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A finite basis is a generating set, so the preceding determinant trick applies with $R=k$ and $M=V$. If $p(t)=\det (A-tI)$, the map $k[t]\to k[\varphi ]$ sending $t$ to $\varphi$ is a ring homomorphism. Since a determinant is a polynomial expression in its entries, it sends $p(t)$ to $\det (A-\varphi I)=0$. ThusWith the more usual monic characteristic polynomial ${\chi }_A(t)=\det (tI-A)=(-1{)}^{\dim V}p(t)$, this reads ${\chi }_A(\varphi )=0$, the Cayley-Hamilton theorem. If the matrix convention uses column coordinates instead of the row convention above, transpose $A$; its characteristic polynomial is unchanged.

Let $S=R[\varphi ]$ be the commutative subring of endomorphisms generated by scalar multiplication and $\varphi$. Commutativity follows because the scalar maps commute with the $R$-linear map $\varphi$, and every element of $S$ is a polynomial in $\varphi$ with scalar coefficients. Thus determinants and the adjugate identity apply to matrices over $S$.

With the convention for the coefficients in the question, the column of generators satisfiesMultiply by $\mathrm{adj}(C)$ over $S$. Since $\mathrm{adj}(C)C=(\det C)I_n$, the endomorphism $\det C$ kills every generator $m_i$. It therefore kills every element $\sum r_i{m}_i$ of the module, by $R$-linearity. Hence it is the zero endomorphism, andThe argument does not require the generating set to be a basis; relations among the generators cause no difficulty. This is the determinant trick.

Use the Pearson chi-squared test of homogeneity for two multinomial samples. Under equal preference distributions, estimate the common probabilities from pooled counts, giving $(1/4,1/4,1/2)$. With row totals forty and sixty, the expected table isThe Pearson chi-squared statistic for contingency tables isThere are $(2-1)(3-1)=2$ degrees of freedom, after fitting the common category probabilities. All expected counts exceed five. ThusAt five percent, do not reject the common preference distribution. These data are consistent with that hypothesis; failure to reject does not prove equality. This conclusion differs from part a because a common nonuniform distribution, especially an overall preference for chocolate, is allowed here.

Under the null hypothesis the three preference probabilities for girls are each $1/3$. Conditional on their total of sixty, the counts have a multinomial distribution, giving expected counts $(20,20,20)$. The Pearson chi-squared goodness-of-fit test statistic isNo probabilities have been fitted, so the asymptotic null distribution is chi-squared distribution with $3-1=2$ degrees of freedom. All expected counts are comfortably large for this approximation. For two degrees of freedom its upper-tail probability is $e^{-X^2/2}$, soAt the conventional five-percent or one-percent significance level, reject equal attractiveness: there is substantial evidence against the null distribution. No level was specified, so reporting the p-value makes the decision rule explicit. The test assumes independent responses and a sampling interpretation appropriate to the children surveyed.

If an open set contains an even integer, invariance under every even translation makes it contain all even integers. Similarly, containing an odd integer makes it contain all odd integers. Conversely each parity class is invariant under those translations. Thus the proposed open sets are exactlyThey form a partition topology. Both the empty and whole sets are included, and any union or finite intersection is again a union of parity classes. Hence this is a topological space. It is not the discrete topology: no singleton is open.

This does not define a topology on $\mathbb{Z}$. Every singleton $\{2k\}$ is declared open, but their union over $k\in \mathbb{Z}$ is $2\mathbb{Z}$, an infinite proper subset. It is therefore not declared open, violating closure under arbitrary unions. This is the example that finite subsets do not define a topology on an infinite set. The empty set, whole space and finite-intersection requirements do hold, but they do not compensate for the failed union axiom. This is different from the cofinite topology, which uses finite complements rather than finite sets.

Use the standard convention that an integral domain has $1\ne 0$. Since $(n)=(-n)$, the sign of a nonzero $n$ makes no difference. For $|n|=p$ prime, every nonzero residue has an inverse by Bézout's identity, so the quotient ring is a field and hence an integral domain. If $|n|=ab$ is composite with $1<a,b<|n|$, the nonzero classes of $a$ and $b$ multiply to zero, giving zero divisors. For $n=\pm 1$, the quotient is the zero ring, excluded by the convention above. Finally, $n=0$ gives $\mathbb{Z}/(0)\cong \mathbb{Z}$, an integral domain. ThusIf a convention restricts this notation to positive moduli, the answer is simply the positive primes.

