Proofs that rely on an infinite matrix If I have an operator $A\in B(\mathcal{H})$ that can be "identified"
with an infinite matrix with countably many entries, is it in any way unrigorous to do actual calculations with the picture we have in mind of finite matrices. i.e. is it wrong to do a calculation like 
$$ Ah = 
\begin{bmatrix}
a & b & \cdots \\ c& d & \cdots  \cdots\\ e& f & \ddots
\end{bmatrix}
\begin{bmatrix} x\\ y \\ z\\ \vdots\end{bmatrix}
= \begin{bmatrix} ax + by +\cdots\\ cx+dy +\cdots \\ ex+fy+\cdots\\ \vdots\end{bmatrix}
$$
Put another way, is this well-defined and always give the same result that we'd expect if we did it in picture-free manner?
 A: Clarification on my earlier comment.

For example, infinite matrix multiplication is not necessarily associative. 

Look at the 3 infinite matrices:
$$\begin{cases}
A = \begin{bmatrix} 1 & 1 & 1 & \dots \end{bmatrix} \\
B = \begin{bmatrix} +1 & -1 & 0 & 0 & \dots \\
                          0 & +1 & -1 & 0 & \dots \\
                          0 & 0 & +1 & -1 & \dots \\
                          0 & 0 & 0 & +1 &  \dots \\ & & \vdots & & \end{bmatrix} \\
C = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \end{bmatrix}
\end{cases}$$
Now consider $X_1 = (AB)C $ and $X_2 = A(BC)$.  A direct evaluation gives $X_1 = \begin{bmatrix} 1 \end{bmatrix}$ but $X_2 = \begin{bmatrix} 0 \end{bmatrix}$.  

The reason is that an infinite sized matrix doesn't have a "last" element, which can make them not exactly the same as the limit of a finite matrix.

Consider if you define 
$$\begin{cases}
A_n \text{ is } 1 \text{ by } n \text{ Matrix} & A_n = \begin{bmatrix} 1 & 1 & 1 & \dots & 1\end{bmatrix} \\
B_n \text{ is } n \text{ by } n \text{ Matrix} & B = \begin{bmatrix} +1 & -1 & 0 & 0 & \dots \\
                          0 & +1 & -1 & 0 & \dots & 0 & 0\\
                          0 & 0 & +1 & -1 & \dots & 0 & 0\\
                          0 & 0 & 0 & +1 &  \dots & 0 & 0\\
                          & & \vdots & & \\
                          0 & 0 & 0 & 0 & \dots & +1 & -1\\
                          0 & 0 & 0 & 0 & \dots & 0 & +1\\
                          \end{bmatrix} \\
C_n \text{ is } n \text{ by } 1 \text{ Matrix} & C = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}
\end{cases}$$
Then define $$X_3 = \lim_{n \to \infty} A_n(B_nC_n)$$
You'll see that here $X_3 = \begin{bmatrix} 1 \end{bmatrix} \ne X_2$.  It's a curious thing to me, since that means that conventional infinite matrix multiplication isn't a direct limit.  It could be defined that way, but maybe it is and maybe it isn't.  You have to be careful.
A: You may be able to get away with this if the operator $A$ is a bounded linear operator defined on all of a separable Hilbert space. However, if the operator is unbounded, the story changes quite radically because there are simple unbounded operators with different domains that have the same diagonal matrix representations with respect to a basis, but which are very different in nature and spectrum.
The simplest example is the differentiation operator $Af = \frac{1}{i}\frac{df}{dt}$ defined its largest natural domain consisting of all $f \in L^{2}[0,2\pi]$ which are equal a.e. to absolutely continuous functions with derivatives in $L^{2}[0,2\pi]$. This is a densely-defined closed linear operator
$$ A : \mathcal{D}(A_{0})\subset L^{2}[0,2\pi]\rightarrow L^{2}[0,2\pi]. $$
The adjoint $A_{0}=A^{\star}$ of this operator is the restriction of $A$ to the functions which vanish at the endpoints of $[0,2\pi]$. The periodic restriction $A_{p}$ of $A$ to functions $f \in \mathcal{D}(A)$ with $f(0)=f(2\pi)$ is selfadjoint. In terms of graph inclusion,
$$
                             A_{0} \prec A_{p}=A_{p}^{\star} \prec A.
$$
So these operators are very different because of small differences in their domains. However, they are all the same when considered on the orthonormal basis of $L^{2}[0,2\pi]$ obtained by normalizing the orthogonal functions $\{ \sin(nx/2)\}_{n=1}^{\infty}$. That's because $\sin(nx/2)$ is the domain of all three operators, and the operators agree on the common domain. But these operators are critically different in character, and obviously those differences cannot be recovered from knowing the matrix entries only.
This ambiguity is greatly amplified when looking at Partial Differential Operators.
