# Find the Inverse of an Infinite Square Matrix

For my mathematics assignment I am using polynomial interpolation to solve certain problems and I end up with the following scenario:

$$\begin{bmatrix}... & 0 &0 & 0 & 1\\... &1^3 & 1^2 & 1 & 1\\ ... &2^3 & 2^2 & 2 & 1\\... &3^3 & 3^2 & 3 & 1\\ \unicode{x22F0} & \vdots & \vdots & \vdots & \vdots\end{bmatrix}^{-1} \unicode{x22c5} \begin{bmatrix}a\\b\\c\\d\\ \vdots \end{bmatrix}$$

Where the value in the rows of the produced single-column matrix correspond to the coefficients of an infinite-polynomial with decreasing powers.

I was wondering is there was a way to either invert this infinite matrix or find what happens to the polynomial as I increase the size of the matrices used (e.g. see if it approaches another function's taylor or power series)

I'm not sure, whether the following is tranferable to your lhs-matrix because I only use this matrix in a mirrored form, calling it "Vandermonde-matrix" $$ZV$$.

Surely the infinite-size version cannot be inverted and sequences of finitely sized matrices with increasing size have increasing and strongly diverging determinant, so I think it is unlikely that you could approximate to a meaningful limit when you actually take matrices of finite size and recompute the inverse with always increasing size.

But I used the LDU-decomposition of my version of the Vandermondematrix arriving at two triangular matrix-factors and one diagonal matrix, whose entries don't change when the size is altered, so a inversion of the factors even when infinite size seems possible and all entries of all three factors and of their principal inverses have a simple structure so that you can formally define series occuring by the dot-products.

So if I take

VZ =
1  0   0    0    0     0 ...
1  1   1    1    1     1 ...
1  2   4    8   16    32 ...
1  3   9   27   81   243 ...
1  4  16   64  256  1024 ...
1  5  25  125  625  3125 ...
...


and do the LDU-decomposition into three matrix-factors

 L*D*U = VZ
L =   (Pascalmatrix)    // the inverse is the same just with signed entries
1  .   .   .  .  .
1  1   .   .  .  .
1  2   1   .  .  .
1  3   3   1  .  .
1  4   6   4  1  .
1  5  10  10  5  1

D = (diagonal of factorials)  // the inverse has reciprocal factorials
1  .  .  .   .    .
.  1  .  .   .    .
.  .  2  .   .    .
.  .  .  6   .    .
.  .  .  .  24    .
.  .  .  .   .  120

U =  (Stirlingnumbers 2nd kind)  // inverse has Stirlingnumbers 1st kind
1  0  0  0  0   0
.  1  1  1  1   1
.  .  1  3  7  15
.  .  .  1  6  25
.  .  .  .  1  10
.  .  .  .  .   1


then your result should be somehow equivalent to

   U^-1 * D^-1 * L^-1 * column([a,b,c,d,e,...] )


where possibly a method of divergent summation should be assigned to the last product

 P= ( D^-1 * L^-1 * column([a,b,c,d,e,...]) )
U^-1 *_(div summation included) P        // this is not always doable


Sometimes (if [a,b,c,...] behave nicely) the parenthese $$P$$ in the latter formula gives a vector with only finitely many nonzero entries, then the formula for the final result gives only finite -and thus exact- sums.

But actually, given your definition of such a row-mirrored Vandermonde-matrix, I don't know whether your problem is even well-defined at all ...

• Thank you! I was able to use the LDU decomposition to find an expression for the components of the inverse matrix. Turns out if I reversed my matrix to match the Vandermonde one it outputs the same results but in reverse which happens to work in my favour anyway. – Toadfuture Sep 29 '18 at 5:14