Let $\mathcal{H}$ be a Hilbert space over the field $\mathbb{K}$ and $T:\mathcal{H}\to\mathcal{H}$ be a bounded positive linear mapping such that $\mathrm{dim}(\mathcal{R}(T))=1$ and $\sum_{k\in\mathcal{I}}\left<T(e_k),e_k\right>=1$ for some Hilbert basis $(e_k)_{k\in \mathcal{I}}$ where the index set $\mathcal{I}$ can be countable or uncountable. I want to show that $T$ is an orthogonal projection onto the span of the single linearly independent element of its range. To be precise,
$$\mathcal{R}(T):=\{T(v)\mid v\in\mathcal{H}\}$$
and as $\mathrm{dim}(\mathcal{R}(T)) = 1$ it follows that $\exists w\in\mathcal{H}:\forall v\in\mathcal{H}:\exists c_v\in\mathbb{K}:T(v) = c_vw$. I have a suspicion that the scalar $c_v$ is given by the inner product $\left<v,w\right>$ with linearity in the first argument. Since $T$ is a bounded positive linear mapping, it suffices to show that $T$ is a projection, i.e. $T^2 = T$ and $T|_S=I$ for some $S\subset\mathcal{H}$.
My problem: Confirming the defining two properties of a projection mapping would be quite easy if we were to know the image of $w$ under $T$. Namely, if we were to know that $T(w) = 1\cdot w = w$, then for any $v := cw,c\in\mathbb{K},$ we would have $T(cw) = cT(w) = cw = v$ showing that $T = I$ on the span of $w$ and $T^2 = T$.
My problem is that I don't know how to a.) conclude that $T(w) = w$ b.) use the assumption $\sum_{k\in\mathcal{I}}\left<T(e_k),e_k\right>=1$ to my advantage. Since the $e_k$s form a Hilbert basis, $w = \sum_{k\in\mathcal{C}}\beta_ke_k,\beta_k\in\mathbb{K}$ for some countable $\mathcal{C}\subset\mathcal{I}$ and $T$ is continuous, I suppose I would have to look at some double series of the form $\sum_{k_1\in\mathcal{I}}\sum_{k_2\in\mathcal{C}}\beta_k\left<T(e_{k_2}),e_{k_1}\right>$ and conclude something from this. But as of now it is not clear to me what I should do.
The operator $T$ (explanation at the end) must be of the form $$Tx=\lambda\langle x,v\rangle v,\quad \lambda>0, \|v\|=1$$ Thus $$1=\sum_{k\in \mathcal{I}}\langle Te_k,e_k\rangle =\lambda\sum_{k\in\mathcal{I}}|\langle v,e_k\rangle |^2=\lambda\|v\|^2=\lambda$$ Hence $Tx=\langle x,v\rangle v$ is the orthogonal projection onto $\mathbb{K}v.$ Explanation Since $T$ is one dimensional it must of the form $$Tx=\alpha\langle x,w\rangle v$$ where $\alpha\in \mathbb{K}$ and $\|v\|=\|w\|=1.$ Namely $v$ is a unit vector spanning the range of $T$ while $v$ is a unit vector spanning $(\ker T)^\perp.$ Mind that due to the continuity of $T$ the space $\ker T$ is closed, of codimension $1.$
$1.$ Assume $\mathbb{K}=\mathbb{C}.$ Then $T$ is self adjoint. Observe that $T^*x=\overline{\alpha}\langle x,v\rangle w.$ Indeed, we have $$\langle T^*x,y\rangle =\langle x,Ty\rangle =\overline{\alpha}\langle x,v\rangle \langle w,y\rangle =\langle \overline{\alpha}\langle x,v\rangle w,y\rangle $$ As $T=T^*$ we may conclude that $w=\overline{\beta} w,$ where $|\beta|=1.$ Hence $$Tx= \alpha\beta \langle x,v\rangle v= \lambda\langle x,v\rangle v,\quad \lambda=\alpha\beta$$ Due to the positivity we get $\lambda>0.$
$2.$ Let $\mathbb{K}=\mathbb{R}.$ Assume for a contradiction that $w\neq \pm v.$ Then $\langle v,w\rangle^2<1.$ For real number $t$ we have $$0\le \langle T(t v+w),t v+w\rangle \\ =\alpha\left \{\langle v,w\rangle t^2+[\langle v,w\rangle^2+1]t+\langle v,w\rangle\right \}$$ If $\langle v,w\rangle =0$ the expression does not have constant sign. For $\langle v,w\rangle \neq 0$ the expression in the brackets is a quadratic function with discriminant equal $$ (1-\langle v,w\rangle^2)^2>0$$ Hence the expression changes sign, also when multiplied by the nonzero real constant $\alpha.$