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A matrix represents a linear projection when applying it twice has the same effect as applying it once. For an orthogonal projection with the standard Hermitian inner product on ${\mathbb{C}}^m$, the additional condition is that the conjugate transpose equals the original matrix:Thus $P$ is both idempotent and a Hermitian matrix. Over the real numbers this says $P^2=P=P^T$. Its image of a linear map is the retained linear subspace, while its kernel of a linear map is the orthogonal complement: vectors in the image are unchanged, and vectors in the kernel become zero.

For a matrix $A$ with full column rank, projecting onto its column space givesThe invertibility of $A^{†}A$ follows from linear independence of the columns. If those columns are an orthonormal set, $A^{†}A=I$ and $P=A{A}^{†}$. These formulas and the conditions $P^2=P=P^T$ in the real case are developed in MIT 18.06SC, Projections onto Subspaces, pp. 1–3.

For the line spanned by $e=(1,2{)}^T$,In particular, $Pe=e$ and $P(2,-1{)}^T=0$. The eigenvalues are $1$ on the retained direction and $0$ on the removed direction. This exposes the meaning of an orthogonal projection as retaining selected components, rather than stretching them.

Symmetry refers to the specified inner product. In the real weighted setting, the formula becomes $P_M=A(A^{T}MA{)}^{-1}{A}^{T}M$ and satisfies $P_M^2=P_M$ and $P_M^{T}M=M{P}_M$. For the preceding weighted line example,It is not symmetric under the ordinary transpose, but it is an orthogonal projection for the weighted Euclidean inner product. The matrix criterion $P=P^{†}$ should therefore be used with the standard Hermitian inner product, not silently transferred to a different geometry.

A finite-dimensional linear subspace $W$ of an inner product space has a unique closest vector to any given vector $x$. This is its orthogonal projection $P_{W}x$. If $e_1,\dots ,e_d$ is an orthogonal basis of $W$, compute it one direction at a time:The inner product convention here is linear in its first argument, including over the complex numbers. For an orthonormal basis, each denominator is one. When $W$ consists only of the zero vector, the empty sum gives $P_{W}x=0$. The finite-dimensional formula is described in Francis Narcowich, Inner Products and Norms: minimization problems.

Take $W=\mathrm{span}\{(1,2{)}^T\}$ and $x=(3,1{)}^T$, using the dot product. With $e=(1,2{)}^T$,The orthogonal projection residual has $r\cdot e=2-2=0$. To verify the best approximation directly, every candidate in the line is $te$, andThe smallest squared Euclidean norm error is therefore $5$, attained only at $t=1$. The diagram connects the perpendicular residual to this one-variable quadratic polynomial.

The general mechanism is the Pythagorean theorem in an inner-product space. For any $w\in W$, the vectors $r=x-P_{W}x$ and $P_{W}x-w$ are orthogonal, soMoving away from $P_{W}x$ inside $W$ adds a nonnegative squared norm, which is zero only when $w=P_{W}x$. This proves both optimality and uniqueness.

The useful feature is separation of preparation from evaluation. Construct an orthonormal basis once with the Gram-Schmidt process, then obtain each new best approximation using inner products instead of a fresh optimization problem. When $W$ is the column space of a matrix, this gives linear least squares; when $W$ is spanned by orthogonal polynomials, it gives a least-squares polynomial in an orthogonal-polynomial basis. The ambient inner product space need not be a Hilbert space for this finite-dimensional construction. In an infinite-dimensional target linear subspace, existence requires additional hypotheses, supplied by orthogonal decomposition by a closed subspace.

An orthogonal projection residual is the difference $r=x-P_{W}x$ between a vector and its orthogonal projection onto a linear subspace $W$. It is orthogonal to every vector in $W$, so $⟨r,w⟩=0$ for all $w\in W$. The Gram-Schmidt process uses such residuals to remove directions already represented by earlier basis vectors.

This is an optimality condition as well as a geometric description. No permitted change within $W$ can remove a further orthogonal component of $r$. In the preceding dot product example, $r=(2,-1{)}^T$ is perpendicular to the line spanned by $(1,2{)}^T$.

