Considere the following transfer function:
$\frac{1}{s^2+1}$
Calculate the Jordan form, real Jordan form and determine if this system is Lyapunov stable?
My approach:
The system's $A$ matrix is:
$\begin{bmatrix}0&-1\\1&0 \end{bmatrix}$
The eigenvalues of this matrix are $0+i$ and $0-i$.
Next we calculate
$\text{dim(ker}(A-\lambda I))$ Which gives:
$\quad \text{dim(ker}(A-(0+i)I)$ $ = 1$ and
$\quad\text{dim(ker}(A-(0-i)I)$ $ = 1$
Which means that each eigenvalue has, at most, one Jordan block.
To see if any of these blocks are of size $\geq 2$ we check:
$\quad\text{dim(ker}(A-\lambda I)^2)- \text{dim(ker}(A-\lambda I))$
For both $\lambda$'s we get that there are no blocks $\geq 2$.
So each eigenvalue with zero real part has only one jordan block and because these Jordan blocks are $1\ $x$\ 1$ the system is Lyapunov stable?
So the Jordan form is:
$\begin{bmatrix}-i&0\\0&i \end{bmatrix}$ and the real Jordan form is $\begin{bmatrix}0&1\\-1&0 \end{bmatrix}$.
Is this correct?
Best Answer
Yes, but you don't need that much work: if an $n\times n$ matrix has $n$ different eigenvalues, then it is diagonalizable (and the Jordan form is the diagonal matrix with the eigenvalues on the diagonal).