Talk:Euclidean distance matrix

Exact rank?

The article currently says:

In dimension m, a Euclidean distance matrix has rank less than or equal to m+2. If the points $x_{1},x_{2},\ldots ,x_{n}$ are in general position, the rank is exactly m+2.

So for example in R² we have m = 2, so n points in general position would give a Euclidean distance matrix with rank of exactly m + 2 = 4. But this is impossible if there are only three points, so the matrix has only 3 rows and columns. Would this be right if there were simply the added restriction that n > 3? Or is something more involved? Loraof (talk) 20:00, 11 February 2015 (UTC)[reply]

Technically, the rank would be exactly min(n, m + 2). I'll put it in, there. — Arthur Rubin (talk) 20:40, 11 February 2015 (UTC)[reply]

Also the converse?

The article states that if n > m+1 points are in general position in m-space, then their Euclidean distance matrix has rank m+2. Is the converse also true?:

Given a matrix satisfying the given properties of a Euclidean distance matrix (non-negative elements, zeroes on the main diagonal, symmetric, and everywhere satisfying the triangle inequality), and with rank = m + 2, does there always exist a set of n points in general position in m-space having the squared distances given by the matrix?

Regardless of whether the converse holds or not, I suggest that this be addressed in the article. Loraof (talk) 17:08, 13 February 2015 (UTC)[reply]