I was given just a distance matrix (I'm talking about distance between cities), and I have to perform a classical multidimensional scaling. But mapping my results, I've noticed that my map is upside down. (For example, Miami is on the left side, Washington is on the right, and so on) Is this normal? My results could get better if I'd had the latitude and longitude?

Here's my R code (as you can see, is not a big deal):

mds<-cmdscale(D,eig=TRUE, k=2)

x <- mds$points[,1] y <- mds$points[,2]

plot(x, y, xlab="Coordinate 1", ylab="Coordinate 2", main="Metric   MDS",   type="n")
text(x, y, labels = row.names(D), cex=.7)

• MDS does nit know where North and West are. You can rotate your map (coordinates) by any angle you like. – ttnphns Oct 15 '16 at 22:17
• @ttnphns, why not make that an official answer? – gung - Reinstate Monica Oct 16 '16 at 0:00
• (Although this is asked in terms of R, the question is driven by a statistical misunderstanding. IMO, this should be considered on topic here.) – gung - Reinstate Monica Oct 16 '16 at 0:01
• Sorry, and how can I rotate the map (using R)? And about my 2nd question, In general, the results could get better if I'd had the latitude and longitude? – Freshman Oct 16 '16 at 17:01

This is actually normal. It boils down to the fact that when we take the orthogonal decomposition of your Gram matrix $$G=U \Lambda U^T$$ where the the i-th column of $$U$$ is the eigenvector $$q_i$$ that satisfies $$Gq_i= \lambda_i q_i$$ where $$\lambda_i$$ is the i-th diagonal element of $$\Lambda$$. We then notice that because $$-1$$ is a scalar, we get $$G(-q_i)= \lambda_i (-q_i)$$, which means $$-q_i$$ is also an acceptable eigenvector of $$G$$. This means taking $$Y$$ to be $$G=(\Lambda^{1/2}U^T)^T(\Lambda^{1/2}U^T)= Y^TY$$ can yield data that will be flipped with respect to each of the axes.