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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.

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    $\begingroup$ It's perfect for me. $\endgroup$ – Bernard Aug 9 '18 at 13:10
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Your argument is correct, but your notation is innapropriate. Where you wrote $\operatorname{rank}(\ker\cdots)$, you should have written $\dim(\ker\cdots)$.

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I like to find the actual Jordan form, especially including the matrices that give $R^{-1}A R = J.$ It is not bad when the eigenvalues are integers, and makes concepts concrete. This begins with choosing a column vector i called $w$ such that $A^2 w \neq 0.$ That becomes the far right column.

$$ \frac{1}{8} \; \left( \begin{array}{cccc} 8&-4&-4&0\\ 0&1&3&0\\ 0&2&-2&0\\ 0&-4&4&8\\ \end{array} \right) \left( \begin{array}{cccc} 0&0&0&1\\ 0&-1&1&3\\ 0&1&-1&-1\\ 0&-1&1&2\\ \end{array} \right) \left( \begin{array}{cccc} 1&2&1&0\\ 0&2&3&0\\ 0&2&-1&0\\ 0&0&2&1\\ \end{array} \right) = \left( \begin{array}{cccc} 0&0&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{array} \right) $$

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