Square root of the negative identity matrix, unique? So I'm looking at an even-dimensional vector space over $\mathbb{R}$, i.e. $\mathbf{V} = \mathbb{R}^{2n}$, and I'm looking at a matrix $\mathbf{A}$ such that $\mathbf{A}^2 = -\mathbf{I}$. Particularly, I want to show that $\mathbf{A}$ uniquely defined by a family of similar matrices.
Obviously $\mathbf{-I}$ has eigenvalue $\lambda = -1$ with multiplicity $2n$, which yields an eigenvalue equation for $\mathbf{A}$ of the form $\lambda^2 = -1$, which doesn't have any solutions in $\mathbb{R}$. I've a feeling that this should uniquely identify the family, but why? I'm guessing there's some theorem I've forgotten in the years since I studied linear algebra, so if anyone could just point me in the right direction or name a theorem that would be great.
 A: Such a matrix is called a complex structure on $V$ (because it acts like multiplication by $i$ and can be used to endow $V$ with the structure of a complex vector space). In detail, define a complex scalar multiplication on $V$ by
$$ (x + iy) \cdot v = xv + yAv $$
and verify that indeed this turns $V$ into a complex vector space. Since $\dim_{\mathbb{R}} V = 2n$, it means that $\dim_{\mathbb{C}} V = n$. Choose some $\mathbb{C}$-basis $v_1, \dots, v_n$ for $V$ and show that $\mathcal{B} = \left( v_1, \dots, v_n, Av_1, \dots, Av_n \right)$ is an $\mathbb{R}$-basis for $V$. With respect to this basis, $A$ (or, more formally, the linear map $v \mapsto Av$) is represented by the block diagonal matrix
$$ J = \begin{pmatrix} 0_{n \times n} & -I_n \\ I_n & 0_{n \times n} \end{pmatrix} $$
which shows that $A$ must be similar to $J$. On the other hand, $J$ satisfies $J^2 = -I$ and any matrix which is similar to $J$ also satisfies this so we have shown that any complex structure on $V$ is similar to $J$.
