If $(I+A)x=0$, take its <dot product> with $x$ and apply (e):
$$
0=x^\top(I+A)x=\|x\|^2+x^\top Ax=\|x\|^2.
$$
Hence $x=0$. A square <matrix> with zero <null space> is an <invertible matrix> by the <rank-nullity theorem>. Therefore \b[$I+A$ is nonsingular]. This proves <invertibility of identity plus a real skew-symmetric matrix>.

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Digital systems, linear algebra and introduction to formal proof / 6 / h / Solution

 Solution (source code)