Let $P$ be a $n\times m$ matrix with linearly independent columns. Then $P$ has a left inverse $P^{-1}$. Let the vector of row-sums of $P$ be given by$$P\boldsymbol{1}_{m}=\left[\begin{array}{c} \sum_{j=1}^{m}P_{1j}\\ \vdots\\ \sum_{j=1}^{m}P_{nj} \end{array}\right]$$ and assume that all row-sums are strictly positive. Is that true that the vector of row-sums of $P^{-1}$ is given by the following?$$P^{-1}\boldsymbol{1}_{n}=\left[\begin{array}{c} \frac{1}{\sum_{j=1}^{m}P_{1j}}\\ \vdots\\ \frac{1}{\sum_{j=1}^{m}P_{mj}} \end{array}\right].$$Note that this is trivially true if $P$ has constant row-sums $P\boldsymbol{1}_{m}=\left[\begin{array}{c} k\\ \vdots\\ k \end{array}\right]$ since in this case$$\left[\begin{array}{c} \frac{1}{k}\\ \vdots\\ \frac{1}{k} \end{array}\right]=\frac{1}{k}P^{-1}P\boldsymbol{1}_{m}=\frac{1}{k}P^{-1}\left[\begin{array}{c} k\\ \vdots\\ k \end{array}\right]=P^{-1}\boldsymbol{1}_{n}.$$Hence I'm specifically interested in the case in which the row-sums of $P$ are not necessarily constant. Thank you in advance!
1 Answer
$\begingroup$
$\endgroup$
0
It is not true. As an example, take the matrix $$P=\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}.$$ It has row sums given by $\begin{pmatrix} 2 \\ 1\end{pmatrix}$; its inverse is $$P^{-1}=\begin{pmatrix}1 & -1 \\ 0 & 1\end{pmatrix},$$ with row sums given by $\begin{pmatrix} 0 \\ 1\end{pmatrix}$.