Given an n-by-n matrix $A$, and using only linear transformation on each of the rows and columns of $A$ (total of $2n$ transformations), is there a way to double-standardize a matrix, so that each row and each column have zero mean and unit variance?
Define centering matrix as follow: $$ C_n = I_n-\frac{1}{n}O_n $$ where $I_n$ is the identity matrix of size $n$ and $O_n$ is an n-by-n matrix of all 1's.
An n-by-n matrix $X$ is called doubly-centered if: $$ X = C_n X C_n $$
A matrix is doubly centered, iff the mean of each row and each column is $0$.
It's a common practice in many fields to standardize data into zero mean and unit variance, given a set of numbers with mean $\mu$ and standard deviation $\sigma$, the following linear transformation will standardize that set: $$ g(x) = \frac{x-\mu}{\sigma} $$
It's easy to double-center a matrix, using only row/column linear transformation, by subtracting the mean of every row/column and adding back the grand mean of the whole matrix.
It's also easy to standardize all rows (or columns) of a matrix using only $n$ linear transformations, one for each row (see $g(x)$ above).
What I'm trying to do is to simultaneously standardize all rows and columns of a matrix in a similar way, is it possible?
This can be solved iteratively, see Olshen & Rajaratnam 2010. The solution is not unique: the outcome depends on whether you start with rows or columns first.