Theorem: If $A_{n \times n}$ is real, skew-symmetric matrix $\rightarrow I+A$ is non-singular.
I have one obstacle, I can't use eigenvalues - that would be fairly easy. I have also received a hint on how to approach this: calculate the angle between $Ax$ and $x$. So I started like this:
$$ cos \alpha = \frac{<Ax,x>}{||Ax||.||x||}$$ $$ -x^TAx=x^TA^Tx=<Ax,x>=<x,Ax>=x^TAx=0$$ That implies that $Ax$ and $x$ are orthogonal. I wanted to prove that $I+A$ is nonsingular, that is equivalent to $dim (N(A))=0$. So we are looking for all nontrivial vectors that satisfy $(I+A)x=0$. Obviously $ Ix=x $ and $Ax$ maps $x$ to a vector orthogonal to $x$. We now want to prove, that $$x+Ax \neq 0 $$ for all non trivial $x$. The geometric idea would be that $x+Ax$ means the "diagonal" in the "rectangle" denoted by $x$ and $Ax$ orthogonal to $x$ (" " is there for the $n$-dimensional space). But this is not rigorous enough for me. How can I prove the last equation?
$I+A$ has trivial null space because $$ \|(I+A)x\|^2 = \|x\|^2+\langle Ax,x\rangle+\langle x,Ax\rangle+\|Ax\|^2 \\ = \|x\|^2+\|Ax\|^2. $$ Therefore, if $(I+A)x=0$, then $x=0$.