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A scalar $\lambda$ is an eigenvalue of an $n\times n$ matrix $P$ if there is a nonzero vector $\mathbf{v}$ such thatAny such vector is an eigenvector corresponding to $\lambda$. The requirement that an eigenvector be nonzero is essential: the zero vector would satisfy the equation for every scalar and would convey no information.

An eigenvalue and eigenvector pair $(\lambda ,v)$ for a square matrix $A$ satisfiesThe scalar $\lambda$ and the coordinates of $v$ belong to the chosen field, usually $\mathbb{R}$ or $\mathbb{C}$. The nonzero condition is essential: the zero vector satisfies the equation for every $\lambda$ and therefore cannot define an eigenvalue.

For the specified row and column, expand the two matrix products and reorder only finite sums:The middle equality uses scalar associativity and distributivity, and the interchange of two finite sums. Together with their equal dimensions, equality of every entry proves $(AB)C=A(BC)$.

For nonzero $x$, let $\varphi$ be its angle to the unit vector $q_1$. The scalar projection on the oriented $q_1$ axis is $\Vert x\Vert \cos \varphi$. Since $\Vert q_1\Vert =1$, the dot product identity gives $q_1^{T}x=\Vert x\Vert \cos \varphi$. Multiplying that signed scalar by $q_1$ places the foot at $p=(q_1^{T}x)q_1$.

The residual is perpendicular to the axis, because $q_1^T(x-p)=q_1^{T}x-(q_1^{T}x)(q_1^T{q}_1)=0$. An obtuse angle gives a negative scalar and a foot on the negative side of the line; $x=0$ has projection zero without needing an angle.

The first panel is a constructed geometric illustration. The second illustrates the directional stretch used later in Question 6.

The initial vector records both initial Fibonacci numbers: $V_1=(U_2,U_1{)}^{\top }=(1,1{)}^{\top }$. For $n\ge 2$,Conversely this vector recurrence forces the second component to be the preceding first component and forces their sum to be the next first component, recovering the scalar recurrence and its initial conditions. Iterating, or using mathematical induction, yields $V_n=A^{n-1}(1,1{)}^{\top }$ for every $n\ge 1$.

Work over the field containing the stated $n$ distinct eigenvalues of $B$; in a real formulation these must be real to obtain real eigenvectors. Choose one eigenvector for each distinct eigenvalue. By independence of eigenvectors for distinct eigenvalues, these $n$ vectors are a basis. Consequently each eigenspace $E_{\lambda }(B)$ has dimension one: in an expansion in this eigenbasis, $Bw=\lambda w$ forces every coefficient with a different eigenvalue to vanish.

Now let $v\ne 0$ be any eigenvector of $B$. By part (d), $Av$ belongs to its one-dimensional eigenspace, so $Av=\mu v$ for some scalar $\mu$. This proves that $v$ is an eigenvector of $A$, including the case $Av=0$, where $\mu =0$. This is the crucial one-dimensional step in simple spectrum gives simultaneous diagonalization.

The intersection of vector subspaces $W_1\cap W_2$ contains zero, so $(0,0)\in U$. If $(u,-u),(v,-v)\in U$ and $a,b$ are scalars, thenbecause $au+bv\in W_1\cap W_2$ by the vector subspace conditions. These pairs belong to $W_1\times W_2$, since both $u$ and $-u$ belong to the required factors. Thus $U$ is a vector subspace of $W_1\times W_2$.

Every output $u_1+u_2$ lies in the sum of subspaces $W_1+W_2$, since it is a linear combination of members of the two vector subspaces. For $u=(u_1,u_2)$, $v=(v_1,v_2)$ and scalars $a,b$,Therefore $g$ is a linear map. Every vector in $W_1+W_2$ is such an output: group the terms of any linear combination from $W_1\cup W_2$ according to their originating vector subspace. Thus the addition map of subspaces is also surjective.

Polynomial evaluation replaces the indeterminate of a polynomial by an input value. For a fixed scalar $a$, the rule $E_a:p\mapsto p(a)$ is a linear functional on a vector space of polynomials, because evaluating a linear combination gives the same linear combination of the values. For $p(t)=1+2t+3t^2$, evaluation at zero gives one and evaluation at two gives seventeen. Polynomial interpolation uses these measurements to reconstruct a polynomial subject to a polynomial degree bound.

