Find the point spectrum of $Ax=\sum_{n=1}^{\infty}\lambda_n\langle x,e_n\rangle e_n$

Question

If a set $\{e_n:n\in\mathbb N\}$ is linearly independent, meaning that any finite number of the $e_n$ are linearly independent, can I conclude that $$\sum_{n=1}^{\infty}\alpha_ne_n=0\implies\alpha_n=0$$ for all $n$? I'm asking because I'm doing a question which asks for the point spectrum of the operator $A:H\to H$ where $H$ is a Hilbert space, $$Ax=\sum_{n=1}^{\infty}\lambda_n\langle x,e_n\rangle e_n$$ where $\lambda_n$ is a non-increasing sequence of non-negative numbers, and $e_n$ is an orthonormal basis for $H$. So if $\mu$ is such that $Ax=\mu x$ for some non-zero $x$, then since $x$ can be written as $\sum_{n=1}^{\infty}\langle x,e_n\rangle e_n$ we get
$$\sum_{n=1}^{\infty}\lambda_n\langle x,e_n\rangle e_n=\mu \sum_{n=1}^{\infty}\langle x,e_n\rangle e_n$$ which we can write as $$\sum_{n=1}^{\infty}(\lambda_n -\mu)\langle x,e_n\rangle e_n=0$$ so if what I first asked were to hold about linear indepedence I could conclude that for each $n$, either $\lambda_n -\mu=0$ or $\langle x,e_n\rangle=0$ which gives me the eigenvalues $\{\lambda_k:k\in\mathbb N\}$ with corresponding eigenvectors $e_k$. I don't think $\mu=0$ is necessarily an eigenvalue. I think this is the right answer I just don't know if it's justified

Answer 1

No, that is not true. Consider, for instance, the Hilbert space $\ell^2$ of all sequences $(a_n)_{n\in\Bbb N}$ such that $\sum_{n=1}^\infty|a_n|^2<\infty$. Let $e_n$ be the sequence $0,0,\ldots,0,1,0,0\ldots$, with the $1$ located at the $n$^th position. And let $e_0$ be the sequence $1,\frac12,\frac13,\frac14,\ldots$ Then $\{e_n\mid n\in\Bbb Z_+\}$ is linearly independent, but$$e_0=\sum_{n=1}^\infty\frac1ne_n.$$