I see that this question has been asked, but no one answered it yet, and I have the same question. Therefore, I will ask it again here and post the link to that very same question: Related to Euclidean and unitary vector space
Let $V$ be a finite-dimensional Euclidean or unitary $K$-vector space. Show or refute the following statements:
(i) $(f + g)^{*} = f^{*} + g^{*}$ for all $f, g ∈ \operatorname{End}(V)$
(ii) $(λf)^{*}$ = $λf^{*}$ for all $ f ∈ \operatorname{End}(V)$ and $λ ∈ K$
(iii) $(f ◦ g)^{*}$ = $f^{*}$ $◦$ $g^{*}$ for all $f, g ∈ \operatorname{End}(V)$
(iv) $f^{-1}$$(U^{⊥})$ = $f^{*}(U)^{⊥}$ for all $f ∈ \operatorname{End}(V)$ and all subspaces $U ⊆ V$
(v) $\operatorname{rk}(f)$ = $\operatorname{rk}(f^{*})$ for all $f ∈ \operatorname{End}(V)$
What are the properties related to the $^{*}$, $◦$, and $^{⊥}$ and $\operatorname{End}(V)$ that may help show or refute the statements?
Extra notes:
$^{*}$ $=$ Hermitian adjoint
$^{⊥}$ $=$ orthogonal complement
$End(V)$ = endomorphism of the unitary space V
unitary space = complex inner product space
Let us denote by $\cdot $ the dot product:
$1-$Let $ x, y $ be elements of $ V $. Note that: $$ (f + g) (x) \cdot y = f (x) \cdot y + g (x) \cdot y = x \cdot f^{*} (y) + x \cdot g^{*} (y) = x \cdot (f^{*}+ g^*) (y) $$ therefore $ (f + g)^* = f^* + g^{*} $.
$2-$ False, $(\lambda f)(x)\cdot y = \lambda (f(x)\cdot y)=\lambda (x\cdot f^{*}(y))=x\cdot (\lambda f^{*}))(y)$, therefore $(\lambda f)^{*}=\lambda f^{*}$ (only true if $K=\mathbb{R}$, otherwise $(\lambda f)^{*}=\alpha f^{*}$, where $\alpha$ is the conjugate of $\lambda$).
$3-$ False, $(g^{*}\circ f^{*})(x)\cdot y=g^{*}(f^{*}(x))\cdot y=f^{*}(x)\cdot g(y)=x\cdot f(g(y))=x\cdot (f\circ g)(y)$, therefore $(f\circ g)^{*}=g^{*}\circ f^{*}$.
$4-$ If $x\in f^{-1}(U^{\perp})$, it is true that $f(x)\in U^{\perp}$, therefore $f(x)\cdot u=0$, for all $u\in U$. Since $0=f(x)\cdot u=x\cdot f^{*}(u)$, hence $x\in (f^{*}(U))^{\perp}$. The reciprocal is clearly true.
$5-$ False. Let $ V $ be the space of polynomials with real coefficients of degree less than $ n $, where $ n> 2 $. Consider the inner product of $ p $ and $ q $ as the integral of $ p (x) q (x) $ over the interval $ (0,1) $, and let $ f $ be the Derivative operator, which is an endomorphism. If $ rk (f) = rk (f ^ {*}) $, then $ (\ker f) ^ {\perp} = rk (f) $ because $ rk (f^{*}) = (\ker f )^{\perp} $ in finite dimension. But $ (\ker f)^{\perp} = L (1)^{\perp} $, where $ L(1) $ is the subspace of constant polynomials. However, the polynomial $p (x) = 2x$ belongs to $V$, and is the derivative of $x^2$ which is also in $V$, that is, $f (x^2) = 2x $, so $ x^{2} \in rk (f) $, but $ x^{2} $ is not in $ L(1)^{\perp}$ because its integral over $ (0, 1) $ is $1/3 \neq 0 $.