5.1.3

• If $\vec{v_1}, \vec{v_2},..., \vec{v_r}$ are eigenvectors corresponding to distinct eigenvalues $\lambda_1, \lambda_2,..., \lambda_r$ of an $n \times n$ matrix $A$, then the set ${\vec{v_1}, \vec{v_2},..., \vec{v_r}}$ is linearly independent.