I have a problem. please help me to solve it.
As we know, the geometric series of a matrix $A_{n\times n} $ is defined as
$\sum_{l=0}^{\infty} A^l$. This series converges to $(1-A)^{-1}$ if and only if $\rho(A) <1$, where $\rho(A)$ is the eigenvalue with maximum magnitude. For every matrix $A$, we can generate a regularized matrix $rA$ where $0 < r < \frac {1}{\rho(A)}$. Know it can be proved that $\rho(rA)<1$ and hence
\begin{equation*}
\sum_{l=0}^{\infty}(rA)^l =\sum_{l=0}^{\infty}r^lA^l = (I-rA)^{-1}
\end{equation*}
Calculating such an inverse is of order $O(n^{2.37})$.
Now suppose that a new matrix $B_{(n+1)\times (n+1)}$ is given such that:
1) $B_{1:n,1:n} = A$, that is $B$ is generated by adding a row and a column to A's ends.
2) $B_{n+1,:} = B_{:,n+1}^T$
The following matrices are examples of $A$ and $B$:
\begin{bmatrix} 1.1 & 2.3 \\ 3.4 & 4.5 \end{bmatrix}
\begin{bmatrix} 1.1 & 2.3 & 4.2 \\ 3.4 & 4.5 & 5.2 \\ 4.2 & 5.2 & 4.7 \end{bmatrix}
Considering that $(I-rA)^{-1}$ (geometric series of regularized $A$) is given, is there any way to compute geometric series of regularized $B$ with any way faster than $O(n+1)^{2.73}$???