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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.

An idempotent matrix is a square matrix $P$ with $P^2=P$. Its linear map fixes every vector in its image and sends its kernel to zero. Every vector splits uniquely as $v=Pv+(v-Pv)$, giving a direct sum of the image and the kernel.

Consequently $P$ is diagonalizable over any field, with only the eigenvalues zero and one. For example, $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$ is idempotent but not symmetric. An idempotent matrix is an orthogonal projection in a real Euclidean inner product precisely when it is also symmetric; idempotence alone gives an oblique projection.

If $Av=\lambda v$ for a nonzero eigenvector, idempotence gives ${\lambda }^{2}v=A^{2}v=Av=\lambda v$. HenceThese are the only possible eigenvalues; either may be absent. In fact $v=Av+(I-A)v$ decomposes every vector into the image and kernel of $A$, so every idempotent matrix is diagonalizable even when it is not an orthogonal projection.

For an idempotent matrix, $A^k=A$ for every integer $k\ge 1$. Thus any scalar polynomial reduces toor equivalently $aI+bA$. This expression describes all polynomials in $A$, although its two coefficients need not be unique if $A=0$ or $A=I$.

If $Mv=\lambda v$ for a nonzero eigenvector, then $M^2=M$ implies $({\lambda }^2-\lambda )v=0$, hence $\lambda \in \{0,1\}$. Conversely, the real-symmetric spectral theorem gives $M=Q\Lambda Q^T$ with orthogonal $Q$. If every diagonal entry of $\Lambda$ is zero or one, then ${\Lambda }^2=\Lambda$ and therefore $M^2=M$. Thus the necessary and sufficient condition isSuch a symmetric idempotent matrix is the orthogonal projection onto its eigenvalue-one eigenspace, with its eigenvalue-zero eigenspace as the kernel. Diagonalizability is essential to the sufficiency argument from eigenvalues alone; symmetry supplies it here.

Using $M^2=M$,Thus $I-M$ is another idempotent matrix. In fact $M(I-M)=(I-M)M=0$: as linear projections, they select the complementary summands in the decomposition $x=Mx+(I-M)x$.

If the idempotent matrix $M$ is invertible, multiply $M^2=M$ on the left by its matrix inverse:Thus a nonidentity idempotent matrix must be singular; the identity itself is both invertible and idempotent.

If $M$ is an $r\times s$ matrix, the product $M^2=MM$ is defined in ordinary matrix multiplication only when $s=r$. Hence the defining equation of an idempotent matrix already requires equal row and column counts:Equivalently, an idempotent linear map is an endomorphism, so the same vector space is both its domain and codomain.

Take determinants of the idempotence equation and use multiplicativity of the determinant:Thus $d=\det M$ satisfies $d(d-1)=0$, provingIf the determinant is one, $M$ is invertible and solution gives $M=I$. Every other idempotent matrix has zero determinant.

Induction from $M^2=M$ gives $M^j=M$ for every integer $j\ge 1$. Separate the zeroth term of the matrix exponential series:Therefore the exponential of an idempotent linear map isThe last form shows its action directly: it multiplies the image of a linear map of $M$ by $e^{\alpha }$ and fixes its kernel of a linear map. No Hermiticity assumption on the idempotent matrix is needed.

Use the residual mean squareTo justify its denominator, let $H=X(X^{T}X{)}^{-1}{X}^T$, the orthogonal projection matrix onto the column space of $X$. Then the residual is $(I-H)ϵ$. The idempotent matrix $I-H$ has rank $n-p$, so diagonalizing it with an orthogonal matrix givesConsequently $\mathbb{E}{s}^2={\sigma }^2$, making $s^2$ an unbiased estimator. If specifically asked for maximum likelihood estimation, maximizing over ${\sigma }^2$ gives the alternative ${\hat{\sigma }}_{\mathrm{ML}}^2=\Vert Y-X\hat{\theta }{\Vert }^2/n$, which is biased downward by the factor $(n-p)/n$.

A normalized pure state has $\rho =|\psi ⟩⟨\psi |$, and thereforeConversely, diagonalize a permissible density operator. Its eigenvalues obey $0\le {\lambda }_j\le 1$ and ${\sum }_j{\lambda }_j=1$, soEquality requires every eigenvalue to be zero or one; exactly one is one. Thus the idempotent matrix condition and purity one both characterize a pure state. A mixed state has at least two positive eigenvalues, ${\rho }^2\ne \rho$ and $\mathrm{Tr}{\rho }^2<1$. In dimension $d$ its purity is at least $1/d$.

Using the unit length of $q_i$,Then ordinary matrix expansion gives $(I_n-P_i{)}^2=I_n-2P_i+P_i^2=I_n-P_i$. Both $P_i$ and its complementary matrix are projectors. The second calculation in fact holds for every idempotent matrix, without requiring the matrix to equal its transpose.

Take $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. It has exactly two nonzero entries, both one, but $A^2=I_2\ne A$. Thus it is not an idempotent matrix. Its being an orthogonal matrix does not make it an orthogonal projector.

Let $r=\mathrm{rank}(A)$. If $r=0$, the matrix is zero and the identity is immediate. Otherwise choose the rank factorization $A=BC$ from part b. The column space of $B$ is the image of $A$.

An idempotent matrix acts as the identity on its image: if $u=Av$, then $Au=A^{2}v=Av=u$. Applying this to each column of $B$ gives $AB=B$. Substituting $A=BC$ yields $B(CB-I_r)=0$. Since $B$ has full column rank, the represented map is injective, so each column of $CB-I_r$ is zero. Thus $CB=I_r$.

Now the rectangular cyclic property of the trace givesThis proves that the trace of an idempotent equals its rank using the supplied factorization method. Orthogonality of the projection is unnecessary: the argument applies to every real idempotent matrix.

For $P=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the off-diagonal equations in $P^2=P$ are $b(a+d-1)=c(a+d-1)=0$. Thus either $b=c=0$ or the trace is one. The condition alone is not sufficient: $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has trace one but is not an idempotent matrix. For a two-by-two matrix of trace one, idempotence is equivalent to zero determinant, by the Cayley-Hamilton theorem.

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.