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The adjoint operator ${\alpha }^{\ast }$ is defined bySince $V$ is a nonzero finite-dimensional complex vector space, $\alpha$ has an eigenvector $v$ with eigenvalue $\lambda$. Normality givesfor every $x$, by expanding both sides and using $\alpha {\alpha }^{\ast }={\alpha }^{\ast }\alpha$. Taking $x=v$ yields

Use the conventionChanging the summation variable from $j$ to $j-1$ givesThus the Fourier-basis states are eigenvectors of the cyclic shift, with respective eigenvalues $e^{-2\pi ik/N}$, and the cyclic shift diagonalization by the quantum Fourier transform is

For $N=4$, write $k=2x+y$ with $x,y\in \{0,1\}$. ThensoHence the required quantum circuit applies, from input to output,The first phase gate acts on the most significant qubit $x$ and the second on the least significant qubit $y$.

The Gershgorin circle theorem states that every eigenvalue $\lambda$ of a complex matrix $A=(a_{ij})$ lies in at least one diskTo prove it, choose a nonzero eigenvector $x$ for $\lambda$ and an index $i$ such that $|x_i|={\max }_j|x_j|>0$. The $i$th component of $Ax=\lambda x$ givesTaking absolute values and using $|x_j|\le |x_i|$ yieldsDivision by $|x_i|$ proves the theorem.

Apply part (b) to the normal matrix $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one hasThus $e$ is an eigenvector of $D^{†}$ with eigenvalue $\overline{\lambda }$.

A real matrix is orthogonal when $Q^{T}Q=I$. If $Qv=\lambda v$, norm preservation gives $|\lambda |=1$. If $Qv=\lambda v$ and $Qw=\mu w$, preservation of the Hermitian inner product gives $(1-\overline{\lambda }\mu )v^{†}w=0$; distinct unit-modulus eigenvalues therefore have orthogonal eigenvectors.

The nonreal eigenvalues of a real $3\times 3$ matrix occur in conjugate pairs. Since their product is one and $\det Q=-1$, the remaining real eigenvalue is $-1$. If $x\cdot n=0$, then $(Qx)\cdot (Qn)=x\cdot n=0$, and $Qn=-n$, so $Qx\cdot n=0$: the plane $\Pi$ is invariant.

The restriction to $\Pi$ is a planar orthogonal map whose determinant is $(-1)/(-1)=1$, hence a rotation through some $\theta$. In an orthonormal basis adapted to $n$,Therefore $\mathrm{tr}Q=-1+2\cos \theta$ and

A linear operator $\alpha$ on a real inner-product space is self-adjoint whenfor all $v,w$.

To prove the finite-dimensional spectral theorem, first note that the continuous quadratic function $⟨\alpha v,v⟩$ has a maximum on the unit sphere. The Lagrange multiplier equation at a maximizing vector $v$ gives $\alpha v=\lambda v$, so $\alpha$ has a real unit eigenvector. Its orthogonal complement is invariant becauseInduction on the dimension supplies an orthonormal eigenbasis of that complement and hence of $V$.

For $f\ne 0$,while symmetry and bilinearity are immediate, so the displayed formula defines an inner product on $P_n$. Sinceintegration by parts givesThe boundary term vanishes for polynomials, proving that $\alpha$ is self-adjoint.

On the monomial $x^k$,Thus the matrix of $\alpha$ in the monomial basis is triangular with diagonalThese are therefore its eigenvalues. For $n=2$, corresponding eigenvectors arefor eigenvalues $0,-1,-2$, respectively. They are the first Laguerre polynomials up to normalization, as described by the Laguerre differential operator on polynomials.

The coefficient matrix has eigenvector $u_{+}=(1,1{)}^T$ with eigenvalue $3$ and eigenvector $u_{-}=(1,-1{)}^T$ with eigenvalue $-1$. WriteSinceand $y(0)=(1,-2{)}^T=-\frac{1}{2}{u}_{+}+\frac{3}{2}{u}_{-}$, the system diagonalises toUsing an integrating factor givesThe unique fastest term is thereforeConsequently $e^{-nt}y(t)$ diverges for $n<3$, tends to zero for $n>3$, and for $n=3$ tends to the finite nonzero vectorThus the only requested integer isThis is an instance of dominant eigenmode in a forced linear system.