Take the integral domain $R=\mathbb{Z}[\sqrt{-5}]$, embedded in $\mathbb{C}$. It exhibits failure of unique factorization in the ring of integers generated by square root of minus five:Here is why these are genuinely distinct factorizations into irreducible elements. The multiplicative algebraic norm $N(a+b\sqrt{-5})=a^2+5b^2$ is a nonnegative integer and equals one only for $\pm 1$, which are therefore the only units. Neither two nor three is a value of $N$: if $b\ne 0$ the value is at least five, and if $b=0$ it is a square. A factorization of $2$ into nonunits would require norms two and two; one of $3$ would require norms three and three; one of $1\pm \sqrt{-5}$ would require norms two and three. All are impossible. Thus all four factors are irreducible. The factors of norms four and nine cannot be associate elements of those of norm six, and $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are not related by a sign. Therefore $R$ is an integral domain which is not a unique factorization domain.

The first column is $a=(2,1,2{)}^T$ with length three. Choose $v=a-3e_1=(-1,1,2{)}^T$, so $v^{T}v=6$. Its Householder reflection isDirect multiplication givesTo eliminate the remaining third-row entry in the second column, reflect the last two coordinates using $v_2=(0,1,-1{)}^T$. ThenandBoth steps are Householder reflections; the second happens to exchange two coordinates. Solve the leading triangular equations $3x_1+3x_2=3$ and $3x_2=-3$ to obtainFor an independent check, $A{x}^{\ast }-b=(-1,-2,2{)}^T$, whose squared norm is nine and which is orthogonal to both columns of $A$. Thus $A^T(A{x}^{\ast }-b)=0$, confirming the least-squares minimizer.

For a nonzero real vector $v$, its Householder reflection isLet $P=v{v}^T/(v^{T}v)$. Then $P^T=P$ and $P^2=P$, soThus $H$ is an orthogonal matrix, and it is also symmetric with $H^{-1}=H$. It sends $v$ to $-v$ and fixes every vector perpendicular to $v$, explaining the reflection in the hyperplane $v^{\perp }$. For a target reduction of a vector $a$ to $\Vert a\Vert e_1$, choose $v=a-\Vert a\Vert e_1$ when this vector is nonzero; an alternative sign avoids cancellation in numerical implementations.

A full QR decomposition of a real $m$ by $n$ matrix with $m\ge n$ writeswhere $Q$ is an orthogonal $m$ by $m$ matrix and $R$ is upper triangular $n$ by $n$. In the full-column-rank case $R$ is invertible. Equivalently the reduced factorization is $A=Q_{1}R$ with $Q_1^T{Q}_1=I_n$. Householder transformations can construct the full factorization.

Let $Q^{T}b=\left(\genfrac{}{}{0ex}{}{c_1}{c_2}\right)$ with $c_1\in {\mathbb{R}}^n$. Since orthogonal transformations preserve the Euclidean norm,Therefore the linear least-squares problem has the full-rank solutionBack substitution solves this triangular system without forming $A^{T}A$. If the matrix is rank deficient, use a rank-revealing pivoted factorization and minimize over its independent columns; the minimizer need not be unique. The original PDF minimizes over $x\in {\mathbb{R}}^n$; the converted TeX's minimization over $b$ is a transcription error.

Set $y=e^{-x/2}v$, the Dirichlet gauge transform for constant drift. Direct differentiation givesThe homogeneous endpoint conditions on $y$ become $v(0)=v(1)=0$. Thus the eigenvalue equation is $v^{\prime \prime }=(\lambda +1/4)v$. To exclude other branches without assuming the eigenvalue real, multiply by $\bar{v}$ and integrate by parts:for a nonzero eigenfunction. Therefore $\lambda +1/4=-k^2$ with $k>0$. The boundary conditions in $v=A\sin (kx)+B\cos (kx)$ give $B=0$ and $k=n\pi$, $n\ge 1$. HenceEquivalently, $L[y]=\lambda y$ is the Sturm-Liouville form $(e^x{y}^{\prime }{)}^{\prime }=\lambda e^{x}y$, with positive weight $e^x$. Its orthogonality relation isIt is the weighted inner product, not the unweighted one, that makes these eigenfunctions orthogonal.

