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Fix a shift $s$ that is not an eigenvalue of $A$. Inverse iteration computesand may use $r({\mathbf{x}}_k)$ as the eigenvalue estimate.

For a real symmetric matrix, suppose that one eigenvalue ${\lambda }_{\ast }$ is uniquely closest to $s$ and that ${\mathbf{x}}_0$ has a nonzero component in its eigenspace. Since the iteration is the power method applied to $(A-sI{)}^{-1}$, the normalized vectors converge, up to sign, to an eigenvector for ${\lambda }_{\ast }$. If ${\lambda }_{\mathrm{next}}$ is the second-closest eigenvalue to the shift, the asymptotic direction-error ratio isThus a shift close to the desired eigenvalue gives rapid convergence. The Rayleigh quotients converge to ${\lambda }_{\ast }$ and, for a simple eigenvalue, their errors are of second order in the vector error. This is convergence of fixed-shift inverse iteration.

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 $\overline{\lambda }$.

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic polynomial. If a $2\times 2$ matrix is diagonalisable, $A=S\mathrm{diag}({\lambda }_1,{\lambda }_2)S^{-1}$; applying $p(t)=(t-{\lambda }_1)(t-{\lambda }_2)$ to the diagonal matrix gives zero and hence $p(A)=0$.

If $B^k=0$, every eigenvalue $\lambda$ satisfies ${\lambda }^k=0$, so all eigenvalues are zero and the characteristic polynomial is $t^n$. Cayley--Hamilton then gives $B^n=0$.

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

True. By Jordan normal form, the characteristic polynomial records the total size of the blocks for each eigenvalue, while the minimal polynomial records the largest block for each eigenvalue. In dimension three this determines every block partition: for algebraic multiplicity three the possibilities $3$, $2+1$, and $1+1+1$ have largest block sizes three, two, and one; multiplicities one and two are equally immediate. Thus the two matrices have the same Jordan form and are conjugate. This is the characteristic and minimal polynomials determine similarity in dimension three phenomenon.

False. At eigenvalue $2$, algebraic multiplicity two and minimal-polynomial exponent two force one block $J_2(2)$. At eigenvalue $1$, the blocks have total size five and largest size two, so there are only two possible partitions,Hence there are exactly two conjugacy classes satisfying the given polynomial data, represented byandThere cannot be three mutually non-conjugate examples.

True. Since $\alpha$ is invertible, every eigenvalue $\lambda$ is nonzero. If $\alpha$ had a nontrivial Jordan block $\lambda I+N$, then for the stated positive integer $N_0$,would still have a nonzero nilpotent part and could not be diagonalizable. Therefore all Jordan blocks of $\alpha$ have size one. This also proves diagonalizability inherited from an invertible power.

False. The real matrixhas the two distinct complex eigenvalues $i$ and $-i$, so it is diagonalizable over $\mathbb{C}$. It has no real eigenvalue and hence no real eigenbasis, so it is not diagonalizable over $\mathbb{R}$.

The spatial wavefunction $\chi (x)$ is the position amplitude of a stationary quantum state, so $|\chi (x){|}^2$ is its probability density after normalization. The real function $U(x)$ is the potential energy, $E$ is the energy eigenvalue, $m$ is the particle mass, and $\hslash$ is the reduced Planck constant.

Taking the Laplace transform of the differential equation and usinggivesHenceWhenever $s$ is not an eigenvalue of $A$, the matrix $sI-A$ is invertible, and thusThis is the Laplace-transform solution of a constant-coefficient vector ODE.

Set $g=0$. The given homogeneous solution is $y(t)=e^{tA}{y}_0$, so the preceding formula saysfor every initial vector $y_0$. Equality on every vector gives the matrix identityon the common domain of convergence and in particular away from the eigenvalues of $A$. This is the Laplace transform of a matrix exponential.

Let $N={\mathbb{Z}}^2$ and let $t$ generate the other factor of ${\Gamma }_B=N{⋊}_B\mathbb{Z}$. For $n\in N$,After abelianization, the image of $N$ is thereforeIf $1$ is not an eigenvalue of $B$, then $\det (B-I)\ne 0$. Hence $(B-I)N$ is a full-rank sublattice of $N$ with finite index of a subgroup $|\det (B-I)|$. The displayed quotient, and therefore the image of the ${\mathbb{Z}}^2$ factor in ${\Gamma }_B^{\mathrm{ab}}$, is finite. In fact the abelianization of a semidirect product by the integers gives

TakeIts eigenvalues areNeither is a root of unity, so $1$ is not an eigenvalue of $C^k$ for any positive $k$.

Let $L$ be a finite-index subgroup of ${\Gamma }_C$, let $N={\mathbb{Z}}^2$, and project ${\Gamma }_C$ onto its $\mathbb{Z}$ factor. Then $N^{\prime }=L\cap N$ has finite index in $N$, while the image of $L$ is $k\mathbb{Z}$ for some $k>0$. Choose $s\in L$ projecting to $k$. Every element of $L$ has a unique expression $n{s}^j$ with $n\in N^{\prime }$, and conjugation by $s$ on $N^{\prime }$ is the restriction of $C^k$. ThusBecause $1$ is not an eigenvalue of $C^k$, part (b), applied to the finite-rank lattice $N^{\prime }$, shows that the image of $N^{\prime }$ in $L^{\mathrm{ab}}$ is finite. The quotient by that finite image is generated by the image of $s$, so $L^{\mathrm{ab}}$ is a finite extension of an infinite cyclic group. It is therefore a virtually cyclic group.

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.

Write $\Phi \alpha ={\sum }_i{\alpha }_i\varphi (X_i)$. By the definition of the kernel matrix,andThe irrelevant positive factor $1/n$ gives exactly the stated constrained optimization.

Let $K{v}_1={\lambda }_1{v}_1$, where ${\lambda }_1>0$ is the largest eigenvalue and $\Vert v_1{\Vert }_2=1$. The generalized Rayleigh quotient is maximized byup to sign and addition of a vector in $\ker K$, which does not change $\hat{u}=\Phi \hat{\alpha }$. Repeated leading eigenvectors give further principal directions.