Find correlation between spectrum of f and f* I was looking through my old algebra taskbook solving most of the tasks with no effort, however this is the one I've struggled with for rather long time:
For $F=\mathbb{C}$ (field $\mathbb{C}$) find any correlation between spectrum (all Eigenvalues) of following linear operators $f$ and $f^{*}$.
Do you have any ideas about it?
 A: First things first: I think we need to get some terms straight. The *-operation you mention is supposed to be the operation that maps a linear map $A$ to its adjoint map $A^*$. For $n \times n$ matrices taking the adjoint just means to take the transpose and complex conjugate all entries (this is a standard result from linear algebra). In your case we have a linear map on a finite dimensional vector space, and from linear algebra we know that if the linear map $f$ is to be interpreted as a $1 \times 1$ matrix, then $f^*$ is just the pointwise complex conjugate of $f$, that is $f^*(x)= \overline{f(x)}$ for all $x$. So $f^*$ is really just the pointwise conjugation $\overline{f}$.
Now the spectrum of a linear map is usually defined to be the set $$\sigma(f) = \{\lambda \in \mathbb{C} \mid f-\lambda I \text{ is not invertible}\}$$ where $I$ denotes the identity map. In the finite dimensional case, the spectrum coincides with the set of eigenvalues, i.e. $\sigma(f)=\{\lambda \in \mathbb{C} \mid \exists v \neq 0 \colon f(v) = \lambda v\}$. 
Getting back to your actual question:

The spectrum of $f^*$ satisfies $\sigma(f^*) = \{\overline{\lambda} \mid \lambda \in \sigma(f)\}$. This holds since if $\lambda \in \sigma(f^*)$, then by definition of the spectrum $f^*-\lambda I = (f-\overline{\lambda} I)^*$ is not invertible. But if $(f-\overline{\lambda} I)^*$ is not invertible, then we also have that $f-\overline{\lambda} I$ is not invertible. Thus, by definition of the spectrum, $\overline{\lambda} \in \sigma(f)$. Analogously $\lambda \in \sigma(f)$ implies $\overline{\lambda} \in \sigma(f^*)$.
  The very easy thing about your map being one-dimensional is that you only have one eigenvalue for $f$, that is $\sigma(f)=\{\lambda\}$. Thus $\sigma(f^*)=\{\overline{\lambda}\}$.

