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The squared Euclidean distance along the line isIts minimum occurs at $t=0$ exactly when its derivative there vanishes, so $a\cdot (b\times c)=0$. Since $b,c$ are independent, their linear span is the two-dimensional orthogonal complement of $b\times c$. Consequentlyfor unique real $\alpha ,\beta$. This is the closest point on the intersection of two planes; the positive quadratic coefficient $|d{|}^2$ proves that the stationary distance is a strict global minimum.

The orthogonality condition is $v_j{u}_j=0$, soEvery vector in the orthogonal complement is fixed by the orthogonal projection.

Under an orthogonal Cartesian change of axes $Q$, the vector components transform as $v_i^{\prime }=Q_{ia}{v}_a$, while $v_k^{\prime }{v}_k^{\prime }=v_a{v}_a$ and ${\delta }_{ij}=Q_{ia}{Q}_{ja}$. Substitution givesThis is the rank-two Cartesian tensor transformation law. The nonzero assumption ensures that the denominator is positive. In matrix form $P=I-v{v}^T/(v^{T}v)$ is the orthogonal projection matrix onto the orthogonal complement of $v$.

Decompose ${\mathbb{R}}^n=\mathrm{span}\{a\}\oplus a^{\perp }$. The rank-one matrix update $a{a}^T$ acts as multiplication by $\Vert a{\Vert }^2$ along $a$ and as zero on its orthogonal complement. ThusThe perpendicular eigenvalue has multiplicity $n-1$ and is absent when $n=1$. Consequently $A^{+}$ is always an invertible matrix, while

An eigenvalue $\lambda$ and associated eigenvector $e\ne 0$ satisfy $Ae=\lambda e$. The characteristic polynomial $\det (A-\lambda I)$ has degree equal to the positive matrix size. The fundamental theorem of algebra gives a complex root, at which $A-\lambda I$ is singular and therefore has a nontrivial kernel. Hence every nonempty square complex matrix has an eigenvector.

A nondiagonalizable example is the Jordan block $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$: its only eigenvalue is zero, and its eigenspace is one-dimensional. A Hermitian matrix satisfies $A^{†}=A$ and has an orthonormal eigenbasis by the spectral theorem. One proof finds an eigenvector, notes that its orthogonal complement is invariant, and repeats. Its eigenvalues are real becausewhere $e^{†}e>0$.

With all eigenvalues negative, $A$ is invertible. Its same complete eigenbasis givesSome powered eigenvalues can coincide, but the original eigenvectors remain a complete basis. Writing $A=U\mathrm{diag}({\lambda }_i)U^{†}$, the convergent matrix exponential series givesThus the matrix inverse from a decaying matrix exponential isThe identity printed in the original PDF is missing a minus sign. Already for the one-by-one matrix $A=(-1)$ its integral is $1$, whereas its inverse is $-1$. Equivalently, integrating $d(e^{tA})/dt=A{e}^{tA}$ gives $A{\int }_0^{\infty }{e}^{tA}dt=-I$, confirming the correction without diagonalization.

A group representation on a vector space $W$ is a group homomorphism $\rho :G\to GL(W)$: $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. An invariant subspace $U$ obeys $\rho (g)U\subseteq U$ for every $g$. A nonzero representation is irreducible if its only invariant subspaces are $0$ and $W$. It is a faithful representation if it is injective, equivalently if its group kernel is $\{e\}$. Its character is ${\chi }_{\rho }(g)=\mathrm{tr}\rho (g)$. For inequivalent complex irreducible representations of a finite group, character orthogonality is

Use right-to-left composition of permutations and assign the matrix labels consistently byThe resulting Cayley table, with row multiplying column, isConjugation relabels the entries of a cycle, preserving its length. Thus the conjugacy classes are

Define the permutation representation by $P_{\sigma }{e}_j=e_{\sigma (j)}$, or $(P_{\sigma }{)}_{ij}={\delta }_{i,\sigma (j)}$. Applying it to a basis vector proves $P_{\sigma }{P}_{\tau }=P_{\sigma \tau }$. The six assignments above produce exactly the listed matrices. They are distinct, and $P_{\sigma }=I$ forces $\sigma (j)=j$ for every $j$, so the representation is faithful.

Every permutation fixes $(1,1,1)$, hence fixes every vector $(c,c,c)$. Therefore $V_0$ is a one-dimensional invariant subspace carrying the trivial representation. Its orthogonal complement issince the inner product with $(c,c,c)$ is $c(v_1+v_2+v_3)$. Permuting coordinates preserves their sum, so this plane is invariant too. Every real vector has the unique decompositionThus the permutation representation is the direct sum of group representations on the line and the plane.

To find the eigenvectors of $A$, note that $A(v_1,v_2,v_3)=(v_3,v_1,v_2)$. With $\zeta =e^{2\pi i/3}$, its three eigenvalues and eigenspaces areThe only real eigenvalue is one, whose eigenline is $V_0$, so $A$ has no nonzero real eigenvector in $V_0^{\perp }$. A proper nonzero real invariant subspace of that two-dimensional plane would be a line, giving such an eigenvector. Hence the plane representation is irreducible over $\mathbb{R}$.

For the later complex dimension-count theorem one should also establish irreducibility over $\mathbb{C}$; real irreducibility alone would not imply it in general. In the complexified plane, the only candidate $A$-invariant lines are its $\zeta$ and ${\zeta }^2$ eigenlines. But $C$ exchanges them:Neither is invariant under the whole group, proving complex irreducibility as well. The required decomposition is thereforewhere $W$ is the standard representation of the symmetric group for three letters. Its character is obtained by subtracting one from the number of fixed coordinates: on the three classes it is $(2,-1,0)$.