The selected inner product matters. Keep $x=(3,1{)}^T$ and the same line, but use the weighted Euclidean inner product with $M=\mathrm{diag}(4,1)$. The projection coefficient and residual becomeNow $4(r_M{)}_1{e}_1+(r_M{)}_2{e}_2=5-5=0$, although $r_M\cdot e=-15/4\ne 0$. The orthogonality is with respect to $M$, so it need not look perpendicular in the ordinary drawing. The squared weighted error isThe weighted optimum reduces the first-coordinate error from $2$ to $5/4$, at the cost of increasing the magnitude of the second-coordinate error from $1$ to $5/2$. The weights express which errors are more costly.

In linear least squares, residual orthogonality to the column space is expressed by the normal equations for linear least squares. With a weighted Euclidean inner product, the corresponding real-matrix equation is $A^{T}M(x-Ac)=0$. This makes the choice of error measurement part of the mathematical model, rather than merely a plotting convention.

A fiber metric on a smooth vector bundle $E$ gives a smooth orthogonal projection $P:E\to F$ onto any vector subbundle. In a local frame of $E$, let $G$ be the metric matrix and $A$ the columns of a frame of $F$. Then $P=A(A^{t}GA{)}^{-1}{A}^{t}G$, proving smoothness. The complementary subbundle $F^{\perp }=\ker P$ maps isomorphically onto the quotient vector bundle. Explicitly $v\mapsto (Pv,[v])$ is a vector bundle isomorphism $E\to F\oplus (E/F)$, with inverse $(f,[v])\mapsto f+(I-P)v$. This inverse is independent of the representative $v$, but the splitting depends on the chosen metric.

Using the Levi-Civita symbol and the contraction of two Levi-Civita symbols, the $i$th component of the vector triple product isThus the triple-product identity is $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.

Apply this with the unit vector $n$ to obtain $\Pi (x)=x-(n\cdot x)n$. Dot products and scalar multiplication are linear in $x$, so $\Pi (sx+ty)=s\Pi (x)+t\Pi (y)$. Its matrix representation isThe fixed-vector equation is equivalent to $(n\cdot x)n=0$. Since $n\ne 0$, all its solutions are $x\in n^{\perp }$. Also $\boxed{\Pi (n)=0}$. Every vector decomposes as $x=\Pi (x)+(n\cdot x)n$, with $\Pi (x)$ orthogonal to $n$. Hence $\Pi$ is the orthogonal projection onto the plane through the origin normal to $n$: it fixes that plane and removes the perpendicular component.

We prove uniform approximation, not just approximation in $L^2$. The analytic facts used are: a continuous square-integrable kernel gives a compact integral operator on $L^2$; the spectral theorem for compact self-adjoint operators decomposes the closure of its range into its finite-dimensional nonzero eigenspaces; and continuous functions are dense in $L^2$ for normalized Haar measure on a compact Lie group. Normalized Haar measure on a compact group is both left- and right-invariant and is preserved by inversion.

Choose a continuous nonnegative function $k$ of integral one, supported in a sufficiently small symmetric neighborhood of the identity, with $k(g^{-1})=k(g)$. Such functions are obtained from a local bump and its inverted bump followed by normalization. Define the convolution operatorIts kernel is continuous on $G\times G$, so $T$ is compact. The symmetry of $k$ makes it self-adjoint. It commutes with the left translations $(L_{g}h)(x)=h(g^{-1}x)$ by a change of variable $y=gz$.

By Cauchy-Schwarz inequality and invariance of Haar measure,The same integral formula and continuity of the kernel show $Th$ is continuous for every $h\in L^2(G)$. Moreover, rewriting $y=x{z}^{-1}$ gives $(Tf)(x)=\int f(x{z}^{-1})k(z)dz$. Uniform continuity on the compact group therefore makes $\Vert Tf-f{\Vert }_{\infty }$ arbitrarily small when the support of $k$ is small. Also $\Vert T{\Vert }_{\infty \to \infty }\le 1$ because $k\ge 0$ and has integral one.

