A given $n\times n$ matrix $B$ of real entries obeys $B^{2}=-I$.
I know that $B$ has only 2 eigenvalues $i,-i$ and that $n$ has to be even. Therefore each eigenvalue must have $n/2$ multiplicity. How do I prove that there are $n$ linearly independent eigenvectors?
This is essentially the same as what @Dbchatto67 says in the comments, but if you know about Jordan normal form (very much worth learning if not), then you can solve this easily by thinking in JNF. A $k\times k$ Jordan block (over $\mathbb{C}$) $$J = \begin{pmatrix} \lambda & 1 & 0 & \dots & 0 & 0 \\ 0 & \lambda & 1 & 0 & \dots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \dots & \lambda \end{pmatrix}$$ does not satisfy $J^2 = -I$ unless $\lambda = \pm i$ (as you already saw) and $k=1$. So the JNF of $B$ over $\mathbb{C}$ is diagonal and there are $n$ linearly independent eigenvectors over $\mathbb{C}$.