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This sounds suspicious right away since the eigenvectors are what they are (nothing to choose here, unless you have degeneracies), but there is much choice for $D$ and we should be able to avoid eigen …

Yes. If $x_n\in D(L)$ is an ONB of $H$ and $Lx_n=\lambda_n x_n$, then the operator $T$ acting in the obvious way on $D(T)=\{ \sum a_n x_n\in H : \sum \lambda_n^2|a_n|^2<\infty\}$ is self-adjoint. It i …

Even after addressing the issues raised in the comments, the matrix coefficients $A_{ij}$ don't give you enough information to find $D(A^*)$. For example, consider $A_j=-d^2/dx^2$ on $L^2(0,1)$, $j=1, …

It is probably best to do this by hand, by looking at the domains described in the usual way as those $f$ with $Lf\in L^2$ that satisfy certain boundary conditions at $x=0,1$ (in fact, there will be n …

Even if you can define $T$ by this limit, this operator need not be closed. Take for example $H=\ell^2$ and $T_n$ as multiplication by $(0,\ldots, 0, n, n+1,\ldots)$ on its natural domain $D$. Note th …

If we ignore possible spectral multiplicity, then we can realize $S,T$ as multiplication by the variable in $H=L^2([0,\infty), \mu)$ and $K=L^2([0,\infty),\nu)$, respectively. That makes $S\otimes T$ …

$L_1+L_f$ is closable. It is perhaps most convenient to take Fourier transforms and consider the operator $$ (Tg)(\xi) = \xi g(\xi) + \left( \int g(\xi)\, d\xi \right) h(\xi) $$ on $L^2(\mathbb R)$. $ …

This will work if you take the assumption that $A,B$ commute in a sufficiently strong sense (commuting resolvents would be enough). Then no extra assumption is needed. There is a version of the spect …

I'm expanding my comment, in response to the OP's comment. Indeed, the case of just the lowest eigenvalue is perhaps not a good illustration of the full argument. In general, let $u_j$ be a normalized …