Jordan Decomposition

Description:

• Diagonalizable Matrix
• Not all matrices are diagonalizable, however, most matrices are, in a sense that any non-diagonalizable matrix is infinitesimally close to a diagonalizable one.
• Symmetric matrices are always diagonalizable, with real eigenvalues and orthogonal eigenvectors
• Any square matrix can be written as $A=VD{V}^{-1}$, where $V$ is a square (possibly complex-valued) invertible matrix, and $D:=\mathrm{diag}\left(J\left({\lambda }_1\right),\dots ,J\left({\lambda }_p\right)\right)$ where each block $J(\lambda )$ is a Jordan Block
  1. ${\lambda }_1,\dots ,{\lambda }_p$ are (possibly equal) eigenvalues of $A$
  2. By convention $J(\lambda )=\lambda$ when the size of that matrix is 1 .
  3. A diagonalizable matrix has Jordan blocks of size 1 only.
  4. The ${\lambda }_i$ ‘s are (possibly equal, possibly complex) eigenvalues.

Applications:

Matrrix powers:

• Assume that $A$ is diagonalizable: $A=VD{V}^{-1}$, with $D$ diagonal. Then any power of $A$ can be easily computed, as $A^k=V{D}^k{V}^{-1},k=1,2,3,\dots$
• Cayley-Hamilton Theorem: For any square matrix $A$, if $p(\lambda ):=\det (\lambda I-A)$ is the characteristic polynomial of $A$, then $p(A)=0$

Matrix exponential:

• For a square matrix we define the exponential as $e^A:=\sum _{k=0}^{+\infty }\frac{1}{k!}{A}^k$
• Using the Jordan decomposition we can prove that the series always converges.
• For diagonalizable matrices $A=VD{V}^{-1}$ with $D=\mathrm{diag}\left({\lambda }_1,\dots ,{\lambda }_n\right)$ and $V$ invertible, we have $e^A=V{e}^D{V}^{-1},e^D:=\mathrm{diag}\left(e^{{\lambda }_1},\dots ,e^{{\lambda }_n}\right)$

Roots of polynomial:

• We can use eigenvalue decomposition to find the zeroes of a (univariate) polynomial: let $p(s):=s^n+a_1{s}^{n-1}+\dots +a_{n-1}s+a_n$
  • where $a\in {\mathbb{R}}^n$ is given.
• Define the $n\times n$ “companion” matrix $A=\left(\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ -a_1& -a_2& -a_3& \cdots & -a_n\end{array}\right)$
• Fact: we have $p(s)=\det (sI-A)$
  • proof: $s$ is root if and only if the vector $v:=(1,s,...,s^{n-1})$ is an eigenvector of $A$, with eigenvalue $s$

Linear discrete dynamical systems:

• Consider the (discrete-time) linear dynamical system $x(t+1)=Ax(t)+Bu(t),t=0,1,2,\dots ,$
• We have for every $T\ge 0:x(T)=A^{T}x(0)+\sum _{t=0}^{T-1}{A}^{t}Bu(t)$
• Asymptotic stability:
  • Assume that $u(t)=0$ for every $t:x(t+1)=Ax(t),t=0,1,2,\dots$
  • The system is said to be asymptotically stable if, from any initial condition, we return to the equilibrium state $x=0$ : that is $\underset{T\to +\infty }{\lim }x(T)=0$
  • Asymptotic stability is equivalent to $A^T\to 0$ as $T\to +\infty$. In turn, this is the same as $\left|{\lambda }_i\right|<1$, for every $i=1,\dots ,n$.
  • Continuous-time case
    • Consider the (continuous-time) linear dynamical system $\frac{d}{dt}x(t)=Ax(t)+Bu(t),t\ge 0$
    • For every $T\ge 0:x(T)=e^{TA}x(0)+{\int }_0^T{e}^{(T-t)A}Bu(t)dt$
    • Proof: we can show that $\frac{d}{dt}{e}^{tA}=A{e}^{tA}$
  • Asymptotic stability is equivalent to $e^{TA}\to 0$ as $T\to +\infty$. In turn, this is the same as $\mathrm{Re}\left({\lambda }_i\right):={\lambda }_i+{\bar{\lambda }}_i<0$, for every $i=1,\dots ,n$.
• examples: structure dynamics
  • Dynamical model of mechanical structure (e.g., bridge) $M\ddot{y}(t)+Ky(t)=f(t)$ where
    • $y(t)\in {\mathbb{R}}^n$ contains position at $n$ discrete locations on bridge
    • $M$, the “mass matrix”, is positive-definite
    • $K$, the “stiffness” matrix, is positive semi-definite
    • $f(t)$ is a forcing function (e.g., wind)
  • Stability of system related to eigenvalues of $A:=\left(\begin{array}{cc}0& I\\ -M^{-1}K& 0\end{array}\right)$