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Given the matrices $X$ and $Y$ in $[0,1]^{n\times m}$, for $n > m > 3$, so that $X1_m=1_m$ and $Y1_m=1_m$, where $1_m$ denotes a $m$-length column vector of ones, find a matrix $Q$ in $R^{m\times m}$ so that $C(X,Y)=Q^TX^TYQ$ satisfies the following tree conditions:

$i$) $C(X,X)$ is diagonal with diagonal elements given by the the sum of rows of $X$

$ii$) The sum of the columns of $C(X,Y)$ equals the sum of rows of $X$

$iii$) The sum of the rows of $C(X,Y)$ equals the sum of rows of $Y$

What can be said about the uniqueness of the matrix $Q$?

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Why is this problem interesting? It seems like an arbitrary list of conditions to me. –  Will Sawin May 4 '12 at 20:10
Small typo.... the dimension of the input and output of X are the same but that doesn't jive with the nxm shape of X. More importantly I can't figure out what the properties of Q are supposed to be –  David Benson-Putnins May 4 '12 at 20:10
Since X and Y do not exist (as n > m), we can take Q to be the matrix of all zeros. Gerhard "One As Good As Another" Paseman, 2012.05.05 –  Gerhard Paseman May 5 '12 at 19:15
Why is the problem interesting? Its solution will have relevance in land cover change detection. Each row of $X$ and $Y$ represents membership vectors of $m$ land cover types. If $X,Y\in\{1,0\}^{n\times m}$, i.e., are binary matrices with exactly one non-zero element per row, then $C=X^TY$ quantifies the changes among land cover types, and $C$ satisfies all three conditions above ($Q=I$). The problem arises when $X,Y\in[1,0]^{n\times m}$, that is partial memberships are allowed. –  silvanmx May 7 '12 at 16:00
BTW, could not figure out the typo. –  silvanmx May 7 '12 at 16:27
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