Fix $\epsilon >0$ and take $k$ so $\Vert Tf-f{\Vert }_{\infty }<\epsilon /3$. ThenLet $P_N$ be the orthogonal projection onto increasing finite sums of the nonzero eigenspaces of $T$. Since $Tf$ lies in the closure of the range, the spectral theorem gives $P_{N}Tf\to Tf$ in $L^2$. Applying the smoothing estimate givesThus some $h=T{P}_{N}Tf$ satisfies $\Vert h-f{\Vert }_{\infty }<\epsilon$.

Each nonzero eigenspace is finite dimensional, invariant under $L_g$, and consists of continuous functions: if $Tv=\lambda v$ with $\lambda \ne 0$, then $v=Tv/\lambda$. The finite sum $M$ containing $h$ therefore carries the finite-dimensional unitary representation $\rho (g)=L_g{|}_M$. It is continuous, since translating each of its finitely many continuous basis functions depends continuously on $g$ in the uniform and hence the $L^2$ norm.

Let $\ell$ be evaluation at the identity on $M$, and let the dual representation on $M^{\ast }$ be $\theta (g)\ell =\ell \circ \rho (g^{-1})$. For the linear functional $L_h$ on $M^{\ast }$ defined by $L_h(u)=u(h)$,The rank-one endomorphism $\alpha =\ell \otimes L_h$ of $M^{\ast }$ satisfies $\mathrm{tr}(\alpha \theta (g))=L_h(\theta (g)\ell )$. ThereforeThis is the requested Peter-Weyl theorem, proved via the convolution proof of uniform Peter-Weyl approximation. The extra convolution after the spectral truncation is what upgrades the $L^2$ estimate to a uniform one.

To deduce a faithful finite-dimensional representation, first note that these representations separate points. For $g\ne e$, choose a continuous function taking different values at $g$ and $e$, and approximate it closely enough by a trace coefficient to retain that difference. The representation appearing in that trace must satisfy $\theta (g)\ne I$.

There is an identity neighborhood $U$ containing no nontrivial subgroup. Here is the Lie groups have no small subgroups argument: choose an injective exponential chart on a Lie-algebra ball of radius $r$, and take $U=\exp (B_{\delta })$ with $0<\delta <r/2$. If a subgroup contained a nonidentity $\exp X$ in $U$, choose the first integer $m$ with $m\Vert X\Vert \ge \delta$. Then $\delta \le m\Vert X\Vert <2\delta <r$, so its power $\exp (mX)$ lies outside $U$, a contradiction. For a zero-dimensional group, take $U=\{e\}$.

For every $g\notin U$, choose a representation nontrivial at $g$. The sets on which these representations are not the identity are open and cover the compact set $G\setminus U$. A finite subcover gives finitely many representations, whose direct sum has kernel contained in $U$. A subgroup contained in $U$ is trivial, so this direct sum is faithful. Averaging its Hermitian form makes it unitary. Thus the finite faithful representation from point-separating representations yieldsUsing the no-small-subgroups neighborhood is essential; $G\setminus \{e\}$ alone need not be compact, so a finite-subcover argument on that punctured set would not be valid.

Work in discounted units with a riskless asset of value one, trivial initial information and finitely many risky assets. Let $Y=S_1-S_0\in L^2(P;{\mathbb{R}}^d)$ be their discounted gains. A terminal attainable claim has the form $v+{\theta }^{T}Y$, whose initial cost is $v$. Their spanis a finite-dimensional closed subspace of a Hilbert space. For $H\in L^2(P)$, one-period least-squares hedging is its orthogonal projection onto $\mathcal{H}$.

Let $m=\mathbb{E}Y$, $\Sigma =\mathrm{Cov}(Y)$ and $b=\mathrm{Cov}(Y,H)$. Remove redundant risky directions so that $\Sigma$ is invertible. Under absence of arbitrage, a gain direction with zero variance is a constant and must be zero, so this reduction loses no genuine trading opportunity. For freely chosen capital and holdings, minimizingfirst gives $v=\mathbb{E}H-{\theta }^{T}m$. The remaining centered error isCompleting the square yieldsThe residual $L=H-v^{\ast }-({\theta }^{\ast }{)}^{T}Y$ satisfies $\mathbb{E}L=0$ and $\mathbb{E}[LY]=0$, the least-squares normal equations. Exact claim replication is possible precisely when this residual is zero. If capital is fixed at $v$, its normal equation instead gives the fixed-capital quadratic hedge in a one-period marketThus fixed-capital and freely optimized hedging are distinct problems.

