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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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What's the definition of a double-centered matrix? – Davide Giraudo Feb 1 '12 at 16:21
It's written above. – user506901 Feb 2 '12 at 10:22
up vote 1 down vote accepted

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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Does that mean if $D$ is orthogonal, then the implication of $Du$ is simply a rotation of $u$? As for the translation: note that if $D$ is centered, so is $Du$, and centering a vector corresponds to adding a constant to its entries, thus translation. Am I wrong? – user506901 Feb 2 '12 at 14:07
@user506901 Ah, that's what you mean by translation (never heard of such a definition for translation, but Ok). But I'm not sure if that is done by a general double-centred matrix. So you mean centering means adding a constant to all its entries? Well in this case, like said, I'm not sure this is achieved by multiplying by a general double-centred matrix. I thought you mean a vector whose element sum is $0$, like in your definition of a double-centred matrix. Maybe you can add this definition to the question (maybe along with what you think is a translation of a vector). – Christian Rau Feb 2 '12 at 14:17
@user506901 And when $D$ is orthogonal (or more precisely orthonormal) it doesn't change the length of a vector and thus corresponds to a rotation (if $\det Q=1$) or a reflection (if $\det Q=-1$) around the origin. – Christian Rau Feb 2 '12 at 14:22
@user506901 Or do you mean adding a different constant to each entry? This sounds more like a translation, but again, this cannot be done with a vector (maybe you mean a vector representing the location vector of a point?). And this cannot be achieved by a matrix multiply, because all summands added to the vector entries depend on the vector itself and cannot be constants. Maybe you are mixing things up here and think of the $\mathbb{R}^{n+1}$ projective space usually used in computer graphics and the like. In this space translation can indeed be realized by a matrix-multiply? – Christian Rau Feb 2 '12 at 14:30
"maybe you mean a vector representing the location vector of a point"...exactly. Thanks. – user506901 Feb 2 '12 at 15:01

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