On a Hilbert- (or otherwise inner-product-) space $\mathcal{H}$ with scalar product $S = \langle.|.\rangle_\mathcal{H}$, a self-adjoint operator is is readily defined as a linear mapping $A : \mathcal{H} \to \mathcal{H}$ with the property
$$
S (A v, w) = S(v,A w) \quad \forall v,w\in \mathcal{H}.
$$
Fair enough, but an inner product is a pretty specific structure on a space. Do we need it to define what self-adjoint means?
It seems that an operator which is self-adjoint with respect to $S$ is also self-adjoint with respect to another scalar product $T$, but I really don't see how one could go about proving this.
If so, would there be a definition of self-adjointness that makes no reference to any particular inner product? A useful candidate would be something like “a self-adjoint operator is one that has a system of eigenvectors $(\psi_i)_i$ which spans the entire space, such that $A(\psi_i) = \lambda_i\cdot \psi_i$ with real $\lambda$”. But is that actually equivalent to the usually given definition? The spectral theorem only goes one way, and makes itself reference to an inner product.
Best Answer
Self-adjointness is the operator-theoretic version of symmetry of bilinear forms. The notion of a symmetric bilinear form can be defined without reference to any extra structure but in order to talk about self-adjoint operators, we need some mechanism to convert operators to bilinear forms. An inner product $g$ allows us to do just that. Namely, let $V$ be a finite dimensional real vector space and let $g$ be an inner product. Given a linear map $T$, we can define the bilinear form associated to $T$ to be $\beta^g_T(v,w) := g(Tv,w)$. This gives us a map (in fact, a linear isomorphism) from $\operatorname{End}(V) \cong V^{*} \otimes V$ to $\operatorname{Bil}(V \times V, \mathbb{R}) \cong V \otimes V$.
The operator $T$ is $g$-self-adjoint if the associated bilinear form $\beta^g_T$ is symmetric. By changing $g$ we get different different maps between $\operatorname{End}(V)$ and $\operatorname{Bil}(V \times V, \mathbb{R})$ so $T$ might be $g$-self-adjoint but not $g'$-self-adjoint. This is best seen by invoking the spectral theorem.
In the real case, we know that $T$ is $g$-self-adjoint if and only if $T$ is $g$-orthogonally diagonalizable. Denote by $\lambda_1, \dots, \lambda_k$ the distinct eigenvalues of $T$. Then $V = \bigoplus_{i=1}^k \ker(T - \lambda_i I)$ is a direct sum orthogonal decomposition of $V$. If we change the inner product $g$ to an inner product $g'$ such that some $\ker(T - \lambda_i I)$ won't be $g'$-orthogonal to some $\ker(T - \lambda_j I)$ (with $i \neq j$) then by the spectral theorem, $T$ won't be self-adjoint. This is always possible as long as $k > 1$. In other words, the only operators $T$ which are self-adjoint with respect to all inner products are the scalar operators.
In fact, a real operator $T$ is $g$-self-adjoint with respect to some inner product $g$ if and only if $T$ is diagonalizable. If $T$ is self-adjoint, it is orthogonally diagonalizable and in particular diagonalizable. For the other direction, if $T$ is diagonalizable we can always take an arbitrary basis of eigenvectors of $T$ and define an inner product $g$ by declaring this basis to be orthonormal - then by the spectral theorem $T$ will be $g$-orthogonal.