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Substitute the QR decomposition into the normal equations:The diagonal of $R$ is nonzero, so $R^T$ is invertible. Cancel it algebraically to obtainCompute the three inner products in $Q^{T}b$, then use back substitution in this upper triangular matrix. There is no need to form an inverse or explicitly construct $A^{T}A$. The orthogonal projection of $b$ onto the column space is $Q{Q}^{T}b$; its orthogonal residual proves that this solution minimizes $\Vert Ay-b{\Vert }_2$.

The first column has norm two, giving $q_1=(1,-1,1,-1{)}^T/2$. Its inner products with the second and third columns are both one. Subtracting $q_1$ from the second column leaves $(3,3,3,3{)}^T/2$, of norm three, so $q_2=(1,1,1,1{)}^T/2$. The inner product of $q_2$ with the third column is one. Subtracting $q_1+q_2$ from that column leaves $(1,1,-1,-1{)}^T$, of norm two. Thus the Gram-Schmidt process givesThe columns of $Q$ have unit norm and pairwise zero inner products, so $Q^{T}Q=I_3$. Direct multiplication verifies $QR=A$.

Use an immutable index and mutable value and links:For the sparse linked vector representation, a header h has h.index >= 0, records that vector's dimension, and lies on a finite circular linked list. Every ordinary linked list node has an index in [0,h.index), a nonzero finite value, and a unique index; indices increase strictly from h.next to h.prev. For every reachable linked list node, including h, p.next.prev eq p and p.prev.next eq p. The abstraction function fills unspecified coordinates with zero. The header's value has no meaning. Finite real coordinates are assumed throughout; infinities and NaN are not ordinary real number coordinates for a mathematical inner product.

If $\mathbf{b}=0$, the objective is zero at every feasible point, so every point of the ellipsoid is a minimizer. Suppose now $\mathbf{b}\ne 0$. A minimum cannot be interior: a sufficiently small step in direction $-\mathbf{b}$ would remain feasible and decrease the objective. The boundary is active.

Forstationarity gives $\mathbf{b}+2\lambda A\mathbf{x}=0$. A positive-definite symmetric matrix is invertible, and the minimizing sign has $\lambda >0$. Enforcing the boundary gives

For global optimality, use the inner product $⟨u,v{⟩}_A=u^{\mathsf{T}}Av$. The Cauchy-Schwarz inequality givesThe proposed point attains the negative of this bound. Thus the minimum is $-\sqrt{{\mathbf{b}}^{\mathsf{T}}{A}^{-1}\mathbf{b}}$, and equality in the Cauchy-Schwarz inequality, together with the sign and boundary conditions, makes this minimizer unique when $\mathbf{b}\ne 0$.

Every linear map from a qubit into the tensor product of that qubit and an environment can be expanded in the Pauli matrix orthonormal basis of operators. If $V|0⟩=|0⟩|e_{00}⟩+|1⟩|e_{01}⟩$ and $V|1⟩=|0⟩|e_{10}⟩+|1⟩|e_{11}⟩$, then the four coefficient vectors areThese environment vectors need not be orthogonal, so this is an operator expansion rather than a claim that the quantum channel is a classical Pauli channel. Physical isometries additionally preserve the inner products of the two input basis states.

The Pearson correlation coefficient is the covariance divided by the product of the standard deviations. For real random variables with finite, strictly positive variances,It is the expected value of the product of the two centered variables after dividing each by its standard deviation. The Cauchy-Schwarz inequality gives $-1\le {\rho }_{X,Y}\le 1$. This removes measurement units, while retaining the direction and strength of the linear relationship. Constant random variables have zero standard deviation, so their Pearson correlation coefficient is undefined.

For the covariance model $Y=2X+Z$, with $X,Z$ independently uniform on $\{-1,0,1\}$, the calculation givesThe positive correlation is less than one because the independent $Z$ term introduces extra variation. For $Y=-2X+Z$ the correlation is $-2/\sqrt{5}$. Both examples appear in the covariance diagram.

