Codex @codex  0

All $n$ singular values are positive, so $\Sigma$ and $V$ are invertible matrices and $A$ has matrix rank $n$. The image of a linear map is thereforewith those columns furnishing an orthonormal basis. Also $Ax=0$ implies $\Sigma V^{T}x=0$ and hence $x=0$. ThusFinally $A^{T}z=V\Sigma U^{T}z=0$ exactly when $U^{T}z=0$. The remaining columns of the full orthogonal matrix span this orthogonal complement:If $m=n$, this last null space also equals $\{0\}$ and has the empty basis.

The orthogonal projection $P=Q{Q}^T$ givesThe first summand has every column in the column space of $Q$, and the second has every column in its orthogonal complement. They therefore have zero Frobenius inner product. The Pythagorean identity, together with $Q^{T}Q=I_r$, gives the stronger exact formulaConsequentlywith equality exactly when $M=Q^{T}A$. Thus solving the linear least-squares problem for each column chooses the best coefficients in the prescribed column space.

At a constrained critical point, the directional derivative of $f$ vanishes along every admissible tangent velocity:For a regular zero locus, $T_{a}M=\ker D{g}_a$. A linear functional vanishing on this kernel lies in the row space of $D{g}_a$. Equivalently there are Lagrange multipliers ${\lambda }_1,\dots ,{\lambda }_k$ such thatGeometrically, $\nabla f$ is orthogonal to the tangent space, so it lies in its orthogonal complement, spanned by the independent constraint gradients. Thus the level set of $f$ has no first-order change in a tangent direction. The equations provide candidates for constrained extrema; they must still be classified and compared, and boundary or singular cases must be treated separately. The condition is necessary for a regular constrained extremum, not a guarantee of a maximum.

If $x\in \mathrm{Null}(A)$, the $i$th entry of $Ax=0$ says $r_i\cdot x=0$. Any $y\in \mathrm{Row}(A)$ is a linear combination $y={\sum }_i{\alpha }_i{r}_i$, so bilinearity of the dot product givesHenceThis proves the orthogonality for every pair of vectors in the two vector subspaces. In fact the null space equals the orthogonal complement of the row space, since being orthogonal to every row is exactly the system $Ax=0$.

The gradient vanishes at $0$. Decompose any vector as $x=tb+w$, with $w\perp b$. Symmetry of $A$ and $Ab=\frac{1}{2}b$ imply $b^{T}Aw=(Ab{)}^{T}w=0$. Since $\Vert b\Vert =1$,If $d\ge 2$, choose a nonzero $w\perp b$. Arbitrarily close to zero, the direction $b$ gives negative values and the direction $w$ gives positive values. Thus $0$ is a saddle point for $d\ge 2$. Equivalently, the Hessian matrix has one negative eigenvalue and all its other eigenvalues positive.

The dimension matters: if $d=1$, the orthogonal complement is zero-dimensional and $g(tb)=-t^2/2$. In that case $0$ is the unique global maximum. This edge case is not excluded by the stated hypotheses. The decomposition is a rank-one downdate along an eigenvector.

For a unit vector $u$ spanning the target eigenspace and a normalized iterate $x_k$, defineThis is the norm of the component in the orthogonal complement of $u$, equivalently the sine of the acute angle to its line. It ignores the arbitrary sign or complex unit factor of an eigenvector. Under power iteration convergence for a diagonalizable matrix, ${\epsilon }_k=O({\gamma }^k)$ in Big O notation. The orthogonal projection defines how we measure direction; it does not assume that the eigenbasis of $A$ is orthogonal. In computation, the projection form avoids subtracting nearly equal numbers inside the square root.

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 orthonormal wavelet refinement filter, ${\sum }_k{h}_k\overline{h_{k-2n}}={\delta }_{n0}$ is equivalent to the displayed identity. A high-pass filter can be chosen with coefficients $g_k=(-1{)}^k\overline{h_{1-k}}$. In frequency, its symbol is $e^{-i\xi }\overline{m(\xi +\pi )}$, up to a constant phase. The two channels split an approximation space into a coarser approximation and its orthogonal complement.

For a unit vector $u$ and $\theta >0$, a rank-one covariance spike is the covariance matrix $I+\theta u{u}^{\top }$. Its eigenvalue is $1+\theta$ in the direction $u$ and one on the orthogonal complement. It arises by adding an independent scalar normal distribution signal $Yu$ to a standard Gaussian random vector.

Let $A$ be a real symmetric positive-definite matrix, and let $b$ be a unit eigenvector with eigenvalue $\lambda >0$. The rank-one matrix update $C=A-\rho b{b}^T$ changes only the eigenvalue along $b$, to $\lambda -\rho$. Symmetry makes the orthogonal complement $b^{\perp }$ invariant, and $Cw=Aw$ there because $b^{T}w=0$.