Finally, the sign representation gives a second one-dimensional irreducible, with value $+1$ on $I,A,B$ and $-1$ on the transpositions. These are exactly the two one-dimensional possibilities: if $r=A,s=C$, a one-dimensional representation sends them to commuting nonzero scalars $v,u$. The relations $r^3=s^2=e$ and $srs=r^{-1}$ imply $v^3=u^2=1$ and $v=v^{-1}$. Hence $v=1$ and $u=\pm 1$, the trivial and sign choices. Together with the two-dimensional $W$, these giveThe sum of squares of irreducible representation dimensions leaves room for no other irreducible. Thus there are exactly two one-dimensional and one two-dimensional complex irreducible representations.

Normalize $e_1$ and put $f_1=M{e}_1/{\mu }_1$. The two vectors are an orthonormal basis of a plane $E$ invariant under $M$, since $M{e}_1={\mu }_1{f}_1$ and $M{f}_1=-{\mu }_1{e}_1$. Its orthogonal complement is also invariant: for $w\perp E$ and $z\in E$,Apply the spectral theorem for real symmetric matrices to the restriction of $M^2$ to $E^{\perp }$. Any remaining negative eigenvalue supplies another orthonormal pair by the same construction. Induction leaves only the kernel. Thus every nonzero eigenvalue of $M^2$ has even multiplicity. The pairs can share an eigenvalue, so this does not assert distinctness of the different $-{\mu }_j^2$.

Use the standard Hermitian inner product $⟨w,v⟩=w^{†}v$. Since $W$ is invariant under every $D_2(g)$, it is in particular invariant underLet $v\in W^{\perp }$ and $w\in W$. Thenbecause $D_1(g{)}^{†}w\in W$ and $v$ is orthogonal to $W$. Thus $D_1(g)v$ is orthogonal to every $w\in W$, provingTherefore $W^{\perp }$ is an invariant subspace for $D_1$. If $D_1$ is irreducible, $W^{\perp }$ is either $\{0\}$ or ${\mathbb{C}}^n$. In finite-dimensional inner product space, $(W^{\perp }{)}^{\perp }=W$, so these alternatives force $W={\mathbb{C}}^n$ or $W=\{0\}$. HenceNo unitarity of $D_1$ was assumed; the inverse-adjoint representation is exactly what makes this orthogonal complement argument work.

The plane spanned by $u_1,v_1$ is an invariant subspace for $M$, sinceIts orthogonal complement is also invariant: for $w$ orthogonal to both, $u_1^{T}Mw=-(M{u}_1{)}^{T}w=0$ and $v_1^{T}Mw=-(M{v}_1{)}^{T}w=0$. The restriction of the real symmetric matrix $M^2$ to this nonempty complement therefore has a unit eigenvector $u_2$, by the spectral theorem for real symmetric matrices.

If its eigenvalue is $-{\mu }_2^2<0$, the argument in part (b) shows that $v_2=M{u}_2/{\mu }_2$ is a unit eigenvector orthogonal to $u_2$. It also lies in the same complement; explicitly,Repeating the construction decomposes every negative eigenspace into orthogonal pairs. The zero eigenspace has dimension $n-2r$ for some number $r$ of pairs, and therefore has the same parity as $n$. This is the paired eigenspaces of the square of a real skew-symmetric matrix property.

For $n=2m$, there can be $m$ distinct negative eigenvalues and no zero, or at most $m-1$ negative ones plus zero, because a nonempty zero eigenspace has dimension at least two. For $n=2m+1$, there are at most $m$ distinct negative ones plus the mandatory zero. HenceThe bound is attained by skew two-by-two blocks with distinct positive parameters, adding one zero direction when $n$ is odd.

The required actions have already been identified:In three dimensions choose a unit vector $w$ perpendicular to both. Its one-dimensional orthogonal complement to the pair is invariant under $M$. On a one-dimensional real space, a skew-symmetric matrix must act as zero: $Mw=\eta w$ would imply $\eta =w^{T}Mw=0$. Thus $Mw=0$.

The columns of $W=(u_1{v}_{1}w)$ form an orthonormal basis, so $W$ is an orthogonal matrix. The displayed actions giveand thereforeIf $M=0$, take ${\mu }_1=0$ and any orthogonal matrix $W$. Otherwise its nonzero two-dimensional block supplies the pair used above.

Choose a real unit eigenvector $e_1$ of the real symmetric matrix $Q$, with eigenvalue ${\lambda }_1$. Its orthogonal complement is an invariant subspace: for any $v$ with $e_1^{T}v=0$,The restriction of $Q$ to this $(n-1)$-dimensional space is again real and symmetric. Applying the same argument inductively produces an orthonormal basis $e_1,\dots ,e_n$ of real eigenvectors. The base case $n=1$ is immediate. This proves the spectral theorem for real symmetric matrices without assuming that distinct eigenvalues are available.

One can also check that eigenvectors with different eigenvalues are automatically orthogonal: ${\lambda }_i{e}_i^T{e}_j=(Q{e}_i{)}^T{e}_j=e_i^{T}Q{e}_j={\lambda }_j{e}_i^T{e}_j$. Within a repeated eigenspace, the Gram-Schmidt process preserves the eigenvector property. Thus repeated eigenvalues cause no obstruction, and $\boxed{Q=OD{O}^T\text{ for an orthogonal }O\text{ and real diagonal }D}$.

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.

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.

The support of a positive operator is the orthogonal complement of its kernel. In finite dimension it is the span of eigenvectors with positive eigenvalues, equivalently the operator's image. Quantum relative entropy is finite only when the first state's support lies in the reference state's support.

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.