Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.