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

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.

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}
• Thank you for your detailled answer. How do I prove that if $\lambda$ is an eigenvalue of $f$ then $(\lambda +i)(\lambda -i)^{-1}$ is an eigenvalue of the linear map $g$?