For positive $\alpha ,\beta$, suppose $p^{\alpha }\le |A|\le p^{1-\alpha }$ and $|B|\ge p^{\beta }$. Repeatedly replace a scalar set $S$ by $3SS-3SS$. Quadratic growth of a sixfold product difference set reaches size at least $p/2$ in a number of steps depending only on $\beta$. The closure bounds for small skew-sumset scalars transfer a putative bound $|A+bA|\le K|A|$ to all these scalars with exponent depending only on the number of steps. Averaging collisions over the resulting large scalar set then forces $K^L\ge \frac{1}{4}\min (|A|,p/|A|)\ge p^{\alpha }/4$. This supplies a positive exponent depending only on $\alpha ,\beta$. The finitely many smaller primes are covered by the strict inequality $|A+bA|>|A|$ for every nonzero $b$ when $1<|A|<p$.

A potential function $V$ assigns a scalar to each point of a configuration or spatial domain. In mechanics it represents potential energy, and its negative gradient gives the corresponding conservative force.

A potential for a conservative vector field $X$ is a function with scalar values $\Phi$ whose gradient is $X$. Potentials differ by a constant on each connected component. The fundamental theorem for line integrals gives ${\int }_{\gamma }X\cdot d\mathbf{x}=\Phi (\gamma (1))-\Phi (\gamma (0))$. In mechanics a conservative force is conventionally written $-\nabla V$, so its potential energy is $V=-\Phi$. This spatial potential differs from the scalar potential $V(\varphi )$ that denotes the derivative-free energy density of a scalar field.

For each unit vector iterate compute the Rayleigh quotient ${\rho }_k=x_k^{\ast }A{x}_k$ and the eigenpair residual $r_k=A{x}_k-{\rho }_k{x}_k$. Use the same multiplication $A{x}_k$ for both diagnostics and the next normalized iterate. For any complex matrix, this Rayleigh quotient minimizes $\Vert A{x}_k-\mu x_k{\Vert }_2$ over complex scalars $\mu$, because $x_k^{\ast }{r}_k=0$. Under power iteration convergence for a diagonalizable matrix, ${\rho }_k\to {\lambda }_1$ and $\Vert r_k{\Vert }_2\to 0$, even when the vectors alternate signs or phases.^[ref]

A dimensionless stopping quantity iswhen the denominator is positive; $\Vert A{\Vert }_2$ is the operator norm, and a known upper bound may replace it if computing it is costly. A small eigenpair residual certifies a small backward error of an approximate eigenpair, but does not establish that the eigenvalue found is dominant.

For a Hermitian matrix with orthonormal eigenbasis $u_j$, write $x_k={\sum }_j{a}_j{u}_j$. Then the independently useful identityshows that some eigenvalue lies within $\Vert r_k{\Vert }_2$ of ${\rho }_k$. If a specific target eigenvalue ${\lambda }_1$ is separated from all the others by ${\delta }_k={\min }_{j>1}|{\lambda }_j-{\rho }_k|>0$, it also gives ${\epsilon }_k\le \Vert r_k{\Vert }_2/{\delta }_k$. This separation is measured from the computed Rayleigh quotient; it differs from the dominant-modulus spectral gap. For a nonnormal matrix, the same orthogonal expansion is unavailable and a small eigenpair residual alone need not imply a comparably small distance to the spectrum.

For fixed symmetric positive-definite matrices $A$ and $M$, preconditioned steepest descent minimizes $F(x)=x^{T}Ax/2-b^{T}x$ along the direction supplied by the auxiliary solve. Given $x_k$, form the linear-system residual $r_k=b-A{x}_k$. If $r_k=0$, stop: $x_k=x_{\ast }$. Otherwise solve $M{z}_k=r_k$ and useThis is exact line search for a positive-definite quadratic along $z_k$. Indeed,whose derivative vanishes at the displayed positive step size. Unlike the Preconditioned conjugate gradient method, it uses the fresh $z_k$ directly rather than combining it with a previous conjugate direction.