For a decision $x$ in the unit ball, conditional expectation under the posterior givesThe symmetric positive semidefinite matrix $C+m{m}^T$ has a Rayleigh quotient maximized over $\Vert x\Vert \le 1$ by any unit eigenvector corresponding to its largest eigenvalue. Bayes decision rule therefore chooses any such eigenvector for the stated utility.

For $\varphi (x)=x$ and centered data, the objective is the Rayleigh quotientThis is the first-direction optimization in principal component analysis. Therefore $\hat{u}$ is any unit eigenvector of the sample covariance matrix $S$ corresponding to its largest eigenvalue.

The Karhunen–Loève expansion has the following Hilbert-space form. Let $X$ be a square-integrable Random element of a Hilbert space $H$, with mean $\mu$ and covariance operator $C$. There are nonnegative eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots$ and orthonormal eigenvectors ${\varphi }_k$ spanning the closure of the range of $C$ such thatin $L^2(\Omega ;H)$. The functional principal component scoressatisfyIf $X$ is a Gaussian random element, the scores are independent normal random variables.

For the proof, $C$ is positive, self-adjoint, and trace-class, withIt is therefore compact, so the spectral theorem for compact Hermitian operators supplies the eigenpairs. Their score covariance isFor the residual $R_m=X-\mu -{\sum }_{k=1}^m{\xi }_k{\varphi }_k$, Parseval identity and the trace formula giveThe centered variable's projection onto $\ker C$ has zero second moment and is therefore zero almost surely, which completes the mean-square expansion. When $C$ has a continuous covariance kernel, Mercer's theorem additionally expands that kernel as $c(s,t)={\sum }_k{\lambda }_k{\varphi }_k(s){\varphi }_k(t)$.

Because $A$ is a real symmetric positive-definite matrix, the finite-dimensional spectral theorem supplies an orthonormal eigenbasis $u_1,\dots ,u_n$ withIts eigendecomposition is also its singular value decomposition. If $e=y^{(\delta )}-y$, thensoThe operator norms satisfy $\Vert A{\Vert }_2={\lambda }_n$ and $\Vert A^{-1}{\Vert }_2=1/{\lambda }_1$. Therefore the worst-case relative perturbation bound isThe ratio ${\kappa }_2(A)$ is the spectral condition number of a positive-definite matrix. A large ratio means that data noise aligned with an eigenvector for the smallest eigenvalue is strongly amplified, so the inverse problem is ill conditioned.

A complex representation of a finite group $G$ is a homomorphismfor a finite-dimensional complex vector space $V$. It is an irreducible representation when its only invariant subspaces are $0$ and $V$, and its degree is $\dim V$. Two representations are isomorphic when there is an invertible linear map intertwining their $G$-actions.

Write the dihedral group asIn an irreducible representation choose an eigenvector $0\ne v\in V$ of $\rho (r)$, which is possible because $r^n=1$. If its eigenvalue is $\lambda$, thenThus $W=\mathrm{span}\{v,\rho (s)v\}$ is invariant under both $r$ and $s$. Irreducibility forces $W=V$, proving $\dim V\le 2$ as in irreducible complex representations of a finite dihedral group.

Let $\omega =e^{2\pi i/n}$. Besides the one-dimensional representations described below, define for each indicated $j$These matrices satisfy the defining relations. Since ${\omega }^j\ne {\omega }^{-j}$, the only one-dimensional $r$-invariant subspaces are the two coordinate axes, and $s$ interchanges them. Hence ${\rho }_j$ is irreducible. Their characters satisfywhose values are distinct in the ranges below, so the ${\rho }_j$ are pairwise nonisomorphic. This is the family of two-dimensional representations of a finite dihedral group.

If $n$ is odd, a one-dimensional character must send $r$ to a scalar $\epsilon$ satisfying ${\epsilon }^n=1$ and $\epsilon ={\epsilon }^{-1}$. Thus $\epsilon =1$, while $s$ may independently map to $1$ or $-1$. These give two one-dimensional irreducibles. Takinggives $(n-1)/2$ two-dimensional irreducibles, for a total of

If $n$ is even, $r$ may map to either $1$ or $-1$, and again $s$ may map independently to either sign. This gives four pairwise nonisomorphic one-dimensional irreducibles. Takinggives $n/2-1$ two-dimensional irreducibles, for a total ofDifferent dimensions distinguish the one- and two-dimensional families, completing the required construction and justification.

