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An orthogonal basis of an inner-product vector space consists of nonzero mutually orthogonal vectors that span the space. The orthogonal projection of $v$ onto a finite-dimensional span is ${\sum }_i⟨v,u_i⟩u_i/⟨u_i,u_i⟩$. Dividing each vector by its norm produces an orthonormal basis. For centered random variables, the inner product is covariance; an orthogonal basis therefore consists of uncorrelated variables, without requiring unit variance.

An orthonormal basis is a basis whose vectors are pairwise orthogonal and each have norm one.

If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, symmetry givesFor $\lambda \ne \mu$, ${\mathbf{u}}^T\mathbf{v}=0$, so the eigenvectors are orthogonal.

Let $H$ be the measured wave scattering response> linear operator, with direct background propagation removed if present. Its time reversal operator is the positive operator $T=H^{\ast }H$. For a normalized emitted signal $h$,The maximizing signal is a right singular vector, equivalently an eigenvector of $T$ with its largest eigenvalue. This is a precise intensity statement independent of a scatterer model.

Under the Born approximation for scalar wave scattering, neglect interactions between point scatterers and writewhere $b_j$ is the receiver response to scatterer $j$, $c_j^{\ast }h$ is its illumination by the source array, and ${\alpha }_j$ is its scattering wave amplitude. Well-resolved point scatterers have approximately orthogonal vectors $b_j$ and $c_j$. In the ideal orthogonal limit,Thus the most reflective, geometrically weighted scatterer gives the largest eigenvalue, and its normalized steering vector $c_j/\Vert c_j\Vert$ gives the corresponding wave focusing signal. If illumination and reception factors are equal for all scatterers, this is precisely the scatterer with largest $|{\alpha }_j{|}^2$. Geometric size by itself is not the quantity being ranked.

Repeated adjoint time reversal acoustics is power iteration on $T$. If its top eigenvalue is simple and the initial signal has a nonzero component along its eigenvector, thenA tied largest eigenvalue leaves a combination in the leading eigenspace, and an initial signal orthogonal to that eigenspace cannot excite it. This is the basis of the DORT method. Mere physical separation is insufficient if the array cannot resolve the scatterers; coherent steering-vector overlap or multiple scattering can mix the modes. Also, if $H$ includes unrestricted homogeneous transmission, its largest eigenvalue need not identify any individual scatterer. The correspondence concerns the resolved wave scattering response>. The ideal correspondence and selective wave focusing are analyzed by Prada and Fink.

Write $a=(2,1{)}^T$, so the data map is $a^{T}u$. The Gaussian likelihood and prior distribution giveCompleting the square, or using Gaussian conjugacy for a normal linear model, identifies a multivariate normal distribution with precision matrix $C_{\delta }^{-1}=I_2+{\delta }^{-2}a{a}^T$. Since $a^{T}a=5$, its posterior mean and covariance matrix areThe parameters areFor example, $(I_2+{\delta }^{-2}a{a}^T)C_{\delta }=I_2$, which checks the square completion.

The orthogonal vectorsare eigenvectors of the covariance matrix. Their respective variances areThey are independent under the posterior distribution, by independence of uncorrelated jointly normal variables. Thus uncertainty is different in the two directions even for positive noise: the data constrain $e_{\parallel }$, whereas $e_{\perp }$ spans the null space of the data map.

As $\delta \to 0$, the posterior distribution converges to the law ofThis is supported on $2u_1+u_2=m$, and its posterior mean is the minimum-norm least-squares solution. The prior distribution still determines the distribution along the unobserved null space; it has not disappeared. This example of a Gaussian posterior in the zero-noise limit distinguishes exact recovery of an observed component from recovery of the whole unknown.

Let ${\theta }_k=k\pi h=k\pi /(m+1)$ and define $q_j^{(k)}=\sin (j{\theta }_k)$. The artificial boundary values are $q_0^{(k)}=q_{m+1}^{(k)}=0$. For the symmetric tridiagonal Toeplitz matrix $H=[\beta ,\alpha ,\beta ]$, the sine addition formula givesThus all such matrices have the same discrete sine transform eigenvectors, with eigenvaluesThe sine vectors are mutually orthogonal vectors and nonzero, so the matrix $Q=(\sin (ij\pi h){)}_{i,j=1}^m$ is invertible. If $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_m)$, then $HQ=QD$ and therefore

A matrix is Hermitian when $A^{†}=A$, and unitary when $U^{†}U=U{U}^{†}=I$.

If $A\mathbf{u}=\lambda \mathbf{u}$ with $\mathbf{u}\ne 0$, thenso every eigenvalue is real. If $A\mathbf{u}=\lambda \mathbf{u}$ and $A\mathbf{v}=\mu \mathbf{v}$, thenDistinct eigenvalues therefore have orthogonal eigenvectors.

The normalized eigenvectors form an orthonormal basis, so the matrix $U=({\mathbf{u}}_1\cdots {\mathbf{u}}_n)$ satisfies $U^{†}U=I$ and is unitary. Withthe equations $A{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ say $AU=UD$, hence $A=UD{U}^{†}$. This is the finite-dimensional unitary diagonalization of a normal matrix.

A unitary conjugate of an arbitrary diagonal matrix need not be Hermitian: $U=(1)$ and $D=(i)$ give $UD{U}^{†}=(i)$. It is Hermitian exactly when all diagonal entries of $D$ are real.

For the displayed matrix, one valid decomposition isThe columns are orthonormal eigenvectors, so direct multiplication gives $UD{U}^{†}=A$.

The supports of ${\phi }_1$ and ${\phi }_2$ are disjoint, so they are orthogonal vectors. Substitution in the covariance formula givesandTherefore the associated eigenvalues are

There are uncountably many physically distinct one-qubit unitaries. Part (ii) would require an orthogonal program state for every one, whereas a $K$-qubit register has dimension only $2^K$. This contradiction is the no-programming theorem, so no deterministic gate satisfying (PROG) exists.