# Jordan form of simple 2x2 matrix

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?

## 1 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).

• Thank you Pedro. This indeed saves me a lot of work. A quick follow up question: Is it safe to assume that each eigenvalue has at least one Jordan block? In other words: If there are 2 distinct eigenvalues for a 2x2 matrix then there are 2 blocks and thus each block is of size 1x1? – user463102 Sep 20 '18 at 12:16