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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?

Squirtle
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  • In particular the indexing is given by $\mathbb{N}$ but would it be wrong if it were indexed by any other countably infinite set? Is it even okay with $\mathbb{N}$. Certainly, we can't do this for an index which is uncountable, because we couldn't even draw the matrix. – Squirtle Oct 25 '14 at 18:24
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    If $\mathcal H$ is a separable Hilbert space, then this is O.K. since such a space has a countable basis. If this basis is $(e_i)$ then $\langle Ae_i,e_j \rangle$ determines $A$ completely. This is even for uncountable bases true if we require that $\langle A\phi, \psi \rangle$ is known for all $\phi,\psi \in \mathcal H$. –  Oct 25 '14 at 18:29
  • The appropriate generalization is the kernel of a linear operator. Essentially, this makes sense if the "infinite matrix" is appropriately absolutely integrable on the space $\mathbb{N} \times \mathbb{N}$. – jxnh Oct 25 '14 at 18:30
  • I advise caution when working with infinitely sized matrices. 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. For example, infinite matrix multiplication is not necessarily associative. – DanielV Oct 25 '14 at 18:30
  • This $\mathcal{H}$ is in fact separable. But is it at all obvious this is "well-defined". I see this sort of thing done a lot in my von Neumann algebras course, its used with caution often in order to speed up a proof when the space is not necessarily countably infinite, but I am just curious if its ever okay? (Comment written before last three comments made, thanks for the input) – Squirtle Oct 25 '14 at 18:32
  • It's not associative? Could you please give an example? – Squirtle Oct 25 '14 at 18:56

2 Answers2

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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.

DanielV
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  • I tried to find a link to where I first saw this example, but my search engine skills didn't turn anything up. If anyone finds a good reference, feel free to edit my answer and add it. – DanielV Oct 25 '14 at 19:21
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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.

Disintegrating By Parts
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