Mixing
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.
matrices matrix-calculus
asked Nov 18 at 17:16
Ignacio Marés
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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.
matrices matrix-calculus
asked Nov 18 at 17:16
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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.
matrices matrix-calculus
asked Nov 18 at 17:16
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
matrices matrix-calculus
asked Nov 18 at 17:16
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 18 at 17:16
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 18 at 17:16
Ignacio Marés
1
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Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 18 at 17:16
Ignacio Marés
1
asked Nov 18 at 17:16
Ignacio Marés
1
1
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Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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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!
answered 2 days ago
Ignacio Marés
1
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I've expanded my answer.
– Doug Chatham
2 days ago
add a comment |
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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.
answered Nov 18 at 18:09
Doug Chatham
1,2041916
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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!
answered 2 days ago
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I've expanded my answer.
– Doug Chatham
2 days ago
add a comment |
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!
answered 2 days ago
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I've expanded my answer.
– Doug Chatham
2 days ago
add a comment |
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!
answered 2 days ago
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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!
answered 2 days ago
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Ignacio Marés
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Ignacio Marés
1
answered 2 days ago
Ignacio Marés
1
1
New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Ignacio Marés is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I've expanded my answer.
– Doug Chatham
2 days ago
add a comment |
I've expanded my answer.
– Doug Chatham
2 days ago
I've expanded my answer.
– Doug Chatham
2 days ago
I've expanded my answer.
– Doug Chatham
2 days ago
add a comment |
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.
answered Nov 18 at 18:09
Doug Chatham
1,2041916
add a comment |
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.
answered Nov 18 at 18:09
Doug Chatham
1,2041916
add a comment |
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.
answered Nov 18 at 18:09
Doug Chatham
1,2041916
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.
answered Nov 18 at 18:09
Doug Chatham
1,2041916
edited 2 days ago
answered Nov 18 at 18:09
Doug Chatham
1,2041916
answered Nov 18 at 18:09
Doug Chatham
1,2041916
answered Nov 18 at 18:09
Doug Chatham
1,2041916
1,2041916
add a comment |
add a comment |
Ignacio Marés is a new contributor. Be nice, and check out our Code of Conduct.
Ignacio Marés is a new contributor. Be nice, and check out our Code of Conduct.
Ignacio Marés is a new contributor. Be nice, and check out our Code of Conduct.
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