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Let $\mathbb{F}$ be a field, $n\in\mathbb{N}_{\geq 1}$ and $A\in M_{2n}(\mathbb{F})$, such that $$A=\begin{pmatrix} 0_n & 0_n \\ B & 0_n \end{pmatrix}$$ with $B\in GL_n(\mathbb{F})$. Show that A is similar to the matrix $$\begin{pmatrix} C & 0_2 & \ldots & 0_2 \\ 0_2 & C & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0_2 \\ 0_2 & \ldots & 0_2 & C \end{pmatrix}$$ where $C=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\in M_2(\mathbb{F})$.

I searched the Internet well enough and found nothing similar.

Thanks in advance!

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3 Answers 3

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Let $i \colon \mathbb{F}^{n} \hookrightarrow \mathbb{F}^{2 n}$ be the inclusion and denote by $\left( e_{1}, \dotsc, e_{n} \right)$ and $\left( E_{1}, \dotsc E_{2 n} \right)$ the canonical bases of $\mathbb{F}^{n}$ and $\mathbb{F}^{2 n}$. Then $\left( i\left( B^{-1} e_{1} \right), E_{n +1}, \dotsc, i\left( B^{-1} e_{n} \right), E_{2 n} \right)$ is a basis of $\mathbb{F}^{2 n}$ and, for all $j \in \lbrace 1, \dotsc, n \rbrace$, we have $A i\left( B^{-1} e_{j} \right) = E_{n +j}$ and $A E_{n +j} = 0_{\mathbb{F}^{2 n}}$.

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Denote by $\Gamma$ the matrix $$\begin{pmatrix} C&0_2&0_2&\dots&0_2\\0_2&C&0_2&\dots&0_2\\0_2&0_2&C&\dots&0_2\\\vdots&\vdots&\vdots&\ddots&\vdots\\0_2&0_2&0_2&\dots&C\end{pmatrix}.$$ Let $P$ denote the permutation matrix associated to the permutation $\sigma\in \mathcal{S}_{2n}$ such that $$\sigma(k)=\left\{\begin{array}{ll} n+\frac{k}{2}&\text{if}\ k\ \text{is even},\\ \frac{k+1}{2}&\text{if}\ k\ \text{is odd}. \end{array}\right.$$ i.e., $P=[p_{i,j}]$, where $$p_{i,j}=\delta_{\sigma(i),j}.$$ Observe that $P^{-1}\Gamma P$ is given by $$J=\begin{pmatrix}0_n &0_n\\ I_n&0_n\end{pmatrix}.$$ Next, we know that $$\begin{pmatrix}I_n&0_n\\0_n&B\end{pmatrix}\begin{pmatrix}0_n&0_n\\I_n&0_n\end{pmatrix}\begin{pmatrix}I_n&0_n\\0_n&B^{-1}\end{pmatrix}=\begin{pmatrix}0_n&0_n\\B&0_n\end{pmatrix}=A.$$ Thus, if $M$ denotes $\begin{pmatrix}I_n&0_n\\0_n&B\end{pmatrix}$, then $$A=MJM^{-1}=M(P^{-1}\Gamma P)M^{-1}=(MP^{-1})\Gamma(MP^{-1})^{-1}.$$

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Let $U=diag(C,\cdots,C)$. Then $A^2=U^2=0$. It remains to show that $dim(\ker(A))=dim(\ker(U))$.

$\ker(A)=[0_n,K^n]^T$ has dimension $n$.

$\ker(U)=span(e_2,e_4,\cdots,e_{2n})$ has dimension $n$ too.

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