mean-deviation form, why orthogonal?

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Why the column of the new design matrix are orthogonal? for example, let say

$A=\begin{pmatrix} 1& 1& 4\\ 1& 2& 0\\ 1& 3& 2 \end{pmatrix}$

how to caclulate the mean-deviation form of matrix A?

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Without the details of how the matrix $\mathbf{A}$ was formed, we are restricted to guessing.

Suppose $\mathbf{A}$ is a collection of column vectors $$ \mathbf{A} = \left[ \begin{array}{ccc} x_{1} & x_{2} & x_{3} \end{array} \right] $$ The transformation would be $$ \tilde{x}_{k} = x_{k} - \bar{x}. $$ The transformed matrix is $$ \tilde{\mathbf{A}} = \left[ \begin{array}{crr} 0 & -1 & 2 \\ 0 & 0 & 0 \\ 0 & 1 & -2 \\ \end{array} \right] $$ Quick check: the columns all have sum $0$.