Finding Eigenvalues
λ is an eigenvalue of A ⟺ λ is a root of d e t ( t I n − A ) = P A ( t ) \lambda \text{ is an eigenvalue of A } \iff \lambda \text{ is a root of } det(t I_n - A) = P_A (t) λ is an eigenvalue of A ⟺ λ is a root of d e t ( t I n − A ) = P A ( t ) PA (t) is known as the **characteristic polynomial. Also, lambda and t can be used interchangeably.
Example
d e t ( λ ( 1 0 0 1 ) − ( 1 1 1 1 ) ) = d e t ( ( λ − 1 − 1 − 1 λ − 1 ) ) = λ 2 − 2 λ = λ ( λ − 2 ) There are two eigenvalues: 0 and 2. det(\lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}) \\
= det(\begin{pmatrix} \lambda - 1 & -1 \\ -1 & \lambda - 1 \end{pmatrix}) = \lambda ^2 - 2\lambda = \lambda (\lambda - 2) \\
\text{There are two eigenvalues: 0 and 2.} d e t ( λ ( 1 0 0 1 ) − ( 1 1 1 1 ) ) = d e t ( ( λ − 1 − 1 − 1 λ − 1 ) ) = λ 2 − 2 λ = λ ( λ − 2 ) There are two eigenvalues: 0 and 2. A = ( 0 6 − 1 5 ) d e t ( λ ( 1 0 0 1 ) − ( 0 6 − 1 5 ) ) = d e t ( λ ( λ − 6 1 λ − 5 ) ) = λ 2 − 5 λ + 6 = ( λ − 3 ) ( λ − 2 ) There are two eigenvalues: 2 and 3. A = \begin{pmatrix} 0 & 6 \\ -1 & 5 \end{pmatrix} \\
det(\lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 0 & 6 \\ -1 & 5 \end{pmatrix}) \\
= det(\lambda \begin{pmatrix} \lambda & -6 \\ 1 & \lambda - 5 \end{pmatrix}) = \lambda^2 - 5\lambda + 6 = (\lambda - 3)(\lambda - 2) \\
\text{There are two eigenvalues: 2 and 3.} A = ( 0 − 1 6 5 ) d e t ( λ ( 1 0 0 1 ) − ( 0 − 1 6 5 ) ) = d e t ( λ ( λ 1 − 6 λ − 5 ) ) = λ 2 − 5 λ + 6 = ( λ − 3 ) ( λ − 2 ) There are two eigenvalues: 2 and 3. A = ( 1 6 5 2 ) d e t ( λ ( 1 0 0 1 ) − ( 1 6 5 2 ) ) = d e t ( λ ( λ − 1 − 6 − 5 λ − 2 ) ) = λ 2 − 3 λ − 28 = ( λ − 7 ) ( λ + 4 ) There are two eigenvalues: -4 and 7. N u l ( 7 I 2 − A ) contains our eigenvectors, as ( 7 I 2 − A ) v → = 0 → → 7 v → − A v → = 0 → N u l ( 7 I 2 − A ) = N u l ( ( 6 − 6 − 5 5 ) ) = N u l ( ( 1 − 1 − 1 1 ) ) = N u l ( ( 1 − 1 0 0 ) ) ( 1 − 1 0 0 ) ( a b ) = ( a − b 0 ) = ( 0 0 ) when a = b Thus, any vectors where a = b are 7-eigenvectors for A, such as ( 1 1 ) N u l ( − 4 I 2 − A ) = N u l ( 4 I 2 + A ) = N u l ( ( 5 6 5 6 ) ) = N u l ( ( 5 6 0 0 ) ) = N u l ( ( 1 6 5 0 0 ) ) ( 1 6 5 0 0 ) ( a b ) = ( a + 6 5 b 0 ) = ( 0 0 ) when a = − 6 5 b Thus, any vectors where a = − 6 5 b are -4-eigenvectors for A, such as ( − 6 5 ) or S p a n { ( − 6 5 ) } A = \begin{pmatrix} 1 & 6 \\ 5 & 2 \end{pmatrix} \\
det(\lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 6 \\ 5 & 2 \end{pmatrix}) \\
= det(\lambda \begin{pmatrix} \lambda - 1 & -6 \\ -5 & \lambda - 2 \end{pmatrix}) = \lambda^2 - 3\lambda - 28 = (\lambda - 7)(\lambda + 4) \\
\text{There are two eigenvalues: -4 and 7.} \\
Nul(7 I_2 - A) \text{ contains our eigenvectors, as } (7 I_2 - A) \overrightarrow{v} = \overrightarrow{0} \rightarrow 7\overrightarrow{v} - A\overrightarrow{v} = \overrightarrow{0} \\
Nul(7 I_2 - A) = Nul(\begin{pmatrix} 6 & -6 \\ -5 & 5 \end{pmatrix}) = Nul(\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}) = Nul(\begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix}) \\
\begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a - b \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \text{ when a = b}\\
\text{Thus, any vectors where a = b are 7-eigenvectors for A, such as } \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\
Nul(-4 I_2 - A) = Nul(4 I_2 + A) = Nul(\begin{pmatrix} 5 & 6 \\ 5 & 6 \end{pmatrix}) = Nul(\begin{pmatrix} 5 & 6 \\ 0 & 0 \end{pmatrix}) = Nul(\begin{pmatrix} 1 & \frac{6}{5} \\ 0 & 0 \end{pmatrix})\\
\begin{pmatrix} 1 & \frac{6}{5} \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a + \frac{6}{5}b \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \text{ when } a = -\frac{6}{5}b\\
\text{Thus, any vectors where } a = -\frac{6}{5}b \text{ are -4-eigenvectors for A, such as } \begin{pmatrix} -6 \\ 5 \end{pmatrix} \text{ or } Span\{\begin{pmatrix} -6 \\ 5 \end{pmatrix}\} \\ A = ( 1 5 6 2 ) d e t ( λ ( 1 0 0 1 ) − ( 1 5 6 2 ) ) = d e t ( λ ( λ − 1 − 5 − 6 λ − 2 ) ) = λ 2 − 3 λ − 28 = ( λ − 7 ) ( λ + 4 ) There are two eigenvalues: -4 and 7. N u l ( 7 I 2 − A ) contains our eigenvectors, as ( 7 I 2 − A ) v = 0 → 7 v − A v = 0 N u l ( 7 I 2 − A ) = N u l ( ( 6 − 5 − 6 5 ) ) = N u l ( ( 1 − 1 − 1 1 ) ) = N u l ( ( 1 0 − 1 0 ) ) ( 1 0 − 1 0 ) ( a b ) = ( a − b 0 ) = ( 0 0 ) when a = b Thus, any vectors where a = b are 7-eigenvectors for A, such as ( 1 1 ) N u l ( − 4 I 2 − A ) = N u l ( 4 I 2 + A ) = N u l ( ( 5 5 6 6 ) ) = N u l ( ( 5 0 6 0 ) ) = N u l ( ( 1 0 5 6 0 ) ) ( 1 0 5 6 0 ) ( a b ) = ( a + 5 6 b 0 ) = ( 0 0 ) when a = − 5 6 b Thus, any vectors where a = − 5 6 b are -4-eigenvectors for A, such as ( − 6 5 ) or Sp an { ( − 6 5 ) } Note : The nullspace associated with an eigenvalue has a special name: Ex = the x-eigenspace of A, where x is replaced by the eigenvalue that represents the eigenspace.
A = ( 1 1 1 2 ) , eigenvalues are 1 2 ( 3 ± 5 ) ( a b ) ∈ E 1 2 ( 3 ± 5 ) A ( a b ) = λ ( a b ) → 1 2 ( 3 ± 5 ) ( a b ) = ( 1 1 1 2 ) ( a b ) = ( a + b a + 2 b ) 1 2 ( 3 ± 5 ) a = a + b , so b = 1 2 ( 1 ± 5 ) E 1 2 ( 3 ± 5 ) is spanned by { ( 1 1 2 ( 1 ± 5 ) ) } A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \text{eigenvalues are } \frac{1}{2} (3 \pm \sqrt{5}) \\
\begin{pmatrix} a \\ b \end{pmatrix} \in E_{\frac{1}{2} (3 \pm \sqrt{5})} \\
A \begin{pmatrix} a \\ b \end{pmatrix} = \lambda \begin{pmatrix} a \\ b \end{pmatrix} \rightarrow \frac{1}{2} (3 \pm \sqrt{5}) \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a + b \\ a + 2b \end{pmatrix} \\
\frac{1}{2} (3 \pm \sqrt{5}) a = a + b \text{, so } b = \frac{1}{2} (1 \pm \sqrt{5}) \\
E_{\frac{1}{2} (3 \pm \sqrt{5})} \text{ is spanned by } \{\begin{pmatrix} 1 \\ \frac{1}{2} (1 \pm \sqrt{5}) \end{pmatrix} \} A = ( 1 1 1 2 ) , eigenvalues are 2 1 ( 3 ± 5 ) ( a b ) ∈ E 2 1 ( 3 ± 5 ) A ( a b ) = λ ( a b ) → 2 1 ( 3 ± 5 ) ( a b ) = ( 1 1 1 2 ) ( a b ) = ( a + b a + 2 b ) 2 1 ( 3 ± 5 ) a = a + b , so b = 2 1 ( 1 ± 5 ) E 2 1 ( 3 ± 5 ) is spanned by { ( 1 2 1 ( 1 ± 5 ) ) }