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Hopelessly false. Consider the one-dimensional case (so the $T_j$ are just numbers). Any function $f$ for which $\sum_j T_j$ convergent implies $\sum_j f(T_j)$ convergent is linear in a neighbourhoo …

Because of the degeneracy in the eigenvalues of $A$, "the eigenvectors matrix of $A$" is far from well-defined. Rather, there is a one-dimensional eigenspace of $A$ for eigenvalue $n$ (spanned by $(1 …

For a $4 \times 4$ example with $0$'s on the diagonal, consider $$ A = \pmatrix{0 & 16 & 2 & 2\cr 2 & 0 & 3 & 2\cr 2 & 2 & 0 & 4\cr 4 & 2 & 2 & 0\cr},\ A' = \pmatrix{0 & 16 + x & 2 & 2\cr 2 - x & 0 …

No. For example, the spectral radii of $$ A = \pmatrix{0 & 1 & 1\cr 1 & 0 & 1\cr 0 & 0 & 0\cr},\ A' = \pmatrix{0 & 0 & 3\cr 1 & 0 & 1\cr 0 & 0 & 0\cr}$$ are $1$ and $0$ respectively.

By orthonormal, I suppose you mean $\text{tr}(C_i C_j) = 0$ for $i \ne j$, $\text{tr}(C_i^2) = 1$? The estimate you gave is tight, in the sense that it is an equality if, for example, all $w_i = \ep …

Yes, Rellich's theorem does not require the eigenvalues to be distinct. See e.g. Reed and Simon, "Methods of modern mathematical physics vol. 4: Analysis of Operators", Chapter XII (in particular Pro …

The eigenvalues of a square matrix $A$ are the roots of the characteristic polynomial, and are analytic except where their multiplicities change. Thus if (in a certain open region of parameter space) …