A dominated martingale measure is a probability measure $Q\ll P$ with integrable gains and ${\mathbb{E}}_{Q}Y=0$. Its density $Z$ obeys $Z\ge 0$, $\mathbb{E}Z=1$, $\mathbb{E}[ZY]=0$. An equivalent martingale measure additionally has $Z>0$ almost surely. Every such measure prices an attainable discounted claim $v+{\theta }^{T}Y$ at $v$, but different measures may give different expectations to an unattainable claim.

The minimal martingale measure in a one-period market is defined by preserving the mean-zero martingale directions orthogonal to the gain innovation $Y-m$. Among square-integrable densities it isIndeed $\mathbb{E}{Z}_{\ast }=1$, $\mathbb{E}[Z_{\ast }Y]=m-\Sigma {\Sigma }^{-1}m=0$, and for $\mathbb{E}L=0$, $\mathbb{E}[L(Y-m)]=0$ we have $\mathbb{E}[Z_{\ast }L]=0$. Conversely preservation of all these orthogonal directions puts $Z$ in $\mathrm{span}\{1,Y-m\}$; normalization and zero gain expectations then determine $Z_{\ast }$. Any other square-integrable signed pricing density differs from $Z_{\ast }$ by a vector in ${\mathcal{H}}^{\perp }$, so $Z_{\ast }$ also has the smallest $L^2$ norm. Its evaluation of a claim isThis is the capital of the optimal quadratic hedge, not automatically an arbitrage-free price. The formula may define only a signed martingale measure: $Z_{\ast }$ is a genuine dominated measure only if it is nonnegative, and equivalent only if strictly positive. For example, gains $(-1,1,2)$ with probabilities $(1/10,4/5,1/10)$ have $m=9/10$, $\Sigma =49/100$ and $Z_{\ast }(2)=-50/49$. Yet probabilities $(11/20,7/20,1/10)$ are all positive and have zero gain mean. Thus absence of arbitrage does not force a positive minimal density.

For the market completeness assertion, assume absence of arbitrage and hence existence of an equivalent martingale measure $Q_0$. In finite states this is the usual fundamental theorem of asset pricing; the one-period version also holds with finitely many assets on a general probability space. Market completeness means that every bounded claim is attainable, equivalently $\mathcal{H}=L^2(P)$ in this square-integrable setting. We now prove completeness and uniqueness of dominated martingale measures.

If the model is complete, replicate every indicator $1_A$. Two dominated martingale measures both evaluate it at its unique claim replication cost, and therefore assign every event $A$ the same probability. They are identical.

Conversely let $k=\dim \mathcal{H}$ and suppose the model is incomplete. There exist $k+1$ disjoint events $A_j$ of positive probability. Otherwise the probability space would have at most $k$ atoms: any non-atomic positive event could be split to increase the number. Its whole payoff space would then have dimension at most $k$, forcing equality with $\mathcal{H}$ and market completeness. Choose a basis $h_1,\dots ,h_k$ of $\mathcal{H}$. Its members are integrable under $Q_0$, because they are combinations of the constant and the gains. The $k+1$ vectorsare linearly dependent. Hence a nonzero bounded $g={\sum }_j{a}_j{1}_{A_j}$ satisfies ${\mathbb{E}}_{Q_0}[gh]=0$ for every $h\in \mathcal{H}$, in particular $h=1,Y_i$. For $0<\epsilon <1/\Vert g{\Vert }_{\infty }$, defineThese strictly positive densities integrate to one, preserve every gain expectation, and give distinct equivalent martingale measures. They are also dominated by $P$. Thus uniqueness is impossible in an incomplete market, proving the equivalence under the stated no-arbitrage hypothesis.