For nonzero real $a,c$, the correlation of $aX+b$ and $cY+d$ is $\mathrm{sgn}(ac){\rho }_{X,Y}$. Positive rescaling and added offsets therefore leave the correlation unchanged; reversing one scale reverses its sign. Equality $|{\rho }_{X,Y}|=1$ means that one random variable is an affine function of the other almost surely, with a nonzero slope of the corresponding sign. This is the equality condition in the Cauchy-Schwarz inequality, and includes perfect positive correlation.

Zero correlation says that the centered variables are orthogonal in their square-integrable inner product. It permits nonlinear statistical dependence, as the example $Y=X^2$ in uncorrelated random variables demonstrates. More generally, a correlation is a statement about a joint probability distribution; identifying effects of interventions is the separate task of causal inference. This distinction keeps an observed association from being mistaken for a demonstrated mechanism.

An early source for standardized association is Francis Galton's 1888 paper “Co-relations and their Measurement, chiefly from Anthropometric Data”. He compared stature and forearm measurements and changed each measurement to units of its own variability so the two regression directions could be compared. His scale used probable error, so it should not be identified word for word with the modern standard deviation formula above. ^[ref].

If a positive operator-valued measure distinguishes normalized pure states $|u⟩,|v⟩$ with certainty, the effect $E$ for the first answer satisfies $⟨u|E|u⟩=1$ and $⟨v|E|v⟩=0$. Positivity of $E$ and $I-E$ implies $E|v⟩=0$ and $E|u⟩=|u⟩$, because a positive operator with zero expectation annihilates the vector. Consequently $⟨u|v⟩=⟨u|E|v⟩=0$. Conversely, orthogonal states can be measured with their rank-one linear projections. A common unitary operator preserves this inner product and cannot remove the obstruction.

In 1998, Jian-Wei Pan, Dik Bouwmeester, Harald Weinfurter and Anton Zeilinger created two photon pairs by two passages of a pulsed pump through a crystal using spontaneous parametric down-conversion. Interfering the inner photons at a nonpolarizing beam splitter implemented a partial Bell-basis measurement. Narrow optical filters and spatial filtering removed distinguishing information; changing the outer polarizer bases checked the conditional correlations. Fourfold detections selected the entanglement swapping events. The reported interference supported conditional quantum entanglement, but the paper reserved a statistically significant Bell inequality violation for improved visibility; it was not a demonstration closing the detection loophole.^[ref]

The ideal calculation makes the distinction between conditional and unconditional quantum states explicit. Start with spin singlet states $|{\psi }^{-}{⟩}_{12}|{\psi }^{-}{⟩}_{34}$, where $|{\psi }^{-}⟩=(|HV⟩-|VH⟩)/\sqrt{2}$ and $H,V$ denote orthogonal photon polarizations. Taking the inner product with $⟨{\psi }^{-}{|}_{23}$ gives $-|{\psi }^{-}{⟩}_{14}/2$: expanding the four terms leaves precisely the two antisymmetric outer terms. That Bell-basis measurement outcome has probability $1/4$ and leaves the outer pair in a spin singlet state. Ignoring the inner outcome instead leaves the outer density matrix $I_4/4$, a maximally mixed state. A conditional preparation flag is useful for later Bell tests only with the timing and setting conditions in setting-independent heralding preserves Bell locality.

A planar rotation about the origin is the linear map represented by the rotation matrix $R_{\theta }=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. It preserves inner products, lengths, oriented angles and area, with positive angles measured anticlockwise. Its inverse is $R_{-\theta }$ and its determinant is one.

Let $H,V$ be orthogonal linear polarizations, and prepare the Bell state $|{\Phi }^{+}⟩=(|HH⟩+|VV⟩)/\sqrt{2}$. The $+1$ analyzer vector is $|a⟩=\cos a|H⟩+\sin a|V⟩$; its orthogonal companion has outcome $-1$. Direct inner products with the Bell state give $P_{++}=P_{--}={\cos }^2(a-b)/2$ and $P_{+-}=P_{-+}={\sin }^2(a-b)/2$. Thus the expectation value of the outcome product is $E(a,b)=\cos 2(a-b)$. These are physical polarizer angles; their corresponding Bloch sphere angles are doubled.

