= Solution
The <adjoint on an inner product space> $\alpha^*:V\to V$ is characterized by
$$
\langle\alpha v,w\rangle=\langle v,\alpha^*w\rangle\qquad(v,w\in V).
$$
Here is a finite-dimensional construction that supplies every required <basis> fact. If $v_1,\ldots,v_n$ is any <basis>, its coordinate <linear functionals> form a <basis> of $V^*$: a <linear functional> is uniquely specified by its values on these $n$ <vectors>. Hence $\dim V^*=n$. The <linear map>
$$
J:V\to V^*,\qquad J(w)(v)=\langle v,w\rangle
$$
is <injective>, because $J(w)=0$ implies $\langle w,w\rangle=0$, and <positive definiteness> implies $w=0$. Equal <dimensions> make $J$ an <linear isomorphism>. If $\alpha'$ is the <dual map>, then
$$
\alpha^*=J^{-1}\alpha'J.
$$
This establishes existence and <linearity>. If two possible <vectors> $u,w$ represent the same <linear functional> through the <inner product>, then $J(u-w)=0$, so $u=w$. Applied to every argument, this proves uniqueness of the <adjoint on an inner product space>.
For <linear maps> $S,T$, two applications of the defining <inner product> identity give
$$
\langle STv,w\rangle=\langle Tv,S^*w\rangle=\langle v,T^*S^*w\rangle.
$$
Uniqueness therefore gives $(ST)^*=T^*S^*$. Likewise $(S+T)^*=S^*+T^*$, $(cT)^*=cT^*$ for real $c$, and $I^*=I$. An <induction> gives $(\alpha^n)^*=(\alpha^*)^n$, and applying additivity to a <polynomial> with real coefficients gives
$$
\boxed{f(\alpha)^*=f(\alpha^*).}
$$
Symmetry of the real <inner product> also gives $(\alpha^*)^*=\alpha$. If $m$ is the <minimal polynomial> of $\alpha$, the displayed identity gives $m(\alpha^*)=0$, so the <minimal polynomial> of $\alpha^*$ divides $m$. Applying the same argument to $\alpha^*$ gives reverse divisibility. Both <polynomials> are <monic>, hence \b[their minimal polynomials are equal].
The distinct <eigenvalues> are exactly the roots of the <minimal polynomial> in the chosen <field>. An <eigenvector> $v$ with <eigenvalue> $\lambda$ satisfies $m(\alpha)v=m(\lambda)v$, so $m(\lambda)=0$. Conversely, if $m(X)=(X-\lambda)q(X)$ and $\alpha-\lambda I$ were a <linear isomorphism>, then $m(\alpha)=0$ would imply $q(\alpha)=0$, contradicting minimal degree. Thus $\alpha-\lambda I$ is not <injective>, giving an <eigenvector>. It follows that $\alpha$ and $\alpha^*$ have \b[the same distinct eigenvalues]. This includes the complex <eigenvalues> after <complexification of a real vector space>; over the real <field> it states equality of their real <eigenvalues>.
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