Proving the map $f-i\cdot \text{id}_v$ is invertible with $f:V\to V$ a self-adjoint linear map

Question

I am struggling with the following question.

Let $V$ be a finite dimensional complex inner product space and let $f:V\to V$ be a self-adjoint linear map. Show that the map $f-i\cdot \text{id}_v$ is invertible and that the linear map $g$ given by $$g=(f+i\cdot \text{id}_v)(f-i\cdot \text{id}_v)^{-1}$$ is unitary by verifying that $g^*g=\text{id}_v$.

I assume that using the inverse of $f-i\cdot \text{id}_v$ is useful in verifying $g^*g=\text{id}_v$. I also imagine that for the second part using the fact that $(fg)^*=g^*f^*$ and $(f^*)^{-1}=(f^{-1})^*$ for linear maps $f$ and $g$ might be useful.

Answer 1

I write $f - \mathrm i$ for $f - \mathrm i \cdot \operatorname{id}_v$. We first show that $f - \mathrm i$ is invertible. Note that by $\dim(V) < \infty$ it therefore suffices to show that $\ker f = \{ 0 \}$. Assume $0 \neq v \in \ker f$. Then we have $fv = \mathrm i v$, which is a contradiction: It would imply that $f$ has eigenvalue $\mathrm i$, but this can't be true since $f$ is self-adjoint and therefore has real eigenvalues.

In order to show that $g$ is unitary, I now use the two properties you already stated and the basic observation that $(f + \mathrm i)$ commutes with $(f - \mathrm i)$. It then follows that

\begin{align} g^\ast g &= ((f - \mathrm i)^{-1})^\ast (f + \mathrm i)^\ast (f + \mathrm i) (f - \mathrm i)^{-1} \\ &=(f + \mathrm i)^{-1}(f - \mathrm i)(f + \mathrm i) (f - \mathrm i)^{-1} \\ &=(f + \mathrm i)^{-1} (f + \mathrm i) (f - \mathrm i) (f - \mathrm i)^{-1} \\ &= \mathrm{id} \circ \mathrm{id} \\ &= \mathrm{id}. \end{align}