# Distinguishing between two possibilities for the Jordan normal form of a matrix

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?

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 $$A$$.