# If we centered the matrix then the rank is at most d-1

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?

$$\bf 1$$ 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.