Given that I know the pseudo-inverse of a matrix(not necessarily a square matrix), how to calculate the pseudo-inverse of the matrix I get by adding a single row/column to the original matrix?
i.e, Is there any way to compute the MP inverse of [A v] if I know the MP inverse of A? (The new matrix is just the original matrix A with an additional column v)
Assume $A\in M_{n,p}(\mathbb{R})$ where $n\leq p$ and $B=[A,v]$. Then, generically, $AA^T$ is invertible (that is, $A$ is full row rank and, consequently, $B$ also) with $A^+=A^T(AA^T)^{-1}$ and $BB^T$ is invertible with $B^+=B^T(BB^T)^{-1}$.
Thus $(BB^T)^{-1}=(AA^T+vv^T)^{-1}=$ (Sherman-Morrison) $(AA^T)^{-1}(I-vu)$ where $u=\dfrac{v^T(AA^T)^{-1}}{1+v^T(AA^T)^{-1}v}$.
Finally $B^+=\begin{pmatrix}A^T\\v^T\end{pmatrix}(AA^T)^{-1}(I-vu)=\begin{pmatrix}A^+(I-vu)\\u\end{pmatrix}$.
EDIT 1. Remark: if $A^+$ and $(AA^T)^{-1}$ are known, then the calculation of the couple $B^+,(BB^T)^{-1}$ is in $O(pn)$. Finally, if we add $3$ columns to $A$, then the calculation is in $3$ steps and the total complexity is again in $O(pn)$.
EDIT 2. Answer to Vinayak Abrol. Assume that $n>p$, $B=\begin{pmatrix}A\\v\end{pmatrix}$ (we add a row) and that $A$ has full column rank; consequently, $B$ has full column rank. Then $A^+=(A^TA)^{-1}A^T$ and $B^+=(B^TB)^{-1}B^T$. Then $(B^TB)^{-1}=(A^TA+v^Tv)^{-1}$ and we apply Sherman-Morrison...
https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula