The following question is from as System Theory test.
Let the system matrix $A$ be given as
$A = \begin{bmatrix} 0&0&0&1\\0&-1&1&3\\0&1&-1&-1\\0&-1&1&2 \end{bmatrix}$
Which has the eigenvalues $\lambda_i=0$ for $i = 1,2,3,4$. Which of the following matrices is the Jordan form of $A$.
$A) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}$
$B) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$
$C) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}$
$D) \quad J=\begin{bmatrix} 0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$
My approach:
The eigenvalues are given so the nullspace of $\left[A-\lambda I\right]$ can be found to be $\begin{bmatrix} 0\\1\\1\\0 \end{bmatrix}$ and $\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}$.
The dimension of this nullspace is $2$ which indicates that there are 2 Jordan blocks. So answer $B$ or $C$.
Also I've found that : $\operatorname{rank}(\ker(A-\lambda I)^2)=3$ and $\operatorname{rank}(\ker(A-\lambda I))=2$. Subtracting these outcomes gives $3-2=1$. So there's one Jordan block of size 2 or larger. And thus only answer $B$ is correct.
Right or wrong? thanks in advance.
Best Answer
Your argument is correct, but your notation is innapropriate. Where you wrote $\operatorname{rank}(\ker\cdots)$, you should have written $\dim(\ker\cdots)$.