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$$C_{n\times n}={\begin{bmatrix} 1 & x_{12} & \dots & x_{1n} \\ {x_{12}}^{-1} & 1 & \dots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ {x_{1n}}^{-1} & {x_{2n}}^{-1} & \dots & 1 \end{bmatrix}}$$

Criteria

$$x_{ab} = \frac{1}{x_{ba}}, \qquad x_{ab}>0, \qquad x_{aa} = 1$$

Alternatively

$$C \cdot C^{T} = \underline1$$

I know that this matrix models currency exchange (without commission or fluctuation). So I'm guessing it's called a currency matrix or a trade matrix. I'm just after a name so I can search it's properties.

I couldn't spot it in the Matrix Reference Manual. http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/special.html

I'm curious about some of it's algebraic properties, for example it has only 1 non-zero eigenvalue. What does this eigenvalue signify?

$$\text{Eigenvalues of} \begin{bmatrix} 1 & 6 & 30 & 210 \\ 1/6 & 1 & 5 & 35 \\ 1/30 & 1/5 & 1 & 7 \\ 1/210 & 1/35 & 1/7 & 1 \end{bmatrix} = 0,0,0,4$$

Any help, ideas, advice greatly welcomed. Thanks

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  • $\begingroup$ I think you mean $C\cdot C^T$ instead of $C\cdot C^{-1}$, and it's $\underline n$ instead of $\underline 1$. $\endgroup$ – Arnaud Mortier Jun 13 '18 at 22:01
  • $\begingroup$ Also you use $j$ both as a dummy variable and the size of the matrix. $\endgroup$ – Arnaud Mortier Jun 13 '18 at 22:02
  • $\begingroup$ you're right I do mean $C^T$! though surely element-wise products of $C$ and $C^T$ would be 1 everywhere? $\endgroup$ – Ben Crossley Jun 13 '18 at 22:03
  • $\begingroup$ Oh I see but product of matrices doesn't work element-wise, that is very unusual. If you mean element-wise it would be better to indicate it. $\endgroup$ – Arnaud Mortier Jun 13 '18 at 22:14
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    $\begingroup$ Some sources call this reciprocal matrices. $\endgroup$ – Algebraic Pavel Jun 13 '18 at 23:47
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$C$ is a positive reciprocal matrix

to be an exchange matrix the additional condition that $a_{ij} \times a_{jk} = a_{ik}$ is required

$C$ is then a Positive Saaty-consistent Reciprocal matrix (PS-cR)

the eigenvalues of an $n\times n$ PS-cR matrix are $n,0,..,0$

Thank you @Algebraic Pavel for the term reciprocal.

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It is laplacian matrix which is normally I-A, where A is adjacency matrix. https://en.m.wikipedia.org/wiki/Laplacian_matrix

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  • $\begingroup$ No. That is false. Have you read Wiki article?? $\endgroup$ – Dog_69 Jun 13 '18 at 22:24
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    $\begingroup$ The example on that page does not fit the criteria. The leading diagonal should be 1 and non-diagonal elements should be the inverse of their "transpose element" $\endgroup$ – Ben Crossley Jun 13 '18 at 22:25
  • $\begingroup$ Sorry, I mean normalised laplacian. Imagine currency as set of point on graph connecect each currency with edge with weight eqals negative of exchange rate. The normalised laplacian of this graph will be your matrix. $\endgroup$ – user569688 Jun 13 '18 at 23:09
  • $\begingroup$ @user569688 even then, $a_{ij} = a_{ji}$ which is not desired. They are supposed to be inverse. Reciprocal Matrices seems to be the correct term. $\endgroup$ – Ben Crossley Jun 14 '18 at 0:36

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