PX153 - K12 - eigenvectors and eigenvalues

if $A \vec{x} = λ \vec{x} A \vec{x} = λ \vec{x}$ , $\vec{x}$ is the eigenvector of $A$ , and $λ$ is its eigenvalue
- 'eigen': proper (german)
generally, $\vec{x} = 0$ , or $λ = 0$ are not acceptable values
restricted to square matrices
the $i^{t h}$ eigenvector is denoted as ${\vec{x}}^{i}$ , and eigenvalue as $λ_{i}$
if $λ_{i}$ are distinct and non-zero, there are $n$ linearly independent eigenvectors for a $n \times n$ matrix
if $λ_{i} = λ_{j}$ , for $i \neq j$ , ie: two eigenvalues are the same, there may or may not be $n$ independent eigenvectors

finding eigenvalues and eigenvectors

\begin{aligned} A \vec{x} & = λ \vec{x} \\ A \vec{x} - λ \vec{x} & = 0 \\ (A - I λ) \vec{x} & = 0 \\ C \vec{x} & = 0 \end{aligned}

if $C \vec{x} = 0$ , $det C = 0$ , so $det (A - I λ) = 0$
think of $C \vec{x} = 0$ as the "magnification" of $\vec{x}$ vectors
if all vectors shrink to $0$ , $det C = 0$
proof: if $det C \neq 0, C^{- 1}$ exists, $C^{- 1} C \vec{x} = 0 ⟹ \vec{x} = 0$
- but $\vec{x} = 0$ is a trivial solution
- therefore, $C^{- 1}$ does not exist and $det C = 0$
always expect at least one free parameter per eigenvector (length of that vector)
because the amplitude is not defined in a basis of vectors
if ${\vec{x}}^{i}$ is an eigenvector, $α {\vec{x}}^{i}$ is also an eigenvector:

A (α {\vec{x}}^{i}) = A {\vec{x}}^{i} = (λ {\vec{x}}^{i}) = λ (α {\vec{x}}^{i})

2D case:
$A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]$ $\begin{align*} \det(A-\lambda\,I) &= (a_{11}-\lambda)(a_{22}-\lambda)- a_{12}a_{21} \\ 0 &= \lambda^{2} - \lambda(a_{11}+a_{22})+ a_{11}a_{22} - a_{12}a_{21} \\$
for eigenvectors, $A \vec{x} = λ_{i} \vec{x}$ , plug in $λ_{i}$
eg:

A = [\begin{matrix} 4 & 1 \\ 3 & 2 \end{matrix}]

$A - λ I = [\begin{matrix} 4 - λ & 1 \\ 3 & 2 - λ \end{matrix}]$
$$\begin{align*}
\det(A-\lambda,I) &= \lambda^{2}- 6\lambda + 5 = 0 \
\lambda_{1}= 5&, ;\lambda_{2}=1
\end{align*}$$
- taking $λ_{2} = 1 :$

\begin{aligned} (A - λ I) \vec{x} & = 0 \\ [\begin{array}{c} 3 & 1 \\ 3 & 1 \end{array}] [\begin{array}{c} x_{1} \\ x_{2} \end{array}] & = [\begin{array}{c} 0 \\ 0 \end{array}] \\ 3 x_{1} + x_{2} & = 0 \\ l e t x_{1} = t & , x_{2} = - 3 t \end{aligned}

- the eigen vector is: $t [\begin{matrix} 1 \\ - 3 \end{matrix}]$ , $t$ is a free parameter
- normalizing:

\begin{aligned} {\vec{x}}^{T} \cdot \vec{x} & = 1 \\ t^{2} (1 + 9) & = 1 \\ ∴ t & = \frac{1}{\sqrt{10}} \end{aligned}

- the normalized eigenvector is: $\frac{1}{\sqrt{10}} [\begin{matrix} 1 \\ - 3 \end{matrix}]$

- taking $λ_{2} = 5 :$

\begin{aligned} (A - λ I) \vec{x} & = 0 \\ [\begin{array}{c} - 1 & 1 \\ 3 & - 3 \end{array}] [\begin{array}{c} x_{1} \\ x_{2} \end{array}] & = [\begin{array}{c} 0 \\ 0 \end{array}] \\ e i t h e r, x_{1} + x_{2} & = 0 \\ o r, 3 x_{1} - 3 x_{2} & = 0 \\ l e t x_{1} = t & , x_{2} = t \end{aligned}

- the eigen vector is: $t [\begin{matrix} 1 \\ 1 \end{matrix}]$ , $t$ is a free parameter
- normalizing:

\begin{aligned} {\vec{x}}^{T} \cdot \vec{x} & = 1 \\ t^{2} (1 + 1) & = 1 \\ ∴ t & = \frac{1}{\sqrt{2}} \end{aligned}

- the normalized eigenvector is: $\frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 1 \end{matrix}]$

- checking:

[\begin{matrix} 4 & 1 \\ 3 & 2 \end{matrix}] \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 1 \end{matrix}] = \frac{1}{\sqrt{2}} [\begin{matrix} 5 \\ 5 \end{matrix}] = \frac{5}{\sqrt{2}} [\begin{matrix} 1 \\ 1 \end{matrix}]

- $5$ is the eigenvalue, $\frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 1 \end{matrix}]$ is the eigenvector