In detail, write $x=tb+w$ with $w\perp b$. ThenConsequently $C$ is a positive-definite matrix for $\rho <\lambda$, singular for $\rho =\lambda$, and has exactly one negative eigenvalue for $\rho >\lambda$. In dimension at least two the last case is indefinite. In dimension one it is negative definite instead: there is no nonzero orthogonal-complement direction. The distinction matters when classifying a stationary point of a quadratic form.

A closed subspace of a Hilbert space reduces a bounded operator when both the operator and its adjoint leave the subspace invariant. Equivalently, the orthogonal projection onto it commutes with the operator: the adjoint invariance makes its orthogonal complement invariant too. For a star-closed algebra of operators, the closed span of the orbit of any vector is reducing for the whole algebra. This enables an orthogonal direct-sum decomposition into cyclic actions when constructing Borel functional calculus for a normal operator.

Let $v$ be a real unit vector in $d$-dimensional Euclidean space. Then $P=v{v}^{\mathsf{T}}$ is a rank-one orthogonal projection: $Pv=v$, while $Pw=0$ for $w$ orthogonal to $v$. Consequently the symmetric matrix $A=\alpha I+\beta P$ has eigenvalue $\alpha +\beta$ on the line spanned by $v$ and eigenvalue $\alpha$ on its orthogonal complement. Any orthonormal basis of that complement completes $v$ to an orthonormal eigenbasis. ThusThe trace and determinant formulas remain valid when one of these eigenvalues is zero. When $\beta =0$, all nonzero vectors have the same eigenvalue, and the chosen splitting is optional.

A unit nowhere-zero section $\sigma$ of a real vector bundle with a fiber metric defines a rank-one trivial vector bundle. Its fiberwise orthogonal complement is a subbundle: locally complete $\sigma$ to an orthonormal frame. The vector bundle isomorphism $(v,t)\mapsto v+t\sigma$ has inverse $w\mapsto (w-⟨w,\sigma ⟩\sigma ,⟨w,\sigma ⟩)$. For an oriented bundle, orient the complement so that its positive frame followed by $\sigma$ is positive.

Let $n\ge 2$ be even and let $E\subseteq {\mathbb{F}}_2^n$ be the vector subspace of even-weight vectors. Its dot product is an alternating bilinear form, since $x\cdot x={\sum }_i{x}_i=0$ on $E$. The orthogonal complement of $E$ in ${\mathbb{F}}_2^n$ is spanned by the all-one vector $1$, which lies in $E$. Consequently the radical of a bilinear form on $E$ is $⟨1⟩$, andis a nondegenerate symplectic vector space of dimension $n-2$. The symmetric group acts by coordinate permutations, preserving this alternating bilinear form.

At $n=6$, this group action is faithful. A weight-two vector and its complement have weights two and four, so a permutation fixing its quotient class must fix that two-element subset itself. Fixing every two-element subset forces the identity. The resulting embedding $S_6\hookrightarrow S{p}_4(2)$ is an isomorphism, because the symplectic group over a finite field order formula gives $2^4(2^2-1)(2^4-1)=720=6!$.

A real orthogonal matrix $T$ of size three with $\det T=-1$ has an eigenvector $n$ of eigenvalue $-1$. Its orthogonal complement is invariant, and the restriction there is an orientation-preserving plane rotation. Thus $T$ is a rotation about the axis $\mathbb{R}n$ composed with a reflection across the plane perpendicular to $n$. Writing the plane angle as $\varphi$, $\mathrm{tr}T=-1+2\cos \varphi$. A pure plane reflection is the case $\varphi =0$; if $\det (T-I)\ne 0$, it cannot be a pure plane reflection.

In an orthonormal basis of localized atomic orbitals, a ring of three equivalent sites with real hopping $t$ has Hamiltonian $H=(E_0-t)I+tJ$, where $J$ is the three-by-three all-ones matrix. The normalized symmetric state $(|1⟩+|2⟩+|3⟩)/\sqrt{3}$ has energy $E_0+2t$. Its two-dimensional orthogonal complement has energy $E_0-t$. Thus the symmetric state is the unique spatial ground state for $t<0$, with an energy difference $3|t|$ to the excited eigenstates. Equivalently, the three discrete Bloch states have energies $E_0+2t\cos (2\pi l/3)$, $l=0,\pm 1$. Nonorthogonal atomic orbitals instead require a generalized eigenvalue problem.

The last-column map has fibre the subgroup fixing the last standard unit vector. Unitary completion gives surjectivity, and compactness gives the quotient homeomorphism. Projecting the other standard vectors onto the orthogonal complement of a nearby unit vector and applying Gram-Schmidt gives a local section, so the projection is a fiber bundle. The bundle $U(n)\to U(n+1)\to S^{2n+1}$ then gives ${\pi }_{i}U(n)\cong {\pi }_{i}U(n+1)$ for $i\le 2n-1$ because both adjacent sphere homotopy groups vanish.