Expand $x{e}^{-x/2}={\sum }_n{b}_n{y}_n$ with the coefficients from part a. Completeness and the nonzero eigenvalues give the Sturm-Liouville eigenfunction expansionThe coefficients are $O(n^{-3})$, so the function series converges uniformly and vanishes at both endpoints. Its first derivative also converges uniformly. Applying the transformed operator in the square-integrable sense gives the part-a expansion of $x$; equivalently $v^{\prime \prime }-v/4=x$ on the interior. Integrating this continuous right-hand side gives the classical twice-differentiable solution there. No nonzero homogeneous solution satisfies both endpoints, so this solution is unique. The mismatch of the forcing's sine series at a single endpoint does not affect the interior equation or these boundary conditions.

For the Fourier sine series on $(0,1)$, orthogonality givesThereforeThe series is that of the odd, two-periodic extension. At $x=0$ its value is zero, agreeing with $f(0)$; at $x=1$ every summand is zero and the series converges to zero, the average of the two one-sided extension limits. Thus it represents the function in the interior and in the square-integrable sense, not the prescribed value one at the endpoint $x=1$. This endpoint issue matters when using sine expansions with homogeneous boundary conditions.

Use the permitted triangularization to write $A=TU{T}^{-1}$ with $U$ upper triangular and diagonal ${\lambda }_1,\dots ,{\lambda }_n$. Products of upper-triangular matrices are upper triangular, and their diagonal entries multiply, soThe result of part b makes every ${\lambda }_j$ zero. Thus $U$ is strictly upper triangular. To see directly that $U^n=0$, a potentially nonzero entry in a product of $n$ factors requires a chainof $n+1$ indices in $\{1,\dots ,n\}$, which is impossible. HenceThis is nilpotence from vanishing traces of powers. It proves nilpotence without substituting an unproved diagonalization for the allowed triangularization.

Suppose $v=(v_1,\dots ,v_r{)}^t$ belongs to the kernel of the displayed matrix $\Lambda$. Define the polynomial $q(z)={\sum }_{k=1}^r{v}_k{z}^k$. Then $q({\lambda }_j)=0$ for all $j$, and $q(0)=0$. These are $r+1$ distinct roots of a polynomial of degree at most $r$. Repeated application of the factor theorem therefore makes $q$ identically zero, so every $v_k=0$. Hence $\Lambda$ is injective and, by finite-dimensional rank-nullity theorem, invertible. Its transpose is invertible with inverse $({\Lambda }^{-1}{)}^t$, proving both kernels are trivial.

For the second matrix, membership of the all-ones vector in the kernel meansGroup the distinct nonzero values as ${\mu }_1,\dots ,{\mu }_r$, with multiplicities $m_1,\dots ,m_r>0$. Zero values contribute nothing to these positive powers. The first $r$ equations areor $\Lambda (\mu {)}^{t}m=0$. The proved transpose-kernel statement implies $m=0$, contradicting its positive entries if $r>0$. Therefore $r=0$ andConversely, all zero values plainly give a zero matrix. This proves the zero power sums force a finite complex multiset to vanish criterion, including repetitions and zero values rather than assuming all eigenvalues distinct.

The matrix trace is the sum of diagonal entries, $\mathrm{tr}A={\sum }_{i=1}^n{A}_{ii}$. For any two square matrices,Apply this cyclic property of the trace with $B=T$ and $C=A{T}^{-1}$:Thus matrix similarity preserves the matrix trace; no assumption that either matrix is diagonalizable is needed.

The map $(x,y)\mapsto (x,y,x^2+y^2)$ is a continuous surjection from the path-connected plane onto this surface. It is a homeomorphism because coordinate projection is its continuous inverse. Thus the paraboloid is path-connected and connected. Explicitly, lifting any straight planar segment by this map produces a path on the surface.

Factorization gives $y=x$ or $y=-x$. Each point can be joined to the origin by a straight segment lying on its own line. Two such segments join any pair of points, so the union is path-connected and hence connected. Alternatively it is the union of two connected lines with a common point.

The equation implies $|z|\ge 1$. Its intersections with $\{z>0\}$ and $\{z<0\}$ are disjoint, nonempty subsets open in the subspace topology, and together cover the entire surface. They therefore form a separation, proving the surface is disconnected. More precisely its two sheets are $z=\pm \sqrt{1+x^2+y^2}$; each is homeomorphic to the connected plane by projection, so these are exactly its two connected components.

The pointwise limit is the constant function $1$. For $0<x<1$, the inequality $1-e^{-u}\le u$ for $u\ge 0$ givesHence the sequence converges uniformly to one on $(0,1)$. Its exact supremum error is $1-e^{-1/n}$, although the supremum is not attained on the open interval. Compactness of the interval is not needed for this uniform convergence bound.