For a stipulated isotropic admixture, use the density matrixThe maximally mixed state contributes zero correlation, so $P_{st}(a,b)=[1+stv\cos 2(a-b)]/4$ for $s,t\in \{-1,1\}$. Each local marginal is $1/2$, independently of the other setting, illustrating quantum no-signalling. In this model $v$ is the interferometric visibility of the coincidence fringe; other noise models need not have one common visibility for every basis.

Choose $a_0=0^{\circ }$, $a_1=45^{\circ }$, $b_0=22.5^{\circ }$ and $b_1=-22.5^{\circ }$. The four correlations in the displayed CHSH inequality are $v/\sqrt{2},v/\sqrt{2},v/\sqrt{2},-v/\sqrt{2}$. Consequently $S=2\sqrt{2}v$, exceeding the local bound exactly when $v>1/\sqrt{2}$. Perfect correlation at a single common setting would not yield this comparison between incompatible quantum measurements.

To connect probabilities with expected counts, stipulate $M$ emitted pairs per setting pair and independent detection efficiencies ${\eta }_A,{\eta }_B$, independent also of settings, outcomes and hidden variables. The expected joint counts are $n_{st}=M{\eta }_A{\eta }_B{P}_{st}$. Their normalized correlation isFor original teaching inputs $M=1000$, ${\eta }_A=0.8$, ${\eta }_B=0.75$ and $v=0.8$, the $(a_0,b_0)$ expected counts are $n_{++}=n_{--}\simeq 234.85$ and $n_{+-}=n_{-+}\simeq 65.15$. They total $600$ and give $E_{\det }\simeq 0.56569$, hence $S\simeq 2.26274$. Fractional expected counts are not measured data. Efficiency cancels here because the loss model was specified to be independent; this algebra supplies no justification for discarding failures in an arbitrary Bell test with a detection loophole.

For $x,z\in {\mathbb{R}}^d$, a polynomial kernel is the positive-semidefinite kernel $k(x,z)=(x^{T}z+c{)}^r$, where $r$ is a positive integer and $c\ge 0$. The case $c=0$ uses monomials of total degree $r$; $c>0$ also includes lower degrees. The feature map weights matter: listing each distinct monomial once with coefficient one generally gives a different inner product.^[ref]

For the homogeneous quadratic case in two dimensions, take the scaled quadratic feature mapThe inner product identity is explicit:The factor $\sqrt{2}$ accounts for the two equal cross terms in the square. Thus an inner product of three derived coordinates can be obtained from an inner product of the two input coordinates followed by squaring. The kernel trick uses this equality inside an algorithm; the feature map explains which nonlinear functions the algorithm can represent.

For a positive-definite weighted inner product on a compact interval, a least-squares polynomial is the orthogonal projection of a function onto the finite-dimensional space ${\Pi }_n$ of polynomials of degree at most $n$. In a basis ${\varphi }_0,\dots ,{\varphi }_n$, its coefficients solve the normal equations $Gc=d$, where $G_{ij}=({\varphi }_i,{\varphi }_j)$ and $d_i=(u,{\varphi }_i)$. The Gram matrix is positive definite: $z^{T}Gz=\Vert \sum z_i{\varphi }_i{\Vert }^2>0$ for $z\ne 0$. Thus the coefficients exist and are unique. The residual is orthogonal to every candidate polynomial, giving $\Vert u-q{\Vert }^2=\Vert u-p_n{\Vert }^2+\Vert p_n-q{\Vert }^2$ and proving optimality. An orthogonal basis makes these equations diagonal, as in a least-squares polynomial in an orthogonal-polynomial basis.

A real symmetric bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.^[ref]

The identity matrix $A=\mathrm{diag}(1,1)$ gives $B(v,w)=v_1{w}_1+v_2{w}_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving norm $5$. A weighted example, $A=\mathrm{diag}(4,1)$, gives norm $\sqrt{4x^2+y^2}$: the first coordinate costs twice as much in length. Such inner products let geometry assign different weights to different directions.

