I have an equation $AX=B$, where $A$ is $\infty \times \infty$ matrix, $X$ is $\infty \times 1$ vector and $B$ is $\infty \times 1$ vector.

$A$ and $B$ are known and I need to determine $X$.

For this, I think that I should calculate inverse of $A$ (if it exists) and obtain $X$ as $A^{-1}B$.

However, I know almost nothing about inverses of infinite matrices, and I do not know when they exist and how to calculate them.

I am an amateur even when it comes to finite matrices, but, the problem that I am trying to solve led me to this equation with infinite matrices and vectors, and, I do not know what to do?

I was thinking that I could find inverse for $A_n$, where $\lim_{n \to \infty}A_n=A$, and then pass to a limit, but I do not know is the limit of inverses an inverse of the limit.

What to do?

  • 3
    $\begingroup$ On the vector space of all polynomials, spanned by the basis $1,x,x^2,\cdots$, the derivative operator is an $\infty\times \infty$ matrix. It is possible to find an an $\infty\times \infty$ matrix which is the left inverse of the $\infty\times \infty$ derivative matrix, and this would correspond to an integration operator. But, nobody achieves understanding of integration or differentiation by doing this. It is rarely if ever useful to think of operators on $\infty$-dimensional vector spaces as $\infty\times\infty$ matrices. $\endgroup$ – Wouter Jun 17 at 8:06
  • $\begingroup$ @Wouter But, this is not some abstract matrix with partial derivatives of functions as members, or something like that, it is just a matrix with numbers that, I think, can be inverted, and I do not know how to calculate an inverse. Something is surely known about such matrices. $\endgroup$ – Grešnik Jun 18 at 8:27
  • $\begingroup$ The $\infty\times\infty$ derivative operator is also just a matrix with numbers. But no: you don't just have an $\infty\times\infty$ matrix of numbers, you have a procedure to find the entries of this matrix. Finding the inverse will involve, at the very least, exploiting regularities, patterns, of that procedure. $\endgroup$ – Wouter Jun 18 at 8:45
  • $\begingroup$ I've exercised with the problem of inverting infinite matrices from time to time and one special interesting case for me I've made an essay of. See go.helms-net.de/math/index.htm and on this the entry "The inverse of this matrix M is the Null-matrix?" with links to an pdf: go.helms-net.de/math/divers/InverseNullmatrix.pdf and a related question at MathOverflow mathoverflow.net/questions/104370/… where my questions were nicely answered. Also you can look for keyword "Carlemanmatrix" for some more. $\endgroup$ – Gottfried Helms Jul 6 at 9:19

I would first look whether a triangular decomposition of $A$ is helpful. For instance by the well known LDU-decomposition. This procedure finds $$ L \cdot D \cdot U = A$$ where $L$ is lower triangular ("row-finite"), $D$ is diagonal, $U$ is upper triangular ("column-finite") and moreover, $L$ and $U$ are normed to have diagonal-entries of $1$. Of course, to invert $D$ we need that no diagonal-entry is zero. Let us assume this for the example computation.

With this you can invert the triangular/diagonal matrices to, say, $K=L^{-1}$, $E=D^{-1}$ and $T=U^{-1}$ to as many rows/columns you want.
After that, you can easily compute $$ \small \begin{array} {} A\cdot X &= (LDU)\cdot X &= B &&\text{the formal decomposition}\\ K \cdot(LDU)X &= DU \cdot X &= K\cdot B && \text {computable because $K$ is rowfinite }\\ E \cdot(DU)X &= U \cdot X &= EK\cdot B && \text {computable because $E$ is diagonal }\\ T \cdot(U)X &= X & \underset{\text{}problem!}= T \overset{\color{red}{???}}\cdot (EK B) && \text {in RHS is the problem because $T$ not rowfinite! }\\ \end{array}$$ In the last row in the RHS occurs that basical problem of matrix-inversion of infinite size in your example-constellation.

  • Sometimes it happens, that $(EKB)$ is actually column-finite, then you are fine and can compute $X=T\cdot(EKB)$ to $(LDU)^{-1}\cdot B = A^{-1}B = X$ as solution.
  • If this is not column-finite you must try for each row-column-dotproduct of $T \cdot (EKB)$ whether you have an analytically determinable expression evaluatable to a finite value.
  • Again with a bit of luck that dotproducts are either convergent or at least Cesaro-, Euler- or Borel-summable and you can have finite entries for $X$.
  • If nothing works, there is no computable solution.

For an example where it luckily came ot to be doable, see my essay on "is the inverse of this infinite matrix the Null-matrix?" A final solution for the still open problem in my essay see here an answer in MathOverflow.


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