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For the symmetric double-delta potential, take an odd state at $E=-{\hslash }^2{\lambda }^2/(2M)$, $\lambda >0$. On $0<x<a$ odd parity selects $B\sinh (\lambda x)$. For $x>a$, decay and continuity give $B\sinh (\lambda a)e^{-\lambda (x-a)}$; extend to negative $x$ by oddness. At $x=a$ the jump condition isFor a nonzero state this means $\lambda \Delta [1+\coth (\lambda a)]=1$, and thereforeOddness makes the jump at $-a$ follow automatically: both its derivative jump and its required wavefunction value have the corresponding opposite signs.

There is a positive solution precisely when $\Delta <a$. To see this, rewrite the equation asThe right side decreases strictly from $a$ to zero as $\lambda$ increases from zero to infinity: its derivative has the sign of $(1+2a\lambda )e^{-2a\lambda }-1<0$. Hence for $0<\Delta <a$ there is exactly one positive solution, while $\Delta =a$ is a nonnormalizable zero-energy threshold and $\Delta >a$ gives no odd bound state. This verifies the stated sufficiently-small-$\Delta$ condition and identifies its exact boundary, consistent with the odd bound state of a symmetric double-delta potential.

For a negative energy write $E=-{\hslash }^2{\lambda }^2/(2M)$, $\lambda >0$. Away from the delta interaction the decaying solutions are $C{e}^{-\lambda x}$ for $x>0$ and $C{e}^{\lambda x}$ for $x<0$, with the same $C$ by continuity. The derivative jump is $-2\lambda C=-C/\Delta$, giving $\lambda =1/(2\Delta )$. Thus the bound state in a delta potential hasThe normalization follows from ${\int }_{\mathbb{R}}{e}^{-2\lambda |x|}dx=1/\lambda$. This is a genuine square-integrable negative-energy state.

The attractive potential has $\Delta >0$. Its time-independent Schrödinger equation isIntegrate from $-\epsilon$ to $\epsilon$. Continuity makes the delta term equal to $-{\hslash }^2\chi (0)/(2M\Delta )$, while the integral of $E\chi$ tends to zero as the interval shrinks. The second-derivative term integrates to the difference of first derivatives. HenceThis expresses the point interaction through a derivative jump while the wavefunction remains continuous.

Use the Möbius transformationThe lens's boundary intersections are $z=\pm 1$. Write $w=u+iv$. Substitution in the inverse gives $\mathrm{Im}z=2v/|w+1{|}^2$, andConsequently the two domain inequalities are exactly $0<v<u$, or $0<\arg w<\pi /4$. This algebra proves that $T$ maps the lens bijectively onto the quarter-sector, not merely that its boundary curves go to rays.

The map $w\mapsto w^4$ maps this sector bijectively onto the upper half-plane: the argument increases from $(0,\pi /4)$ to $(0,\pi )$, and the positive modulus map is bijective. Its inverse is the fourth-root branch with argument in $(0,\pi /4)$. Therefore the quarter-sector mapping of a circular lens isIts derivative is nonzero in the lens, since $T^{\prime }(z)=2/(1-z{)}^2$ and $T(z)\ne 0$ there. Finally the Cayley transform $\zeta \mapsto (\zeta -i)/(\zeta +i)$ maps $H$ bijectively to the unit disc, because $|\zeta -i|<|\zeta +i|$ exactly when $\mathrm{Im}\zeta >0$. A required conformal bijection is thusAll poles and branching endpoints of the construction lie outside the open domain.

Let two regular differentiable paths through $z_0$ have nonzero tangent vectors $v_1,v_2$. The chain rule gives tangent vectors $f^{\prime }(z_0)v_1$ and $f^{\prime }(z_0)v_2$ for their images. Multiplication by a nonzero complex number is a positive scaling and rotation: writing $f^{\prime }(z_0)=r{e}^{i\theta }$ scales lengths by $r>0$ and adds the same angle $\theta$ to both tangents. Equivalently,Thus the oriented angle between the paths is preserved, proving $f$ is conformal at $z_0$. The nonzero derivative also gives local invertibility; a zero derivative could multiply angles through branching instead.

Take $V={\mathbb{R}}^3$, $\alpha =\mathrm{diag}(0,1,0)$ and $W=\mathrm{span}(e_1,e_2)$. Then $W$ is a proper invariant subspace, and the restriction has matrix $\mathrm{diag}(0,1)$. In each case $X(X-1)$ annihilates the operator, while any annihilator must vanish at both zero and one. ConsequentlyThis same example works over any scalar field.

Invariance of $W$ means every power of $\alpha$ restricts to the corresponding power of $\beta$. Therefore $m_{\alpha }(\beta )$ is the restriction of $m_{\alpha }(\alpha )$, and it is zero. Divide $m_{\alpha }=q{m}_{\beta }+r$ with $\mathrm{deg}r<\mathrm{deg}{m}_{\beta }$. Evaluation at $\beta$ gives $r(\beta )=0$. By minimality of $m_{\beta }$, $r=0$, provingThe zero-subspace case has $m_{\beta }=1$ and satisfies the same conclusion.

If $\alpha v=\lambda v$ with $v\ne 0$, then $m_{\alpha }(\alpha )v=m_{\alpha }(\lambda )v$. The left side vanishes, so $m_{\alpha }(\lambda )=0$.