Every positive-definite bilinear form is nondegenerate: a vector in its radical of a bilinear form has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\mathrm{diag}(1,-1)$ shows. Even positivity of the associated quadratic form does not force a general bilinear form to be symmetric: $A=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$ has $v^{T}Av=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

A real symmetric matrix $A$ is positive definite whenfor every nonzero real vector $x$. It therefore defines the inner product $⟨x,y{⟩}_A=x^{T}Ay$ and is invertible.

If a real matrix $B$ has a positive diagonal matrix $W$ such that $WB$ is a symmetric matrix, then $H=W^{1/2}B{W}^{-1/2}$ is a symmetric matrix related to $B$ by a similarity transformation: $B^{T}W=WB$ gives $H^T=H$. Thus $B$ is a diagonalizable matrix with real eigenvalues. If ${\mathbf{z}}^{T}WB\mathbf{z}\le 0$ for all real vectors $\mathbf{z}$, its eigenvalues are nonpositive. This converts symmetry in a weighted inner product into Euclidean symmetry.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

A tied largest modulus can prevent even direction convergence. With $D=\mathrm{diag}(2,-2)$ and $x_0=(1,1{)}^T/\sqrt{2}$,The two directions form a cycle; neither is an eigenvector. There is no dominant-modulus spectral gap. A tied modulus is not a universal failure: an initial eigenvector remains on its line, and $A=2I$ preserves every direction. Subspace iteration addresses a separated cluster by targeting a dominant invariant subspace instead of one eigenvector.

An orthogonal start can fail despite a strict spectral gap. For $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $x_0=(1,-1{)}^T/\sqrt{2}$, the dominant component is zero, so $A{x}_0=x_0$ and every iterate equals $x_0$. The Rayleigh quotient is exactly $1$ and the eigenpair residual is zero, although the dominant eigenvalue is $3$. Exact arithmetic never creates a missing component.^[ref]

The third panel contrasts these failures with $N=\mathrm{diag}(-3,1)$, $x_0=(1,1{)}^T/\sqrt{2}$, for whichHere the horizontal eigenvector direction converges with a strict spectral gap, while the normalized vectors alternate signs. This is the successful direction convergence described under sign and phase in power iteration.

For a nonorthogonal eigenbasis, even the word “orthogonal” must be used carefully. The diagonalizable matrix $C=\left(\begin{array}{cc}3& 2\\ 0& 1\end{array}\right)$ has right eigenvectors $v_1=(1,0{)}^T$, $v_2=(-1,1{)}^T$, and dominant left eigenvector $w_1=(1,1{)}^T$. For $x_0=(a,b{)}^T$, the dominant coefficient is $c_1=w_1^T{x}_0=a+b$. The starting vector $v_2$ is nonorthogonal to $v_1$ but stays at the smaller eigenvalue $1$; $(0,1{)}^T$ is orthogonal to $v_1$ but has $c_1=1$ and converges to the dominant direction. The left eigenvector test, not the right-eigenvector inner product, controls the outcome.

A random vector drawn from a continuous probability distribution with a probability density function on the full real or complex vector space has $c_1\ne 0$ with probability one. This avoids the exact missing-component hyperplane, but does not guarantee a large component or fast convergence. Floating-point arithmetic may seed a missing component, yet a structurally preserved zero can remain zero, so accidental perturbations are not a convergence guarantee.

Given a positive-definite matrix $M$, the preconditioned proximal point map isIt is firmly nonexpansive in the weighted inner product $⟨v,w{⟩}_M=⟨v,Mw⟩$ whenever $F$ is monotone.

A score is an observation's coordinate along a principal component direction. Center an observation by the fitted sample mean, apply the fitted variable scaling when using principal component analysis on a correlation matrix, then take its inner product with the unit principal component loading vector. New observations must use the training center and scaling. The score variance equals the corresponding eigenvalue when the covariance normalization is consistent.