Let $T : H \to H$ be a compact self adjoint operator on a Hilbert space with spectrum $\sigma(T) = \{\lambda_n\}_{n \in F}$ with $F$ countable (the latter guaranteed by the spectral theorem).
I want to show that the definition of the operator $$ \widetilde{T}x := \sum_{n \geq 1}\alpha_nP_{E_{\lambda_n}}(x), \quad \alpha \in \ell^\infty $$
does not depend on the ordering of the eigenvalues (here $P_{E_{\lambda_n}}$ denotes the eigenspace projection).
I've argued as follows. Since $H = \bigoplus_{\lambda \in \sigma(T)}E_\lambda$, by definition we have that the space generated by the eigenspaces is dense on $H$ and these are pairwise orthogonal. Thus, is suffices to see that any reordering
$$ \widetilde{T}_\rho \ x := \sum_{n \geq 1}\alpha_{\rho(n)}p_{E_{\lambda_\rho(n)}}(x) $$
with $\rho \in S(F)$ agrees with $\widetilde{T}$ in each eigenspace (and this is immediate from the definition of both operators).
Does this seems sound? That is, can we always that if $H$ is the hilbert sum of certain subspaces $(E_n)_{n \geq 1}$, then two operators $T,S \in \mathcal{L}(H)$ are equal if and only if they agree on each space $E_i$?
This seems true to me in the subcase of an orthonormal basis, but I'm not sure if I'm missing something when generalizing.
If $\mathcal{H} = \bigoplus_\alpha M_\alpha$ in the Hilbert space sense (the subspaces are orthogonal and the algebraic sum generates $\mathcal{H}$), then if $T,S$ are bounded operators that agree on $\forall \alpha \ M_\alpha$, then since a general element of $x \in \mathcal{H}$ can be expressed as $x = \sum_\alpha a_\alpha v_\alpha$ where the sum is at most countable, Since the operators are bounded you have $$T(x) = \sum_\alpha a_\alpha T(v_\alpha) = \sum_\alpha a_\alpha S(v_\alpha) = S(x)$$
So yeah, if the sum is in the Hilbert space sense, agreeeing on subspaces means agreeing on the whole space. In particular this applies to your case.