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Let $A=\begin{pmatrix} 2&2&0&0 \\ -2&-2&0&0 \\ 0&0&-1&1\\0&0&-1&1 \end{pmatrix}$. I noticed that $A^2=0_4$, and I deduced that the characteristic polynomial is $p(t)=t^4$ and the minimal polynomial is $m(t)=t^2$, so the largest order of the Jordan blocks relative to $\lambda_0=0$ is 2. Am I right that at this point I can only say that the JNF must either be $J_1=\begin{pmatrix} 0&1&0&0 \\ 0&0&0&0 \\ 0&0&0&0\\0&0&0&0 \end{pmatrix}$ or $J_2=\begin{pmatrix} 0&1&0&0 \\ 0&0&0&0 \\ 0&0&0&1\\0&0&0&0 \end{pmatrix}$ ?

Then I found that $\dim\ker A=2,$ thus $\operatorname{rk} A=2$ is the number of $1$'s in the JNF, which must then be $J_2$. Was there another way?

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An alternative way I could suggest is, since $A$ is a block diagonal matrix, to compute the Jordan normal forms of the blocks $$A_1 = \begin{pmatrix} 2 & 2 \\ -2 & -2 \end{pmatrix}, \quad A_2 = \begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix} $$ of $A$ which are both $$ J' = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ since $A_1, A_2$ are non-diagonalizable matrices with unique eigenvalue $0$.

Let $S_1, S_2$ be basis change matrices such that $S_i^{-1} A_i S_i = J'$ for $i=1,2$. Then we obtain $S^{-1} A S = J$ where $$ \begin{align} S &= \begin{pmatrix} S_1 & 0 \\ 0 & S_2 \end{pmatrix} \quad \left(\Longrightarrow S^{-1} = \begin{pmatrix} S_1^{-1} & 0 \\ 0 & S_2^{-1} \end{pmatrix} \right), \\ A &= \begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix} \text{ and } \\ J &= \begin{pmatrix} J' & 0 \\ 0 & J' \end{pmatrix}, \end{align} $$ our desired Jordan normal form of $A$.

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