## Catalan numbers as a convergent of iterative matrix inversion, INVERT transform.

I have here left out the indices of the elements for easier typing but hopefully this can still be understood.

${A=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\\ x&x&x&x&x&x&1&0\\ x&x&x&x&x&x&x&1\end{array}\right)$

Shift down the elements in matrix ${A}$ one step to get matrix ${B}$.

${B=}\left(\begin{array}{cccccccc}0&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ x&1&0&0&0&0&0&0\\ x&x&1&0&0&0&0&0\\ x&x&x&1&0&0&0&0\\ x&x&x&x&1&0&0&0\\ x&x&x&x&x&1&0&0\\ x&x&x&x&x&x&1&0\end{array}\right)$

Add a diagonal of ones to the main diagonal in matrix ${B}$ to get matrix ${C}$.

${C=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ x&1&1&0&0&0&0&0\\ x&x&1&1&0&0&0&0\\ x&x&x&1&1&0&0&0\\ x&x&x&x&1&1&0&0\\ x&x&x&x&x&1&1&0\\ x&x&x&x&x&x&1&1\end{array}\right)$

Change the sign of all the elements below the main diagonal in matrix ${C}$ so that you get matrix ${D}$.

${D=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -x&-1&1&0&0&0&0&0\\ -x&-x&-1&1&0&0&0&0\\ -x&-x&-x&-1&1&0&0&0\\ -x&-x&-x&-x&-1&1&0&0\\ -x&-x&-x&-x&-x&-1&1&0\\ -x&-x&-x&-x&-x&-x&-1&1\end{array}\right)$

Calculate the matrix inverse of matrix ${D}$ to get matrix ${E}$

Iterate by replacing matrix ${A}$ with matrix ${E}$.

Matrix ${E}$ should then converge to the Catalan numbers in all columns.

Keywords: Catalan numbers, INVERT transform, matrix inverse.