# Recovering coordinates by eigendecomposition without double-centering

Suppose an Euclidean distance $D\in\mathbb{R}^{n\times n}$ matrix between a set of $n$ objects is given. To obtain inner-products (which will be further be used to recover coordinates), entries of $D$ are squared, and the matrix is double-centered and scaled, ie., $K=-\frac{1}{2}JD^{(2)}J$, where matrix $J=I-\frac{1}{n}11^T$ defines the origin wrt which the inner-products are formed. So, the aim is to reconstruct coordinates $X$ that give rise to inner products $K$, $$K=-\frac{1}{2}JD^{(2)}J=JXX^TJ.$$ This is done by eigendecomposition of $K$.

Given the above equation, I wonder if one could use a shortcut $-\frac{1}{2}D^{(2)}=XX^T$, ie, obtain coordinates $X$ without double-centering $-\frac{1}{2}D^{(2)}$ (matrix $J$ is removed from the left and from the right).

Note that you can't cancel $J$ since it is singular. In general, $Ju=Jv$ implies $u=v$ if and only if $J$ has a trivial kernel, i.e., iff the rank of $J$ equals the number of columns of $J$.
• Note that I'm mainly interested in a "technical reasoning". Wouldn't a solution $X$ to i) $-\frac{1}{2}D^{(2)}=XX^T$ be also a solution to ii) $-\frac{1}{2}JD^{(2)}J=XX^T$, up to origin position. Basically, why could not one solve ii) to obtain a solution to i) (just cancel $J$ from the sides of both sides?) Note that diagonal entries would remain zero. – usero Sep 18 '12 at 13:22
• No. $X$ will usually have complex entries, and is useless. - Multiplication by $J$ on both sides makes the diagonal positive when the distance matrix is Euclidean. Note that ypu can't cancel $J$ since it is singular. – Arnold Neumaier Sep 18 '12 at 14:56
• Your last sentence answered the question and many related ones. So, in case some $J$ is invertible (general case), one could cancel it. However, suppose a system i) $C^TACx=C^Tb$ is given, for $A\in\mathbb{R}^{n\times n}$ and a rectangular $C\in\mathbb{R}^{n\times m}$, $m<n$. In what manner could one rationalize that the solution to ii) $ACx=b$ is not a solution to i) (now, $C^T$ is rectangular, hence the singularity does not apply) – usero Sep 18 '12 at 15:10
• @usero: $C^Tu=C^Tv$ implies $u=v$ if and only if $C^T$ has a trivial kernel, i.e., iff the rank of $C$ equals the number of rows. – Arnold Neumaier Sep 18 '12 at 17:32