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I have found that centering is defined as: We let $\bar{x}$ be the in-sample mean vector of the input data so $\bar{x}=\frac{1}{N}\sum_{n=1}^Nx_n$ and in matrix notation can be written as $\bar{x}=\frac{1}{N}X^T1$, $1$ is the column vector of N 1's. Then we substrate the mean from each point $z_n=x_n-\bar{x}$.

By calculation we have that $Z=X-1\bar{x}^T$ hence we have that: $$\bar{z}=\frac{1}{N}Z^T1=\frac{1}{N}X^T1-\frac{1}{N}\bar{x}1^T1$$ And use definition of $\bar{x}=\frac{1}{N}X^T1$ and that $1^T1=N$ and get: $$=\bar{x}-\frac{1}{N}\bar{x}N=0$$ Thus, the transformed vectors are ‘centered’ in that they have zero mean, as desire. But now I have to use this definition of centering, to show that if we have $dxd$ matrix denoted S, then if we centered the matrix then the rank of the resulting matrix at most $d-1$. I'm a bit confused how to prove that. I'm not sure about centering matrix, but I think if we get that the result matrix is $d$ or higher then we do not have that the transformed vectors are ‘centered’ and they will not have zero mean? But how can I prove that? Can anyone help?

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$\bf 1$ is a left kernel vector of $S$, as you computed, ${\bf 1}^TS=0$. Thus $Z$ can not have full rank, as you have $S$ given as square matrix.

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