# Linear state space equations and system stability

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• dim null $T$ + dim range $T$ = dim $V$
• Linear equation: $Ax=b$
• Solution (if it exists): $u+$ null$(A)$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space
• Eigenvalue $\lambda$ and Eigenvector $v$: $Tv=\lambda v$
• Distinct eigenvalues have independent eigenvectors
• Basis of eigenvectors results in a diagonal matrix for $T$
• Every linear map has an upper triangular matrix representation

## System state and evolution

System state variables at time $k=0,1,\ldots$

$x_k=(x_{k1},x_{k2},\ldots,x_{kn})$

Evolution from time $k$ to $k+1$

$x_{k+1} = A x_k$, where $A$: $n\times n$ matrix

From time $0$ to $k$

$x_k = A^k x_0$, where $x_0$: initial state

Bounded-input, bounded-output stable

If $x_0$ is bounded, $x_k$ is bounded for all $k$.

## Eigenvalues and instability

$\lambda$: eigenvalue of $A$ with eigenvector $v$

$Av=\lambda v$

Initial state: $x_0=v$

$x_1 = A x_0 = \lambda v$

$x_2 = A^2 x_0 = \lambda^2 v$

$\vdots$

$x_k = A^k x_0 = \lambda^k v$

Unstable if $|\lambda|>1$

## $A$: diagonalizable

Basis of eigenvectors for $A$: $\{v_1,\ldots,v_n\}$

Eigenvalues: $\lambda_1,\ldots,\lambda_n$

Initial state in eigenbasis: $x_0=\tilde{x}_{01}v_1+\cdots+\tilde{x}_{0n}v_n$

$x_1 = Ax_0=\tilde{x}_{01}\lambda_1 v_1+\cdots+\tilde{x}_{0n}\lambda_n v_n$

$x_2 = A^2x_0=\tilde{x}_{01}\lambda^2_1 v_1+\cdots+\tilde{x}_{0n}\lambda^2_n v_n$

$\vdots$

$x_k = A^k x_0=\tilde{x}_{01}\lambda^k_1 v_1+\cdots+\tilde{x}_{0n}\lambda^k_n v_n$

Stable if $|\lambda_i|< 1$ for $i=1,\ldots,n$

## $A$: non-diagonalizable - $2\times 2$, $3\times 3$ examples

$A = \begin{bmatrix} \lambda&1\\ 0&\lambda \end{bmatrix}$

Eigenvalues: $\lambda,\lambda$; Eigenvector: $(1,0)$

$A^k=\begin{bmatrix} \lambda^k&k\lambda^{k-1}\\ 0&\lambda^k \end{bmatrix}$

Proof: by induction

$A = \begin{bmatrix} \lambda&1&0\\ 0&\lambda&1\\ 0&0&\lambda \end{bmatrix}$

Eigenvalues: $\lambda,\lambda,\lambda$; Eigenvector: $(1,0,0)$

$A^k=\begin{bmatrix} \lambda^k&k\lambda^{k-1}&\dfrac{(k-1)k}{2}\lambda^{k-2}\\ 0&\lambda^k&k\lambda^{k-1}\\ 0&0&\lambda^k \end{bmatrix}$

Proof: by induction

## $A$: non-diagonalizable - $5\times 5$ example

$A = \begin{bmatrix} \lambda&1&0&0&0\\ 0&\lambda&1&0&0\\ 0&0&\lambda&1&0\\ 0&0&0&\lambda&1\\ 0&0&0&0&\lambda \end{bmatrix}$

Eigenvalues: $\lambda,\lambda,\lambda,\lambda,\lambda$; Eigenvector: $(1,0,0,0,0)$

$A^k=\begin{bmatrix} \lambda^k&k\lambda^{k-1}&O(k^2)\lambda^{k-2}&O(k^3)\lambda^{k-3}&O(k^4)\lambda^{k-4}\\ 0&\lambda^k&k\lambda^{k-1}&O(k^2)\lambda^{k-2}&O(k^3)\lambda^{k-3}\\ 0&0&\lambda^k&k\lambda^{k-1}&O(k^2)\lambda^{k-2}\\ 0&0&0&\lambda^k&k\lambda^{k-1}\\ 0&0&0&0&\lambda^k\\ \end{bmatrix}$

Proof: by induction

## $A$: non-diagonalizable - general case

Jordan form for any matrix (in a suitable basis)

$A\leftrightarrow\begin{bmatrix} A_1&0&\cdots&0\\ 0&A_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&A_m \end{bmatrix}$, $n\times n$

Form of $A_i$, $l_i\times l_i$, $n=l_1+\cdots+l_m$

$\lambda_i$ ($l_i=1$) or $\begin{bmatrix} \lambda_i&1&0&0&\cdots&0\\ 0&\lambda_i&1&0&\cdots&0\\ \vdots&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&\cdots&0&\lambda_i&1&0\\ 0&0&\cdots&0&\lambda_i&1\\ 0&0&0&\cdots&0&\lambda_i \end{bmatrix}$ ($l_i\ge2$)

Values of $A^k$: $O(k^{l_i-1})\lambda^{k-l_i+1}_i$

If $|\lambda_i|<1$, values of $A^k$ tend to 0 as $k\to\infty$ $\Rightarrow$ stable