5
$\begingroup$

Given a matrix A, I want to complete Multidimensional Scaling by hand, instead of using any given R functions.

As such, I have calculated the centered matrix B with the following code:

 n<-nrow(A)
 id<-diag(n)
 e<-diag(id)
 H <- id - (1/n)*e %*% etranspose
 B <- (-1/2)* H %*% A %*% H

My question is: how can I use my B matrix to complete multidimensional scaling on my A matrix, without the cmdscale function or anything along those lines?

$\endgroup$
1
$\begingroup$

If you're interested in the classic MDS algorithm, it is spelled out very nicely on the Wikipedia page:

Classical MDS uses the fact that the coordinate matrix can be derived by eigenvalue decomposition from $B=XX'$. And the matrix $B$ can be computed from proximity matrix $D$ by using double centering.

  1. Set up the squared proximity matrix $D^{(2)}=[d_{ij}^{2}]$
  2. Apply double centering: $B=-{\frac {1}{2}}JD^{(2)}J$ using the centering matrix $J=I-{\frac {1}{n}}11'$, where $n$ is the number of objects.

  3. Determine the $m$ largest eigenvalues $\lambda _{1},\lambda _{2},...,\lambda _{m}$ and corresponding eigenvectors $e_{1},e_{2},...,e_{m}$ of $B$ (where $m$ is the number of dimensions desired for the output).

  4. Now, $X=E_{m}\Lambda _{m}^{1/2}$, where $E_{m}$ is the matrix of $m$ eigenvectors and $\Lambda _{m}$ is the diagonal matrix of $m$ eigenvalues of $B$.

I can't really tell what your $A$ matrix is, but if each entry is a measure of the distance from the $i^{th}$ to $j^{th}$ entry, then that would be your proximity matrix.

Also, $11'$ is an n-by-n matrix of all 1's. This should be fairly self-explanatory, however most of these terms can be Googled if you're unsure. It really shouldn't be any more than about 10 lines of code.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.