Let $X$ be a separable Hilbert space, endowed with an inner product $\langle \cdot \,, \cdot \rangle_X$. Let $z_1, z_2, \ldots , z_N$ be fixed elements of $X$. Consider, \begin{equation*} V \stackrel{\text{def}}{=} \text{span} \{z_j \colon 1 \leq j \leq N\}, \end{equation*} and denote $\text{dim}(V) \stackrel{\text{def}}{=} d < \infty$. Define the linear map $\mathcal{S} \colon X \rightarrow X$ by \begin{equation*} S x \stackrel{\text{def}}{=} \sum_{j=1}^N \langle x \,, z_j \rangle_X z_j \end{equation*} This map is bounded, self-adjoint and compact (since its rank is finite). Consequently we may deduce from the Hilbert-Schmidt Theorem, that there exists an orthonormal basis $\{\psi_k\}_{k=1}^\infty$ of $X$ and $\{\lambda_k\}_{k=1}^\infty$ such that \begin{equation*} \mathcal{S}\psi_k = \lambda_k \psi_k \quad \text{and} \quad \lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_k \geq \ldots . \end{equation*}
Denote the number of positive eigenvalues by $p$, meaning that \begin{equation} \lambda_1 \geq \ldots \geq \lambda_p > 0 = \lambda_{p+1} = \lambda_{p+2} = \ldots \end{equation}
I am trying to understand why/if the number $p$ of positive eigenvalues of $\mathcal{S}$, is equal to the dimension $d$ of $V$.
What I tried: On the one hand, for any $k \in \{1 , \ldots , p\}$, the eigenvector $\psi_k$ rewrites as \begin{equation*} \psi_k = S\big(\frac{\psi_k}{\lambda_k}\big) \in \text{Ran}(\mathcal{S}) \end{equation*} hence $\text{span}\{\psi_k\}_{k=1}^p \subset \text{Ran}(\mathcal{S})$ and $p \leq \text{rank}(\mathcal{S})$ (the eigenvectors being independent in $X$). On the other hand, take $y \in \text{Ran}(\mathcal{S})$, then $y= \mathcal{S}x$ for some $x \in X$. Since $\{\psi_k\}_{k=1}^\infty$ is an orthonormal basis of $X$, the vector $x$ is equal to the the limit in $X$ of a sequence $(x_n) \subset \text{span}\big( \{\psi_k\}_{k=1}^\infty \big)$. The map $\mathcal{S}$ being bounded, \begin{equation*} \mathcal{S}x_n \xrightarrow[n \rightarrow \infty]{} \mathcal{S}x \quad \text{in }X. \end{equation*} However for any $n$ $\mathcal{S}x_n$ belongs to $\text{span} \{\psi_k\}_{k=1}^p$ which is closed (since finite dimensional), hence $\mathcal{S}x \in \text{span} \{\psi_k\}_{k=1}^p$. As a consequence, $\text{Ran}(\mathcal{S}) \subset \text{span} \{\psi_k\}_{k=1}^p$, and $\text{rank}(\mathcal{S}) \leq p$. Thus \begin{equation*} \text{rank}(\mathcal{S}) = p. \end{equation*}
Consequently, to answer my question, it would be sufficient to show that \begin{equation*} \text{rank}(\mathcal{S})=d. \end{equation*}
Since the range of $\mathcal{S}$ is included in $V$, $\text{rank}(\mathcal{S})\leq d$. But I don't know how to understand why/if $\text{rank}(\mathcal{S}) = d$. Could someone tell me if it is true, and if so, give an idea to show it?