That hypothesis is essential. On three positive-probability states, the gain $(0,1,2)$ has the unique dominated martingale measure, concentrated on the first state, but its attainable span has dimension only two. The market is incomplete and admits arbitrage. This explains why uniqueness among merely dominated probabilities must not be used without an equivalent pricing measure or a no-arbitrage assumption.

For a bounded linear operator $T:E\to F$, write the weak square norm of a finite family asIt is a 2-summing operator if $({\sum }_j\Vert T{x}_j{\Vert }^2{)}^{1/2}\le C{w}_2(x_1,\dots ,x_m)$ for all finite families, and ${\pi }_2(T)$ is the least such $C$. This specializes the definition of an absolutely p-summing operator.

First let $E=H_1$, $F=H_2$ be Hilbert spaces. For the column operator $A:{\ell }_2^m\to H_1$, $A{e}_j=x_j$, the Riesz representation theorem gives $w_2(x_1,\dots ,x_m)=\Vert A^{\ast }\Vert =\Vert A\Vert$. If $T$ is a Hilbert-Schmidt operator, the Parseval identity applied to orthonormal bases givesThus ${\pi }_2(T)\le \Vert T{\Vert }_{\mathrm{HS}}$. Conversely test the summing inequality on any finite orthonormal family in $H_1$. Its weak square norm is $1$, by the Bessel inequality, with equality by choosing one of its vectors. Hence ${\sum }_j\Vert T{e}_j{\Vert }^2\le {\pi }_2(T{)}^2$. Taking the supremum over finite subsets of an orthonormal basis proves the Hilbert-Schmidt operator condition and the reverse norm inequality. These basis sums work even for nonseparable Hilbert spaces, being understood as suprema of finite subsums. Therefore

For an $n$-dimensional normed vector space $E$, again associate a family $(x_j)$ with $A:{\ell }_2^m\to E$. By the Hahn-Banach theorem and scalar Euclidean duality,Let $Q$ be the orthogonal projection onto $(\ker A{)}^{\perp }$, whose dimension is at most $n$. Since $A=AQ$, the trace of $Q$ givesThis proves ${\pi }_2(I_E)\le \sqrt{n}$ for every norm, without comparing that norm to an arbitrary Euclidean norm.

To obtain equality and the projection bound, choose a maximum-volume centered ellipsoid inside the unit ball $B_E$ and use coordinates in which it is the Euclidean ball $B_2$. In the real case this is the John ellipsoid; in the complex case maximize among Hermitian ellipsoids. There are unit contact vectors $u_r\in \partial B_E\cap \partial B_2$ and positive weights $c_r$ with the John contact decompositionFor completeness, the maximum-volume argument gives this identity. If $I_E$ were outside the closed convex cone of the contact matrices, finite-dimensional separation would supply a symmetric, or Hermitian, matrix $D$ with $\mathrm{tr}D>0$ and $u^{\ast }Du\le 0$ at every contact. The cone is closed because its compact generators have trace one. Subtract a sufficiently small multiple of the identity to make all contact inequalities strict while retaining positive trace. The support function of $(I+tD)B_2$ decreases near the contact directions for small $t>0$; away from them the strict gap between the support functions of $B_E$ and $B_2$ permits the same small deformation. Thus this deformed ball still lies in $B_E$. Its volume increases, since the real determinant is $1+t\mathrm{tr}D+O(t^2)$ in the real case and $|{\det }_{\mathbb{C}}(I+tD){|}^2=1+2t\mathrm{tr}D+O(t^2)$ in the complex case. This contradicts maximality. The finite-dimensional convex-cone representation gives finitely many contacts, and taking traces gives the sum of the weights.

At a contact $u_r$, a supporting hyperplane to $B_E$ also supports $B_2$. The Euclidean ball has a unique supporting direction there, so the corresponding functional is ${\varphi }_r(x)=u_r^{\ast }x$ and has dual norm one on $E$. In the complex case balancedness of $B_E$ turns the real support bound into $|{\varphi }_r(x)|\le \Vert x{\Vert }_E$. Also $B_2\subseteq B_E$ gives $\Vert x{\Vert }_E\le |x{|}_2$. For any dual-unit functional $\varphi$, this containment implies that its Euclidean norm is at most one. Apply the summing definition to $x_r=\sqrt{c_r}{u}_r$:The weak square norm is exactly one, since any contact functional attains that Euclidean norm. Hence ${\pi }_2(I_E)\ge \sqrt{n}$. Together with the upper estimate, the norm isThis proves the 2-summing norm of a finite-dimensional identity formula.

