Suppose that $M_1$ is the residual maker for a unity vector (i.e. a vector made of $n$ 1's).
I am told that this matrix, when premultiplying a variable, transforms the variable "into deviations from its sample mean". What does this mean?
Suppose that $M_1$ is the residual maker for a unity vector (i.e. a vector made of $n$ 1's).
I am told that this matrix, when premultiplying a variable, transforms the variable "into deviations from its sample mean". What does this mean?
Let's consider the context of linear regression, where the least square estimator in matrix form is:
$$\hat{\mathbf{\beta}}= \begin{bmatrix} \hat{\beta_{0}} \\ \hat{\beta_{1}} \\ \vdots \\ \hat{\beta_{m}} \end{bmatrix} = (X'X)^{-1}X'\mathbf{y}$$
Let's call $P=X(X'X)^{-1}X'$ the prediction maker matrix.
For the residuals you simply do $\mathbf{y}-\hat{\mathbf{y}}$, which is equal to $(I-P)\mathbf{y}$. Hence, we call $M=(I-P)$ the residual maker matrix.
In your specific example, instead of a predictor matrix, you have a matrix that computes the mean of a variable, let's call it $A$. In this case, for a vector variable $Z$ you'd have $(I-A)Z=Z-\overline{Z}$, which is indeed $Z$ expressed as deviations from its mean value.