N by N matrix of order 1











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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.










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    up vote
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    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.










    share|cite|improve this question
























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      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.










      share|cite|improve this 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.







      matrices matrix-calculus






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      asked Nov 18 at 17:16









      Ignacio Marés

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          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!






          share|cite|improve this answer





















          • I've expanded my answer.
            – Doug Chatham
            Nov 20 at 0:35


















          up vote
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          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.






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            2 Answers
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            2 Answers
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            active

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            up vote
            0
            down vote













            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!






            share|cite|improve this answer





















            • I've expanded my answer.
              – Doug Chatham
              Nov 20 at 0:35















            up vote
            0
            down vote













            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!






            share|cite|improve this answer





















            • I've expanded my answer.
              – Doug Chatham
              Nov 20 at 0:35













            up vote
            0
            down vote










            up vote
            0
            down vote









            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!






            share|cite|improve this answer












            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!







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 19 at 22:14









            Ignacio Marés

            1




            1












            • I've expanded my answer.
              – Doug Chatham
              Nov 20 at 0:35


















            • I've expanded my answer.
              – Doug Chatham
              Nov 20 at 0:35
















            I've expanded my answer.
            – Doug Chatham
            Nov 20 at 0:35




            I've expanded my answer.
            – Doug Chatham
            Nov 20 at 0:35










            up vote
            0
            down vote













            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.






            share|cite|improve this answer



























              up vote
              0
              down vote













              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.






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                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.






                share|cite|improve this answer














                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.







                share|cite|improve this answer














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                edited Nov 20 at 0:35

























                answered Nov 18 at 18:09









                Doug Chatham

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