Proof of invertibility of $I+S$ such that $S^* = -S$

Question

We're given a linear transformation $S:V \rightarrow V$ on a finite dimensional inner product space $V$ such that its adjoint is negative $S$ (i.e. $S^* = -S$). We are able to prove that $(v+Sv,v+Sv)=(v,v) + (Sv,Sv)$ and that the kernel of $(I+S) = \{\mathbf{0}\}$.

Using these facts how can we deduce that the linear transformation $I+S$ is invertible?

Any help would be much appreciated!

Answer 1

If $T \colon V \rightarrow V$ if a linear map on a finite dimensional vector space with trivial kernel then $\dim \ker T + \dim \operatorname{Im} (T) = \dim V$ implies that $\dim \operatorname{Im}(T) = \dim V$ and so $T$ is also onto and invertible.

In your case, if you have shown that $I + S$ has trivial kernel it implies that $I + S$ is invertible.

Answer 2

The fact is true even if $V$ is not finite dimensional. Note that by the same reasoning, $(I+S)^*=I+S^*$ has trivial kernel, but in general, the kernel of $T^*$ is the orthogonal complement of the image of $T$ (this is easy to check using the obvious fact that $v$ is zero iff for all $w$ we have $(v,w)=0$).

If you apply this fact to $T=(I+S)$, it immediately follows that it must be onto, and since it has trivial kernel as well, we can deduce that it is invertible.