Hamiltonian of oscillators quantized proof https://docs.google.com/open?id=0BxrBcN1-BZWUOXNxR1l4S0l2MjQ
http://www.2shared.com/complete/Qjy1_uzp/Quantum_Mechanics_in_Simple_Ma.html 
(I uploaded a pdf file that contains the parts of the textbook that I am having some difficulty understanding. Seven pages.) (two links point to the same pdf.)
A matrix $N$ has all entries except diagnoal ones zero, and first row, first column also zero, other diagonal entries depending on its column, row place; for example, second row, second column has one as its entry, third row, third column has two as its entry and so on.
It is known that $H = \hbar\omega (N + \frac{1}{2}) $.
The question is, my textbook says that if there are a continuous range of possible values, then for some states, $\langle (N - \langle N \rangle )^2 \rangle \leq [\frac{1}{2} (n - \langle N \rangle)]^2$ , where $n = 0,1,2,3,...,$ then progresses to say that since $\langle (N - \langle N \rangle )^2 \rangle \geq (n - \langle N \rangle)^2 (\langle I_0 \rangle + \langle I_1\rangle + ...)$ then $\langle (N - \langle N \rangle )^2 \rangle \geq (n - \langle N \rangle)^2$. 
Then it says that this proves that there cannot be a continuous range of possible values, as $(0-\langle N \rangle)^2$, $(1 - \langle N \rangle)^2$ and so on cannot be smaller than $(n - \langle N \rangle)^2$.
I am not sure how one gets $\langle (N - \langle N \rangle )^2 \rangle \geq (n - \langle N \rangle)^2 (\langle I_0 \rangle + \langle I_1\rangle + ...)$. Can anyone show me the proof?
Also, how does one get the part - $(0-\langle N \rangle)^2$, $(1 - \langle N \rangle)^2$ and so on cannot be smaller than $(n - \langle N \rangle)^2$?
 A: The likely interpretation:
we assume there is a continuous range of possible values, then we are able to measure it so that we get some particular value of $\langle N \rangle$ and particular uncertainty, but still then $N$ is not exactly determined. In this context, for some states, $\langle (N - \langle N \rangle^2 \rangle \leq [\frac{1}{2}(n-\langle N \rangle)]^2$, where the left part is uncertainty. And I think the aforementioned part is where you got confused.
The next part will come as obvious.
A: I) It seems that the question(v4) is essentially 

Prove the that the point spectrum of the infinite-dimensional matrix 
  $$ N~:= \left(\begin{array}{ccccc} 
0 & 0 & 0 & 0 & \ldots  \\ 
0 & 1 & 0 & 0 & \\
0 & 0 & 2 & 0 & \\
0 & 0 & 0 & 3 & \\
\vdots &   &   &  &  \ddots\\
\end{array}\right) $$ 
  is the non-negative numbers 
  $${\rm Spec}_{\rm pt}(N)~=~ \mathbb{N}_0~:=~\{0,1,2,3,\ldots\}.$$ 

II) Now the (linear) operator $N$ is an unbounded operator, so let us for simplicity (in order to avoid various technical issues related to unbounded operators) instead ask

Prove the that the point spectrum of the bounded operator 
  $$ e^{-N}~:= \left(\begin{array}{ccccc} 
e^0 & 0 & 0 & 0 & \ldots  \\ 
0 & e^{-1} & 0 & 0 & \\
0 & 0 & e^{-2} & 0 & \\
0 & 0 & 0 & e^{-3} & \\
\vdots &   &   &  &  \ddots\\
\end{array}\right) $$ 
  is the numbers 
  $${\rm Spec}_{\rm pt}(e^{-N})~=~\{e^{-n}\mid n \in\mathbb{N}_0 \} ~=~\{e^{0},e^{-1},e^{-2},e^{-3},\ldots\}.$$ 

III) Or more generally

If a bounded operator $A$ is diagonal in an orthonormal basis $(|i\rangle)_{i\in I}$ for a Hilbert space 
  $$H~=~{\rm span}\{|i\rangle | i\in I\}$$ 
  with eigenvalues $\lambda_i\in \mathbb{C}$, show that the point spectrum is
  $$ {\rm Spec}_{\rm pt}(A)~=~ \{ \lambda_i | i\in I\}~\in \mathbb{C}. $$

This is almost a triviality. A sketched proof could be as follows:


*

*Since $A$ is a normal operator, the eigenspaces for different eigenvalues are pairwise orthogonal
$$ \forall i,j\in I:~~\lambda_i~\neq~\lambda_j\qquad \Rightarrow \qquad  
{\rm ker}(A-\lambda_i{\bf 1}) ~\perp~ {\rm ker}(A-\lambda_j{\bf 1}).$$

*Since $A$ is diagonal, the eigenspaces corresponding to the diagonal elements $\{ \lambda_i | i\in I\}$ span the full Hilbert space
$$H~=~\oplus_{i\in I} {\rm ker}(A-\lambda_i{\bf 1}). $$

*Hence there are no room for eigenvalues $\lambda$ that are not diagonal elements $\{ \lambda_i | i\in I\}$,
$${\rm ker}(A-\lambda{\bf 1}) ~\subseteq~ H^{\perp}~=~ \{0\}. $$ 