Finally retain the preceding finite-dimensional hypothesis when $E\subseteq X$. Extend each ${\varphi }_r$ to a dual-unit functional ${\tilde{\varphi }}_r$ on $X$ using the real or complex Hahn-Banach theorem. DefineThe contact identity gives $P{|}_E=I_E$, hence $P^2=P$ and its range is $E$. The synthesis operator $(a_r)\mapsto {\sum }_r\sqrt{c_r}{a}_r{u}_r$ has Euclidean operator norm one, because its product with its adjoint is the identity. ThereforeThus the linear projection satisfies $\boxed{\Vert P\Vert \le \sqrt{n}}$, the Kadec-Snobar projection bound. If $n=0$, the zero projection gives the same conclusion. No assertion about arbitrary infinite-dimensional subspaces is required.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

Let $a_1,a_2,a_3$ denote the columns. The Gram-Schmidt process begins with $\Vert a_1\Vert =2$, soThe second column has $r_{12}=q_1^{\mathsf{T}}{a}_2=4$ and residualThis residual has norm $2$, giving $q_2=\frac{1}{2}(-1,1,-1,1{)}^{\mathsf{T}}$ and $r_{22}=2$.

For the third column the orthogonal projections are $r_{13}=6$ and $r_{23}=4$. Subtracting them givesIts norm is $2$, so $q_3=\frac{1}{2}(1,1,-1,-1{)}^{\mathsf{T}}$ and $r_{33}=2$. Thus the requested thin QR decomposition isThe three displayed columns have pairwise zero inner products and unit norms; direct multiplication gives $QR=A$.

Write $1$ for the all-ones vector and setThe centering assumption makes $P_0{P}_X=0$. Thus the full normal linear model has fitted-value orthogonal projection $P_0+P_X$, and its residual sum of squares is $R=Y^{\mathsf{T}}(I-P_0-P_X)Y$. Assume $\nu >0$, as is needed to estimate the unknown error variance from residuals.

For the null that all slopes vanish, the reduced fitted value is $\bar{Y}1$, with residual sum of squares $R_0=Y^{\mathsf{T}}(I-P_0)Y$. The F-test isReject for an upper-tail value exceeding the $1-\alpha$ quantile of this F-distribution. The required theorem is Cochran's theorem: for an isotropic normal distribution, projections onto orthogonal subspaces are independent, and the squared norm on a rank-$r$ subspace divided by ${\sigma }^2$ has a chi-squared distribution with $r$ degrees of freedom. Here the null makes the projected mean in $\mathrm{col}(X)$ zero, so $(R_0-R)/{\sigma }^2\sim {\chi }_p^2$, independently of $R/{\sigma }^2\sim {\chi }_{\nu }^2$.

Let $p_1,p_2$ be the numbers of columns in the two predictor blocks. To test the first block while retaining the second as nuisance, formThe partial F-test for nested linear models isAgain reject in the upper tail. The difference of the nested fitted-value projections has rank $p_1$, is orthogonal to the full residual projection, and annihilates the null mean, which gives the same independent chi-squared argument. If a block has no columns, the corresponding test is vacuous.

The phrase orthogonal statistical parameters refers to zero cross-block Fisher information, not to the numerical parameter vectors being perpendicular. In this model the cross-block information is $X_1^{\mathsf{T}}{X}_2/{\sigma }^2$, so the orthogonal coefficient blocks in a centered normal linear model condition isThen the two fitted subspaces are orthogonal, the block least-squares estimators have zero covariance and are independent because they are jointly Gaussian, and each block's extra sum of squares is unchanged by including the other block. Centering also makes both blocks orthogonal to the intercept.