For diagonal scaling for a two-dimensional linear system, start at $x_0=0$. Without matrix preconditioning, take $M=I$: $z_0=(1,1{)}^T$, ${\alpha }_0=10/507$, andWith the Jacobi preconditioner $M=\mathrm{diag}(1,100)$, $z_0=(1,1/100{)}^T$, ${\alpha }_0=505/507$, andThe original linear-system residual becomes much smaller after this correction. Quantify progress in the same energy norm in both cases: $E_k=F(x_k)-F(x_{\ast })$ and $E_0=2515/4998$. Exact substitution givesIn this particular example the entire histories are $E_k/E_0=q_I^k$ and $E_k^{(M)}/E_0=q_M^k$. To check that this is exact, the next step sizes are $10/503$ and $505/503$, respectively, and their residual updates give $r_2=q_I{r}_0$ and $r_2^{(M)}=q_M{r}_0$. Multiplying a linear-system residual by a nonzero scalar leaves the next step size unchanged; every pair of steps repeats the same directions at a smaller scale. The two first-step energy ratios then establish these formulas for all $k$.

The curves come from exact rational updates of the specified linear system, converted to decimal values for plotting. The left panel counts iterations. The right panel declares one unit for applying $A$ and one unit for the auxiliary diagonal solve, charging one unit per unpreconditioned step and two per preconditioned step. It excludes setup and vector operations, so it compares a declared cost model rather than measured runtime. Scaling is advantageous even with that extra charge.

These curves compare steepest descent method iterations. On this two-dimensional linear system, both the conjugate gradient method and the Preconditioned conjugate gradient method terminate within two steps in exact arithmetic. The plot therefore establishes no CG iteration-count advantage for this example; the benefit of scaling for larger problems depends on their eigenvalues and auxiliary-solve costs.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

For two projectively equivalent connections that are Levi-Civita connections, contraction of their difference gives $(n+1)V_c={\bar{\Gamma }}^a{}_{ac}-{\Gamma }^a{}_{ac}$. The Jacobi formula gives ${\Gamma }^a{}_{ac}={\partial }_c\log \sqrt{|\det g|}$. Subtraction proves the displayed expression. The two determinant densities transform by the same coordinate Jacobian factor, so their ratio is a scalar and its derivative a covector. The formula is a necessary consequence of projective equivalence, not a test asserting that arbitrary metric tensors have identical geodesics.

For $S\subseteq {\mathbb{F}}_p$ with $|S|\ge 2$, put $R=(S-S)/(S-S)$ using nonzero denominators. If $R$ is proper, some $r\in R+1$ lies outside $R$, since invariance under addition of one would give all of the prime field. The map $(x,y)\mapsto x+ry$ on $S^2$ is injective. Clearing the denominator in a representation of $r-1$ puts a dilate of its image inside $3SS-3SS$. If $R$ is the full field, averaging the collision count of $S+rS$ gives some $r$ with $|S+rS|\ge \frac{1}{2}\min (p,|S{|}^2)$. Clearing its denominator puts a dilate inside $2SS-2SS\subseteq 3SS-3SS$. Here $SS$ is a product set, and the prefactors denote repeated sumsets, not scalar multiplication.

A nonzero vacuum value of the scalar quark density breaks the massless nonsinglet chiral symmetry to its diagonal vector subgroup. Indeed, the nonsinglet axial charge algebra of quark densities gives $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨\bar{q}q⟩$, whereas invariance of the vacuum under every conserved axial charge would make these expectations vanish. The three broken generators produce three massless Goldstone bosons in the exact chiral limit. Small equal quark masses explicitly break that symmetry and turn the pions into pseudo-Goldstone bosons, with squared masses linear in the quark mass at leading order.

A rank-one matrix update changes a matrix $A$ by an outer product $u{v}^T$. For real vectors $u,v\in {\mathbb{R}}^n$, its action on an input vector $x$ isThe dot product $v^{T}x$ reads one scalar from the input, and the correction moves the output along $u$. All correction outputs therefore lie in one linear span. If either vector is zero, the update is zero; otherwise the correction is a rank-one matrix. The full updated matrix need not have rank one.

For example, chooseThe update sends $x=(x_1,x_2{)}^T$ to the correction $(3x_1-x_2)(1,2{)}^T$. Although all four entries change, there is only one direction of correction. Adding it to the identity matrix gives $B=\left(\begin{array}{cc}4& -1\\ 6& -1\end{array}\right)$.

This special structure allows existing work to be reused. The matrix determinant lemma updates the determinant using one scalar, and the Sherman–Morrison formula updates the matrix inverse using an outer product. Such formulas are useful when a linear system changes slightly after a new observation or a parameter adjustment: the unchanged part can retain its LU decomposition. The term rank-one describes the rank of the correction, not the size of the entry changes. ^[ref].

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]