proof of Invertible matrix Let $U$ be a n by n matrix, satisfy $U^T=-U$
Prove that $U-I$ is invertible.
So, I have some hints, $\vec{v}^T\vec{v}$ for $\vec{v} \in Null(U-I)$
 A: If $v \in \text{Null}(U-I)$, then $Uv=v$. Then
$$v^\top v = v^\top (U v) = (U^\top v)^\top v = \cdots$$
Can you use the fact $U^\top = -U$ and finish from here?
A: If $v$ is a vector such that $(U-I)v=0$ then $Uv=Iv=v$, and taking the transpose $v^\mathsf tU^\mathsf t=-v^tU=v^t$. Multiplying by $v$ on the right we get $v^\mathsf tUv=-v^\mathsf tv=-\|v\|^2$. Since $Uv=v$ we get $v^\mathsf tv=-\|v\|^2$, so $\|v\|^2=-\|v\|^2$, which is only possible if $v=0$. Therefore $\text{null}(U-I)=\{0\}$, so $U-I$ is invertible.
A: I assume $U$ is a real matrix.
Note that
$(I + U)(I - U) = I - U^2; \tag 1$
given
$U^T = -U, \tag 2$
we have
$U^2 = -U^T U = -UU^T, \tag 3$
whence
$(I + U)(I - U) = I + U^TU; \tag 4$
now if $I - U$ were not invertible, there would be a vector $v \ne 0$ such that
$(I - U)v = 0, \tag 5$
whence
$(I + U^TU) v = (I + U)(I - U)v = 0; \tag 6$
but
$\langle v, (I + U^TU) v \rangle = \langle v, Iv \rangle + \langle v, U^TUv \rangle$
$= \langle v, v \rangle + \langle Uv, Uv \rangle = \Vert v \Vert^2 + \Vert Uv \Vert^2  \ge \Vert v \Vert^2 > 0, \tag 7$
since $v \ne 0$, in contradiction to (6); thus there is no $v$ such that (5) holds, and thus $I - U$ is invertible, as is
$U - I = -(I - U). \tag 8$