Conversely, if $m_{\alpha }(\lambda )=0$, write $m_{\alpha }(X)=(X-\lambda )q(X)$. Then $(\alpha -\lambda I)q(\alpha )=0$. If $\alpha -\lambda I$ were invertible, multiplying by its inverse would imply $q(\alpha )=0$, contradicting the minimal degree. Thus $\alpha -\lambda I$ is not invertible. In finite dimension this means it has a nonzero kernel, furnishing an eigenvector with eigenvalue $\lambda$. Hence the eigenvalues are exactly the roots of the minimal polynomial, with the scalar-field convention stated in part (a).

After choosing a basis of $V$, a linear endomorphism is specified by an $n$ by $n$ matrix, with $n^2$ freely chosen coefficients. ThereforeThe $n^2+1$ endomorphisms $I,\alpha ,\dots ,{\alpha }^{n^2}$ are linearly dependent. Some nonzero coefficient list gives a nonzero polynomial $p$ satisfying $p(\alpha )=0$.

For nonzero $V$, the minimal polynomial $m_{\alpha }$ is the monic polynomial of least degree annihilating $\alpha$. It is unique: subtracting two monic least-degree choices would give a lower-degree annihilator. More generally division of any annihilator $p$ by $m_{\alpha }$ gives $p=q{m}_{\alpha }+r$, with $\mathrm{deg}r<\mathrm{deg}{m}_{\alpha }$, and evaluation implies $r(\alpha )=0$. Minimality forces $r=0$, so every annihilator is divisible by $m_{\alpha }$. If $V=\{0\}$, the convention $m_{\alpha }=1$ handles the degenerate case.

The arguments work over the underlying scalar field. In part (b), roots in that field correspond to eigenvalues there; to discuss other roots one extends scalars, for example by complexifying a real vector space.

Put $m=(p-1)/2$, so the requested modulus is $pm$. The factorial $(p-1)!$ is divisible by $m$, and so is $p-1=2m$. Also, Wilson's theorem gives $(p-1)!\equiv -1\equiv p-1(\mathrm{mod}p)$. Because $m$ and $p$ are coprime integers, their product divides the difference. Thus the Wilson factorial congruence modulo a triangular number isThe displayed residue lies in the standard range even at $p=3$.

For each $1\le j\le p-1$, Fermat's little theorem gives $j^p\equiv j(\mathrm{mod}p)$. Since $p$ is odd,

In the set of all $n!$ permutations, let $A_i$ consist of those fixing $i$. For a prescribed collection of $k$ fixed points, its intersection has $(n-k)!$ members. Applying the inclusion-exclusion principle to the complement of ${\bigcup }_i{A}_i$ gives the number of derangements:The convergent power series for the exponential function at $-1$ therefore gives $\boxed{d_n/n!⟶e^{-1}}$.

Take the finite ambient set $S={\bigcup }_i{A}_i$ and the additive set function $f(A)=|A|$. Disjoint subsets have additive cardinality, so (a) applies and givesThis is the finite inclusion-exclusion principle.

First, finite additivity of a set function gives $f(\varnothing )=0$, since $f(\varnothing )=2f(\varnothing )$. Put $w_s=f(\{s\})$. Decomposing any subset into its singleton elements and using induction gives $f(A)={\sum }_{s\in A}{w}_s$. Fix $s\in S$ and let $r$ be the number of sets $A_i$ containing it. Its total coefficient in the alternating intersection sum is zero if $r=0$, and otherwise isby the binomial theorem. These are exactly the coefficients in $f({\bigcup }_i{A}_i)$. Summing against $w_s$ proves the required inclusion-exclusion principle for the additive set function, including signed weights.

For any $c\in C$, the surjectivity of $g\circ f$ supplies $a\in A$ with $g(f(a))=c$. Therefore $f(a)\in B$ is a preimage of $c$ under $g$, proving surjectivity of $g$.

For the requested counterexample, take $A=C=\{0\}$, $B=\{0,1\}$, $f(0)=0$, and $g(0)=g(1)=0$. The function composition is the identity bijection of the singleton, but $f$ is not surjective and $g$ is not injective.

If $f(a)=f(a^{\prime })$, applying $g$ gives $(g\circ f)(a)=(g\circ f)(a^{\prime })$. The injectivity of $g\circ f$ implies $a=a^{\prime }$, so $f$ is injective.

Suppose $(g\circ f)(a)=(g\circ f)(a^{\prime })$. The injectivity of $g$ implies $f(a)=f(a^{\prime })$, and the injectivity of $f$ then implies $a=a^{\prime }$. Hence $g\circ f$ is injective.

Let $c\in C$. Since $g$ is surjective, choose $b\in B$ with $g(b)=c$. Since $f$ is surjective, choose $a\in A$ with $f(a)=b$. Then $(g\circ f)(a)=g(b)=c$. Thus every element of $C$ has a preimage, proving that $g\circ f$ is surjective.

Take the half-turn about the $z$-axis, represented by the rotation matrixIt is an orthogonal matrix with determinant $1$, and it fixes the $z$-axis while reversing both perpendicular coordinates. Direct substitution gives $Ma=b$, $Mb=a$, $Mc=d$ and $Md=c$. A linear map that permutes the four vertices also preserves their convex hull, which is the solid tetrahedron. Hence this rotation matrix leaves $T$ invariant. Its axis joins the midpoints of the two opposite edges $ab$ and $cd$.

The three known vertices have pairwise squared distances $8$, so the regular tetrahedron must have every squared edge length equal to $8$. Comparing the squared distance from $d$ to $b$ with that from $d$ to $a$ givesEquality forces $x=-1$. Similarly,so $y=-1$. The resulting point has squared distance $8$ from each of the other three vertices. Therefore