Projection (linear algebra, $P^2=P$)

A linear projection is an idempotent linear operator: $P^2=P$. Every vector decomposes as $x=Px+(I-P)x$, into the image of a linear map and the kernel of a linear map. Those two vector subspaces form a direct sum. An orthogonal projection additionally uses the orthogonal complement in an inner product space.

For an idempotent linear map $P$, $I-P$ is also idempotent andFor the first identity, $P(I-P)=0$ proves one inclusion. If $Pv=0$, then $v=(I-P)v$, proving the reverse inclusion. Apply the same argument to $I-P$ for the second identity. Consequently the ambient vector space is the direct sum of the image and kernel of a linear map of $P$. Orthogonality is not required; it is a separate property of an orthogonal projection.

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.

For an idempotent linear map $P$, induction gives $P^j=P$ for all integers $j\ge 1$. Summing the matrix exponential series therefore proves the displayed identity. The direct sum of its kernel of a linear map and image of a linear map is fixed and scaled by $e^{\alpha }$, respectively. The result requires no Hermitian matrix hypothesis and is valid for real or complex $\alpha$.

Let ${\ell }_0,{\ell }_1$ be norm-one extensions to $L^{\infty }((0,1))$ of endpoint point evaluations on $C([0,1])$. For continuous functions $e_0,e_1$ with $e_i(j)={\delta }_{ij}$, the linear projection $Pf={\ell }_0(f)e_0+{\ell }_1(f)e_1$ has the displayed operator norm, independently of which extensions are chosen.

The triangle inequality proves the upper bound. At a point $x_{\ast }$ maximizing $|e_0|+|e_1|$, choose unit complex numbers $a_j$ with $a_j{e}_j(x_{\ast })=|e_j(x_{\ast })|$. The continuous function $f(x)=(1-x)a_0+x{a}_1$ has supremum norm one and ${\ell }_j(f)=a_j$, so $(Pf)(x_{\ast })=|e_0(x_{\ast })|+|e_1(x_{\ast })|$. The L-infinity norm of the continuous output equals its supremum norm, proving equality even at an endpoint.

For a finite-dimensional real or complex linear projection $P$, every vector splits as $Px+(I-P)x$ in the image of a linear map and the kernel of a linear map, respectively. Their intersection is zero, so a basis adapted to this direct sum represents $P$ as $\mathrm{diag}(I_r,0)$. Its trace and rank both equal $r$. In particular, a projection's determinant is zero unless it is the identity. Hermiticity is unnecessary for this algebraic identity; a Hermitian matrix projection is additionally orthogonal.

Every finite-dimensional vector subspace $E$ of a Banach space $X$ is the range of a bounded linear projection with the displayed norm bound. In maximal inscribed Euclidean or Hermitian ellipsoid coordinates, write the John contact decomposition as ${\sum }_r{c}_r{u}_r{u}_r^{\ast }=I_E$. Its contact functionals ${\varphi }_r(x)=u_r^{\ast }x$ have norm one on $E$. Extend them to $X$ by the Hahn-Banach theorem and set $Px={\sum }_r{c}_r{\tilde{\varphi }}_r(x)u_r$. This restricts to the identity on $E$. The contact synthesis operator has Euclidean norm one, the weights sum to $\dim E$, and the Euclidean unit ball is contained in the norm unit ball; these facts give $\Vert Px{\Vert }_E\le \sqrt{\dim E}\Vert x{\Vert }_X$.

For an idempotent linear map $P$ on a finite-dimensional real vector space, take any positive definite inner product on that space. The displayed new inner product is positive definite since its squared norm vanishes only if both $Pu$ and $(I-P)u$ vanish. It makes the kernel of a linear map and image of a linear map orthogonal. Since $P^2=P$, one has $⟨Pu,v{⟩}_P=⟨u,Pv{⟩}_P$; thus $P$ becomes an orthogonal projection.

If $T^2=T$, every vector splits as $v=(v-Tv)+Tv$ in $\ker T\oplus \mathrm{im}T$. On these summands $T$ is respectively zero and the identity. Its matrix in an adapted basis is therefore $\mathrm{diag}(0,I_r)$, where $r$ is its rank of a linear map. Two idempotent linear maps on the same finite-dimensional vector space are similar matrices exactly when their ranks agree.

If two closed subspaces of a Hilbert space give $H=V\oplus W$, the projection onto $V$ along $W$ fixes $V$ and vanishes on $W$. It need not be an orthogonal projection. The bounded inverse theorem applied to the addition map $(v,w)\mapsto v+w$ on the complete component spaces proves boundedness of the component maps. With both complementary subspaces nonzero, its operator norm and that of $I-Q$ equal $\sec {\theta }_{V,W^{\perp }}$, with the directed subspace angle convention. Indeed, for fixed $v\in V$, the smallest possible norm of $v+w$ over $w\in W$ is $\Vert P_{W^{\perp }}v\Vert$, which proves the formula for $\Vert Q\Vert$. Equality with $\Vert I-Q\Vert$ follows from the block representation $Q=\left(\begin{array}{cc}I& B\\ 0& 0\end{array}\right)$ on $V\oplus V^{\perp }$: both squared operator norms are $1+\Vert B{\Vert }^2$. If a summand is zero, $Q$ is zero or the identity operator, and one complementary operator norm is zero; those cases must be treated directly.

For equal finite-dimensional closed subspaces of a Hilbert space $T,S$, positivity of $\cos {\theta }_{T,S}$ makes $P_S{|}_T$ invertible. The displayed reconstruction is the unique element of $T$ whose orthogonal projection onto $S$ matches that of $f$. It is the oblique projection onto $T$ along $S^{\perp }$. Its operator norm is the value of the secant function at the directed subspace angle, and its error is at most that secant function value times the best orthogonal projection error. The lower error bound follows from the Pythagorean identity. Zero-dimensional cases are handled directly.