 Solution

The adjoint on an inner product space ${\alpha }^{\ast }:V\to V$ is characterized byHere is a finite-dimensional construction that supplies every required basis fact. If $v_1,\dots ,v_n$ is any basis, its coordinate linear functionals form a basis of $V^{\ast }$: a linear functional is uniquely specified by its values on these $n$ vectors. Hence $\dim V^{\ast }=n$. The linear mapis injective, because $J(w)=0$ implies $⟨w,w⟩=0$, and positive definiteness implies $w=0$. Equal dimensions make $J$ an linear isomorphism. If ${\alpha }^{\prime }$ is the dual map, thenThis 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 giveUniqueness therefore gives $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Likewise $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, $(cT{)}^{\ast }=c{T}^{\ast }$ for real $c$, and $I^{\ast }=I$. An induction gives $({\alpha }^n{)}^{\ast }=({\alpha }^{\ast }{)}^n$, and applying additivity to a polynomial with real coefficients givesSymmetry of the real inner product also gives $({\alpha }^{\ast }{)}^{\ast }=\alpha$. If $m$ is the minimal polynomial of $\alpha$, the displayed identity gives $m({\alpha }^{\ast })=0$, so the minimal polynomial of ${\alpha }^{\ast }$ divides $m$. Applying the same argument to ${\alpha }^{\ast }$ gives reverse divisibility. Both polynomials are monic, hence 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 ${\alpha }^{\ast }$ have 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.