The calculation is principal component analysis on a correlation matrix, rather than on an unscaled covariance matrix. For each of the eleven rating variables, subtract its column mean and divide by its column standard deviation. If $Z$ is the resulting $43\times 11$ matrix, use the divisor convention of the fitted object to writeHere the scaling standard deviations use divisor $43$, consistently with this convention. Using sample standard deviations with divisor $42$ gives the same correlation matrix and directions, with the corresponding score scale. Every diagonal entry of $R$ is one, so ${\sum }_k{\lambda }_k=\mathrm{tr}R=11$.

A principal component is the linear combination of the standardized ratings along $a_k$. Its principal component score for judge $i$ is $t_{ik}=z_i^T{a}_k$, and its variance is ${\lambda }_k$. The displayed component standard deviations are $\sqrt{{\lambda }_k}$; the explained variance of a principal component is ${\lambda }_k/11$. In particular,The first component explains about $92.122\%$ of the standardized variation, and the first two jointly explain $95.982\%$. Components three, four and five add about $2.321\%$, $0.830\%$ and $0.339\%$. The cumulative line adds these proportions successively; it is not the proportion explained by that component alone. Retaining two components discards about $4.018\%$ of the total standardized squared variation.

The displayed principal component loadings are the unit eigenvector coefficients $a_{jk}$. The first direction has positive, broadly similar weights for all ratings. It measures an overall favorable-rating tendency: a judge high on most ratings has a high first score. The second direction compares integrity and demeanour, with large positive coefficients, against such attributes as decision promptness, case-flow management and physical ability, with negative coefficients. Thus among similarly rated judges, a positive second score indicates relatively stronger integrity/demeanour compared with those operational ratings. This is a description of a data direction, not a causal interpretation. Multiplying a whole eigenvector and its scores by minus one reflects the display but changes nothing substantive.

These principal component loadings are not themselves variable-component correlations. With standardized variables,For example, the integrity correlation with component one is approximately $3.1833\times 0.289\simeq 0.920$. Blank loadings in the printed second column indicate suppressed small entries, not exact zeros. The complete unrounded loading vectors in the fitted object must be used to calculate scores.

A two-dimensional representation plots the 43 pairswith each point labeled by its judge. In the fitted software object these are the first two columns of the score matrix. Nearby points have similar projections of their standardized ratings. The rank-two reconstruction ${\hat{z}}_i=t_{i1}{a}_1+t_{i2}{a}_2$ is the orthogonal projection onto the leading principal plane and minimizes the total squared reconstruction error over all two-dimensional linear subspaces. To see this, for any orthonormal pair of directions the captured variance is the sum of its two Rayleigh quotients, at most ${\lambda }_1+{\lambda }_2$ by diagonalizing $R$; the remaining variance is the total minus that captured sum. Distances in the plot omit the smaller components, so a close projected pair need not be equally close in every original rating. The actual 43 coordinates cannot be recovered from the summary alone; the supplied data matrix or fitted score matrix is needed.

The printed descriptive list omits INTG, although the loading table includes it. INTG denotes integrity and supplies the eleventh variable. The standard data set also contains a contacts variable; the stated eleven-rating analysis uses the ratings, not contacts.

Put $\mu ={\mu }^0$, $\lambda ={\kappa }^0-2\mu /3$, and $L=\lambda +2\mu ={\kappa }^0+4\mu /3$. Assume positive bulk modulus and shear modulus. For a unit vector $\xi$, direct contraction of the isotropic elastic stiffness tensor gives the acoustic tensor and its inverse:These are its transverse and longitudinal orthogonal projections. Symmetrizing the definition of the directional elastic strain Green operator consequently givesSubstitution into the spherical representation is already the isotropic specialization. We can also evaluate its angular integrals as distributions. For $r=|x|>0$, direct integration with the polar axis parallel to $x$ givesThe first identity holds as a locally integrable distribution. Since the second derivative of $|s|$ is $2\delta (s)$, taking two derivatives in the first identity and four in the second givesThus an explicit complete isotropic elastic strain Green kernel isAll derivatives here are distributional derivatives, which retain the contribution at the origin.