Endomorphisms, unitary map and composition

Question

I have trouble figuring out if the following proposition is true or false.

Let $f\in End(V)$ be normal and let $g\in U(V)$ (where $U(V)$ denotes the unitary group). Then the function composition $g^{-1} \circ f \circ g$ ist normal.

Thanks in advance, Hofmusicus.

Answer 1

Let $(V,\left< \cdot, \cdot \right>)$ be a complex inner product space. The map $f \mapsto f^{*}$ (from $\operatorname{End}(V)$ to itself) defined using the inner product satisfies the following useful properties:

1. For all $a,b \in \mathbb{C}$ and $f,g \in \operatorname{End}(V)$ we have $(af + bg)^{*} = \overline{a}f^{*} + \overline{b}g^{*}$.
2. For all $f,g \in \operatorname{End}(V)$ we have $(f \circ g)^{*} = g^{*} \circ f^{*}$.
3. For all $f \in \operatorname{End}(V)$ we have $\left( f^{*} \right)^{*} = f$.

Note that properties $(2)$ and $(3)$ above imply that

$$ \operatorname{id}^{*} \circ f = \left( f^{*} \circ \operatorname{id} \right)^{*} = \left( f^{*} \right)^{*} = f, \\ f \circ \operatorname{id}^{*} = \left( \operatorname{id} \circ f^{*} \right)^{*} = \left (f^{*} \right)^{*} = f $$

for all $f \in \operatorname{End}(V)$ and so $\operatorname{id}^{*} = \operatorname{id}$. In addition, if $f$ is invertible then property $(2)$ and the observation above imply that

$$ \operatorname{id} = \operatorname{id}^{*} = (f \circ f^{-1})^{*} = \left( f^{-1} \right)^{*} \circ f^{*}, \\ \operatorname{id} = \operatorname{id}^{*} = (f^{-1} \circ f)^{*} = \left( f \right)^{*} \circ \left( f^{-1} \right)^{*}, $$

and so $f^{*}$ is also invertible with inverse $\left( f^{*} \right)^{-1} = \left( f^{-1} \right)^{*}$.

---

Using the properties above and the fact that $g \circ g^{*} = g^{*} \circ g = \operatorname{id}$ (and so $g^{-1} = g^{*}, \left( g^{*} \right)^{-1} = g$) and $f \circ f^{*} = f^{*} \circ f$, we have

$$ (g^{-1} \circ f \circ g)^{*} \circ (g^{-1} \circ f \circ g) = g^{*} \circ f^{*} \circ \left( g^{-1} \right)^{*} \circ g^{-1} \circ f \circ g = g^{-1} \circ f^{*} \circ \left( g^{*} \right)^{*} \circ g^{*} \circ f \circ g \\ = g^{-1} \circ f^{*} \circ g \circ g^{*} \circ f \circ g = g^{-1} \circ f^{*} \circ f \circ g = g^{-1} \circ f \circ f^{*} \circ g \\ = g^{-1} \circ f \circ g \circ g^{*} \circ f^{*} \circ \left( g^{-1} \right)^{*} = (g^{-1} \circ f \circ g) \circ \left( g^{-1} \circ f \circ g\right)^{*} $$

which shows that $g^{-1} \circ f \circ g$ is normal.