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Suppose an arbitrary double-centered matrix $D\in \mathbb{R}^{n\times n}$ and an unit vector $u\in \mathbb{R}^{n}$ are given. What happens to the vector after applying $Du$? Does the vector change completely, or just translate, rotate, scale? The application $Du$ should yield a centered vector.

To remind you, double centered matrix is a matrix with all entries in one row summing to zero, for all rows, and with all entries in one column summing to zero, for all columns.

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    It's written above.2012-02-02

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Since the matrix $D$ doesn't need to be orthogonal, the vector doesn't need to be rotated only. And since $u$ doesn't need to be an eigenvector of $D$, the vector doesn't need to be scaled only. And translation, well you cannot translate a vector, anyway. So in this sense, the vector can indeed change completely if nothing else than double-centredness is known about $D$.

But you're right in that $D\mathbf{u}$ should be centred (in case with a centred vector you mean one whose element sum is $0$). This can be seen quite easily (using $\mathbf{d}_i$ to denote the $i$th row of $D$):

$\mathbf{v} = D\mathbf{u} = (\langle\mathbf{d}_i,\mathbf{u}\rangle)_{i=1}^n$

$\sum_{i=1}^n{v_i}=\sum_{i=1}^n{\langle\mathbf{d}_i,\mathbf{u}\rangle} = \left\langle\sum_{i=1}^n{\mathbf{d}_i},\mathbf{u}\right\rangle=\langle\mathbf{0},\mathbf{u}\rangle=0$

So in fact only $D$'s columns need to be centred in order to make $D\mathbf{u}$ centred.

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    @ChrisianRau Since there is at least one zero eigenvalue associated with the double-centered matrix, is the corresponding eigenvector $1_n=[1~\dots~1]^T\in\mathbb{R}^n$?2012-03-14