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I am currently working on a Homework and cant solve this.

Let J be a 2x2 matrix such that J²=-1, show that ther existis a invertible matrix G such that $ G^{-1}JG= \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} $.

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Pick any non-zero $v \in \mathbb{R}^2$ and consider the list $B = (v, Jv)$. If this list were linearly dependent, then $Jv = \lambda v$ for some $\lambda \in \mathbb{R}$ (important: $v \neq 0$), but then $$-v = J^2 v = \lambda^2 v \implies \lambda^2 = -1,$$ which contradicts $\lambda \in \mathbb{R}$. Thus $B$ is linearly independent, and being of length $2$, also a basis. If we compute the matrix is for $J$ under the basis $B$, \begin{align*} Jv &= 0v + 1Jv, \\ J(Jv) &= -1v + 0Jv, \end{align*} hence the matrix becomes $$\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}.$$ This is the same operator $J$ under a different basis, hence the above matrix is similar to the standard matrix of $J$.

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Your matrix cannot have real eigenvalues, because if $\lambda\in\mathbb R$ and $Jv=\lambda v$ for some $v\neq0$, then $-v=J^2v=J.(Jv)=\lambda^2v$, which is not possible.

So, the roots of the characteristic polynomial of $J$ are complex conjugate non-real numbers. And their product is $-1$. Therefore, they are $i$ and $-i$. Let $(z,w)\in\mathbb{C}^2$ be such that $J.(z,w)=(iz,iw)$. Then $J.\left(\overline z,\overline w\right)=-i\left(\overline z,\overline w\right)$. So,$$J.\left((z,w)+\left(\overline z,\overline w\right)\right)=\left(iz-i\overline z,iw-i\overline w\right),$$but$$(z,w)+\left(\overline z,\overline w\right),\left(iz-i\overline z,iw-i\overline w\right)\in\mathbb{R}^2.$$On the other hand$$J.\left(iz-i\overline z,iw-i\overline w\right)=-\left((z,w)+\left(\overline z,\overline w\right)\right).$$

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