Centering matrix ($P=I-d^{-1}11^{\mathsf{T}}$)

The centering matrix subtracts the arithmetic mean from every component of a real vector. For a positive integer $d$, let $1\in {\mathbb{R}}^d$ have every component equal to one and defineFor $x\in {\mathbb{R}}^d$, write $\bar{x}=d^{-1}{1}^{\mathsf{T}}x$. Then $Px=x-\bar{x}1$, so $1^{\mathsf{T}}Px=0$. The output lies in the subspace of vectors whose components sum to zero.

Its orthogonal projection properties follow directly. Set $J=11^{\mathsf{T}}/d$. Since $1^{\mathsf{T}}1=d$, we have $J^2=J$ and henceThus $P$ is a symmetric idempotent matrix, the orthogonal projector onto $1^{\perp }$. It kills constant vectors and leaves zero-sum vectors unchanged. Its kernel of a linear map is $\mathrm{span}\{1\}$ and its rank is $d-1$; for $d=1$ it is the zero matrix. The eigenvalues are zero on the constant direction and one on its orthogonal complement.

For an original example, $x=(1,2,4,9{)}^{\mathsf{T}}$ has mean $4$, giving $Px=(-3,-2,0,5{)}^{\mathsf{T}}$. The centered sum of squares isFor these four observations, the usual sample variance is $38/(4-1)=38/3$. Centering removes the mean; it does not itself divide by a standard deviation.

For a data matrix $X$ with observations in its $d$ rows, $PX$ centers each column. When $d>1$, its sample covariance matrix is $X^{\mathsf{T}}PX/(d-1)$. Also $P(x+c1)=Px$ for any scalar $c$, which proves invariance to adding the same offset to every observation. These identities make the centering matrix useful in least squares and principal component analysis.

A weighted centering matrix subtracts the same weighted mean from every component of a vector. Let $a\in {\mathbb{R}}^n$ satisfy $a_i\ge 0$ and $a^{\mathsf{T}}1=1$, where $1$ has every component equal to one. DefineThen $P_{a}x=x-(a^{\mathsf{T}}x)1$ and $a^{\mathsf{T}}{P}_{a}x=0$. The centered vector has zero weighted mean; its unweighted component sum need not vanish.

The projection properties follow directly from normalization:Also $P_{a}1=0$, so its kernel of a linear map is $\mathrm{span}\{1\}$, its range is $\ker a^{\mathsf{T}}$ and its rank is $n-1$. It is an idempotent matrix. For unequal weights it is generally not symmetric: it projects onto the zero-weighted-mean subspace along the constant-vector direction, an oblique projection in ordinary Euclidean geometry.

For an original example, $a=(1/4,3/4{)}^{\mathsf{T}}$ and $x=(4,2{)}^{\mathsf{T}}$ give $a^{\mathsf{T}}x=2.5$ and $P_{a}x=(1.5,-0.5{)}^{\mathsf{T}}$. The weighted mean is zero, while the ordinary mean is $0.5$. Euclidean orthogonal projection of $x$ onto the same line instead gives $(3,-1{)}^{\mathsf{T}}$.

If every $a_i>0$, put $D=\mathrm{diag}(a)$. Then $D{P}_a=P_a^{\mathsf{T}}D=D-a{a}^{\mathsf{T}}$, so this same map is orthogonal for the inner product $⟨u,v{⟩}_D=u^{\mathsf{T}}Dv$. It is the residual-making map for minimizing ${\sum }_i{a}_i(x_i-c{)}^2$ over a fitted constant $c$, the intercept-only case of weighted least squares.^[ref] Zero weights preserve the algebraic projection but make this proposed inner product degenerate. When all weights equal $1/n$, the construction reduces to the usual centering matrix.