Left & Right Eigenvectors and Eigenvalues

Eigenvalue

\( \chi (\lambda) = \textbf{det}(\lambda\textit{I}-A)=0 \) 에서,

\( \lambda \in \mathbb{C} \) : \( A \in \mathbb{C}^{n \times n} \) 의 eigenvalue

 

 

(Right) eigenvector

\( \exists\ nonzero \ v \in \mathbb{C}^n \)

s.t. \(( \lambda I-A)v = 0 \), i.e., \(Av = \lambda v \)

\(v \) : A의 eigenvector

 

(Left) eigenvector

\( \exists\ nonzero \ w \in \mathbb{C}^n \)

s.t. \( w^{T}(\lambda I-A) = 0 \), i.e., \(w^{T}A = \lambda w^{T} \)

\(v \) : A의 left eigenvector

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Dynamic interpretation

먼저, matrix exponential에 대한 이해가 필요하다.

 

모든 \(M \in  \mathbb{R}^{n \times n} \)에 대하여

\(e^{M}=I+M+\frac{M^{2}}{2!}+\cdots\)

 

이 부분은 지수함수의 테일러 급수로 표현하는 방법(\(e^{x}=1+x+\frac{(ta)^{2}}{2!}+\cdots\))을 참고하면 된다.

 

\(A \in \mathbb{R}^{n \times n}\)이며 상수일 때, \(\dot{x}=Ax\)의 해는

\(x(t)=e^{tA}x(0)\)

 

이때 \(e^{tA}\)는 state transition matrix(\(\Phi (t)\))라 한다. 시간 t만큼 전진시키는 matrix로서 생각할 수 있다.

 

 

\( \exists\ Av=\lambda v, \ v!=0 \)

if \(\dot{x}=Ax, x(0)=v\), then \(x(t)=e^{\lambda t}v\)

\begin{align*}
x(t) &=e^{tA}v=(I+tA+\frac{(tA)^2}{2!}+\cdots)v\\&=v+\lambda tv+\frac{(\lambda t)^2}{2!} v+\cdots\\&=e^{\lambda t}v
\end{align*}

 

• initial state를 eigenvector \(v\)라 두면, resulting motion이 매우 간단해짐(\(v\)를 span한 형태로 존재)
• \( x(t)=e^{\lambda t}v \) : 시스템 \(\dot{x}=Ax\) 의 mode ;복잡한 해를 간단한 원소로 표현
• \(\lambda \in \mathbb{R}, \lambda < 0\), mode contracts or shrinks
• \(\lambda \in \mathbb{R}, \lambda > 0\), mode expands or grows

Dynamic interpretation : left eigenvectors

\( \exists\ w^{T}A=\lambda w^{T}, w!=0\)

\(\frac{d}{dt}(w^{T}x)=w^{T}\dot(x)=w^{T}Ax=\lambda(w^{T}x)\)

 

이에 따라, \(w^{T}x(t)=e^{\lambda t}w^{T}x(0)\)

 

• even if trajectory x is complicated, \(w^{T}x\) is simple
• if \(lambda \in \mathbb{R}, \lambda < 0\), halfspace \(\left\{z | w^{T} \leq a\right\} \) is invariant (for \(a \geq 0\))
• for \(\lambda=\sigma +iw \in \mathbb{C}\), \((\Re w)^{T}x\) and \((\Im w)^{T}x\) both have form, \(e^{\sigma t}(\alpha cos(wt)+\beta sin(wt))\) 

right eigenvectors : initial conditions from which resulting motion is simple

left eigenvectors : linear functions of state that are simple, for any initial condition

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Stability

if \(e^{tA} \to 0\) as \(t \to \infty \), \(\dot{x} = Ax\) is stable

 

• state \(x(t)\) converges to 0, as \(t \to \infty\), no matter what \(x(0)\) is
• all trajectories of \(\dot{x}=Ax\) converge to 0 as \(t \to \infty \)

fact : \(\dot(x)=Ax\) is stable if and only if all eigenvalues of A have negative real part

\(\Re \lambda_{i} < 0, \ i=1, \cdots, n\)

 

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Diagonalization

 

\begin{align*}
A^{k} &= \left (T \Lambda T^{-1}  \right )^{k} \\ &=\left (T \Lambda T^{-1}  \right ) \cdots \left (T \Lambda T^{-1}  \right ) \\ &= T \Lambda^{k} T^{-1} \\ &=T \textbf{diag}(\lambda_{1}^{k}, \cdots, \lambda_{n}^{k})T^{-1}
\end{align*}