The characteristic polynomial of $A$ isA scalar $\lambda \in \mathbb{C}$ is an eigenvalue when there is a nonzero vector $v$ such thatSuch a $v$ is an eigenvector belonging to $\lambda$. Equivalently, $\lambda$ is a root of ${\chi }_A$ and the corresponding eigenvectors are the nonzero elements of $\ker (A-\lambda I)$.

The characteristic polynomial ${\chi }_A(t)=\det (tI-A)$ has degree $n$. By the fundamental theorem of algebra, it has a complex root $\lambda$. Henceso $\lambda I-A$ has a nonzero kernel. Any nonzero $v$ in that kernel satisfiesThus $\lambda$ is an eigenvalue and $v$ is a corresponding eigenvector.

First note that if $A$ is normal, thenfor every $v$. Hence an eigenvector $v$ with $Av=\lambda v$ also satisfiesNormalize $v$ and extend it to an orthonormal basis. Let $U$ have these basis vectors as columns. The first column of $U^{†}AU$ is $(\lambda ,0,\dots ,0{)}^T$. For every $w\perp v$,so the first row also has no off-diagonal entries. ThereforeComparing the two block products in the normality identity shows that $B{B}^{†}=B^{†}B$, so $B$ is normal.

The result is trivial in dimension one. Applying the induction hypothesis to $B$ and adjoining the eigenvector $v$ gives an orthonormal eigenbasis for $A$. This proves the unitary diagonalization of a normal matrix.

The proof is by induction on $n$. Choose a unit eigenvector $v$ of $A$, which exists by part (a), and extend it to an orthonormal basis. In that basis,By induction, a unitary change of basis on the last $n-1$ coordinates makes $C$ upper triangular. Extending that change by the identity on the first coordinate leaves the displayed zero block intact. Their product is unitary and makes all of $A$ upper triangular. This proves Schur triangularization.

The characteristic polynomial isTo prove triangularizability, use induction on $n$. The result is immediate for $n=1$. Over $\mathbb{C}$, ${\chi }_A$ has a root $\lambda$, so $A$ has an eigenvector $v_1$. Extend it to a basis. In this basis,By induction, a change among the remaining $n-1$ basis vectors makes $B$ upper triangular. Thus the triangularization over an algebraically closed field gives

The minimal polynomial $m_A$ is the monic polynomial of least degree satisfying $m_A(A)=0$. To establish existence without quoting the Cayley-Hamilton theorem, let $T$ be an upper-triangular matrix similar to $A$, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$, and putThenThe factors $T-{\lambda }_{k}I$ commute, so applying all of them successively lowers the invariant flag to zero:Similarity gives the same polynomial identity for $A$. Hence a nonzero monic annihilating polynomial of degree $n$ exists, and a least-degree one exists.

If $m$ and $\tilde{m}$ were two monic annihilating polynomials of the same least degree, then $m-\tilde{m}$ would be an annihilating polynomial of smaller degree unless it were zero. Thus the minimal polynomial is unique, and the minimal polynomial bound from a triangular invariant flag gives

Proceed by induction on $\dim V$. Over $\mathbb{C}$, $\alpha$ has an eigenvector $v$, say $\alpha v=\lambda v$. The normal operator $\alpha -\lambda I$ satisfies the norm equality from part (c), soThus ${\alpha }^{\ast }v=\overline{\lambda }v$. Part (a) now shows that $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. The restriction of $\alpha$ to $v^{\perp }$ is normal. By induction it has an orthonormal eigenbasis, and adjoining the normalized vector $v$ proves the finite-dimensional spectral theorem for normal operators.

Hence

A shift-invariant basis is a common eigenbasis of all cyclic shift operators $U(\gamma )$. For the displayed Fourier states, orthonormality follows from the root-of-unity filter:There are $N$ vectors in this orthonormal set in the $N$-dimensional Hilbert space, so they form a basis. Reindexing the finite sum givesThus every $|{\xi }_{\alpha }⟩$ is simultaneously an eigenvector of every shift, with eigenvalue ${\omega }^{\alpha \gamma }$ for $U(\gamma )$.