Clarification on point spectrum of an operator From my understanding, if $\lambda$ is in the point spectrum, then $\lambda$ is a complex number such that it satisfies the equation $(T - \lambda I) x = 0$.
My confusion arises from problems like the following: 
Suppose $T:l^{2} \to l^{2}$ and it is defined as $Tx = y$ where $x = (\xi_{j}), y = (\eta_{j})$ and $\eta_{j} = \frac{\xi_{j}}{j}$. 
To find the point spectrum: 
$(T - \lambda I)x = 0 \Rightarrow Tx - \lambda x = 0 \Rightarrow  \left(\frac{\xi_{j}}{j}\right) - \lambda (\xi_{j}) = 0$. Therefore $\frac{1}{k} \in \sigma_{p}(T)$ for $k = 1,2, ..$
To me, this seems to say that the eigenvalue is not a single complex number but a sequence. If I were to just choose $1$ or $\frac{1}{2}$ for $\lambda$, it wouldn't be true that the equation $(T - \lambda I) x = 0$ would be satisfied. So is $\lambda$ a sequence of complex numbers? 
If it is a sequence, then it seems that for a problem like: 
$T:l^{2} \to l^{2}$ defined as $Tx = y$ where $x = (\xi_{j}), y = (\eta_{j}), \eta_{2j} = \xi_{2j}$ and $\eta_{2j-1} = 0$
I should be able to choose the sequence $\lambda = (0,1,0,1,..)$ for the spectrum since $Tx = (0,\xi_{2},0,\xi_{4},0,...)$. 
 A: The following discussion holds for any complex Hilbert space $H$, in particular your example of $H = l^2$.
It's important to remember that the non-zero complex number $\lambda$ is in the point spectrum if there exists a vector $x_\lambda\neq 0$ such that the equality $(T-\lambda I)x_\lambda = 0$ holds.  Each eigenvalue ($\lambda$) is always a single complex number, i.e. an element of the "base field" of your vector space, while $x_\lambda$ is a vector, i.e. an element of the vector space $H$, which in your case is a sequence $x = (x_1,x_2,\ldots)$.  So for your example of $(Tx)_j = \frac{x_j}{j}$, you have shown that each number $\lambda$ of the form $\lambda = 1/k$ for some integer $k>0$ is an eigenvalue, because there exists an eigenvector for that value (you should work out what the eigenvector is for each $\lambda$!)
I think the confusion comes from saying "the eigenvectors take the form $\lambda_k = 1/k$", which makes them look like a sequence.  Sure, you could think of them as a sequence, but there is no reason to think so - the spectrum is simply a subset of $\Bbb{C}$, not an element of your vector space.  It's unfortunate that in your example, the vector space is a sequence space and the point spectrum also looks like a sequence.  This won't always happen, for instance you can have your vector space be $L^2$, a function space, and your point spectrum can still be a "discrete" sequence of numbers.
Hope that helps.
A: No. $\lambda$ is a number. First of all, definition of eigenvalues is "$\lambda\in\sigma_p(T)$ if there is a non zero vector $x$ such that $(T-\lambda I)x=0$".
In your case, $\lambda\in\sigma_p(T)$ iff there is an $x=(x_1,x_2,...)\neq 0$ such that $\frac{x_j}{j}=\lambda x_j$ for all $j$. Then, using that $x\neq 0$, there is at least one $j$ such that $x_j\neq 0$, and thus $\lambda=1/j$. But if there are more than two such $j$, say $j_1$ and $j_2$, then $\lambda=1/j_1$ and $\lambda=1/j_2$. Thus, it can't happen that $x$ has two non zero "coordinates". Therefore $x=(0,0,...)$ except for only one coordinate.
Conclusion: $\sigma_p(T)=\{1/j:j\in\mathbb{N}\}$. 
