# How is the rank of a matrix affected by centering the columns of a matrix?

For some $$n$$ by $$p$$ matrix $$X$$, I'm trying to figure out how the rank of $$X$$ is affected if each column in $$X$$ is centered by the mean of that column (call the centered design matrix $$Z$$).

If $$p < n$$ and $$X$$ is full column rank, $$Z$$ is full column rank if multicollinearity is not present.

If $$p = n$$ and $$X$$ is full rank, $$Z$$ has rank $$n-1$$ due to the constraint from centering the variables, regardless of whether multicollinearity is present or not.

If $$p > n$$ and $$X$$ is full row rank, $$Z$$ has rank $$n-1$$ due to the constraint imposed from centering the variables

Which means rank of $$Z \leq$$ rank of $$X$$. I'm wondering if these observations are correct, and if so, if there's a technical way to show them, especially a).

Your centered matrix is given by $$Z= PX$$ where $$P:=I-\frac{1}{n}\mathbf{11}^T$$.

Your 1st statement holds iff the ones vector is not in the column space of $$X$$. I.e. if $$X\mathbf y = \mathbf 1$$ then $$PX\mathbf y = \mathbf 0$$ and the kernel has dimension (at least) 1. Otherwise for any $$X\mathbf y\neq \mathbf 1$$ you have $$P(X\mathbf y) = \mathbf 0$$ iff $$(X\mathbf y) = \mathbf 0$$ which occurs iff $$\mathbf y = \mathbf 0$$ since $$X$$ has full column rank, so the kernel dimension is at most 1 as well. Again the key issue is whether $$\mathbf 1$$ is in the column space of $$X$$.

As for your 2nd and 3rd points
having full row rank and at least as many rows as columns means the columns of $$X$$ span your space ($$X$$ is surjective) so the ones vector is in the column space of $$X$$ in both cases. By the above argument $$P$$ acting on $$X$$ increments the kernel by 1 when we select $$\mathbf y$$ such that $$X\mathbf y =\mathbf 1$$ so $$PX\mathbf y = \mathbf 0$$.

To tighten this up, consider that
since $$X$$ is surjective it has a right inverse $$M$$ such that $$XM = I_n$$, then
$$\text{rank}\big(PX\big) = \text{rank}\big(P(XM)X\big)\leq \text{rank}\big(P(XM)\big) = \text{rank}\big(P\big) \leq \text{rank}\big(PX\big)$$
so $$\text{rank}\big(PX\big) =\text{rank}\big(P\big)=n-1$$