N by N matrix of order 1

Question

I am looking at a paper by Slawomir Jarek called "REMOVING INCONSISTENCY IN PAIRWISE COMPARISON MATRIX IN THE AHP" (http://cejsh.icm.edu.pl/cejsh/element/bwmeta1.element.cejsh-fdb88af9-ba25-435f-9c85-3dcedcc7be57/c/mcdm16_11__5.pdf).

I do not see how a NxN matrix can have an order of 1. Thank you for your help.

Answer 1

It looks like a mistranslation. I think that ''order'' should be ''rank'', which is the number of linearly independent rows (or columns) in the square matrix.

The matrix that the paper says has order 1 is an $n \times n$ matrix $W$ where, for each $1 \leq i,j \leq n$, the element $W_{ij}$ in the $i^{th}$ row and $j^{th}$ column is equal to $\frac{w_{i}}{w_{j}}$, where $[w_{1} \ldots w_{n}]$ is a given vector of $n$ elements. The first row of $W$ is $[ \frac{w_{1}}{w_{1}} \frac{w_{1}}{w_{2}} \ldots \frac{w_{1}}{w_{n}}]$. For each $2 \leq k \leq n$, the $k^{th}$ row of $W$ is $[ \frac{w_{k}}{w_{1}} \frac{w_{k}}{w_{2}} \ldots \frac{w_{k}}{w_{n}}]$, which is $\frac{w_{k}}{w_{1}}$ times the first row. (So (row $k$) + ($-\frac{w_{k}}{w_{1}}$ times row 1) is the zero row.) So any collection of two or more rows of $W$ is linearly dependent. Any linearly independent set of rows of $W$ has at most 1 element. Hence, $W$ has rank 1.

Answer 2

Thank you very much for your answer! Can you please explain in a bit more detail the following step: Thus, for each 1≤k≤n, the kth row of W is wk/w1 times its first row. Therefore there is only one linearly independent row in W, and W has rank 1.

Thank you again!