Lagrange's theorem states that if $H$ is a subgroup of a finite group $G$, thenwhere the index of a subgroup $[G:H]$ is the number of distinct left cosets.

Every $g\in G$ lies in $gH$, since $g=ge$. By i, distinct left cosets are disjoint, so these sets form a partition of a set $G$. Each has cardinality $|H|$: the function $h\mapsto gh$ from $H$ to $gH$ is a bijection, with inverse $gh\mapsto h$, by cancellation in a group. Counting the elements in this finite partition of a set gives $|G|=[G:H]|H|$, proving the theorem.

For a subgroup $H$ of $G$, the left coset represented by $g$ isSuppose two left cosets intersect. There are $h_1,h_2\in H$ such that $g_1{h}_1=g_2{h}_2$, soBecause $H$ is a subgroup, $h_2{h}_1^{-1}\in H$. Thus, for every $h\in H$, $g_{1}h=g_2(h_2{h}_1^{-1}h)\in g_{2}H$, proving $g_{1}H\subseteq g_{2}H$. Interchanging the two representatives gives the reverse inclusion. Therefore two left cosets are equal or disjoint.

$A$ is not an orthogonal matrix. Its first column has squared Euclidean normso $A^{T}A\ne I$. The reason this is compatible with ii is that $1,x,x^2$ is not an orthonormal basis for the given inner product. Its Gram matrix isThe correct condition on this matrix representation of a linear map iswhich follows either by direct multiplication or by the inner product preservation just proved. This is the rule for an orthogonal transformation in a nonorthonormal basis.

Put $p=(3x^2-2)/\sqrt{2}$ and use the orthonormal basis $\mathcal{B}=(1,\sqrt{3/2}x,p)$ from iii. The columns of the given matrix giveConsequentlyThe images of $\mathcal{B}$ are therefore $(p,\sqrt{3/2}x,-1)$, again an orthonormal basis. By i, $T$ is an orthogonal transformation. Its matrix in this orthonormal basis is

If $T$ is an orthogonal transformation and $e_1,e_2,e_3$ is an orthonormal basis, thenThe images form an orthonormal set, hence are linearly independent and form an orthonormal basis in dimension three.

Conversely, suppose the images $T{e}_1,T{e}_2,T{e}_3$ of one chosen orthonormal basis form an orthonormal basis. Write $u={\sum }_i{a}_i{e}_i$ and $v={\sum }_j{b}_j{e}_j$. By linearity of $T$ and bilinearity,Thus mapping one orthonormal basis to an orthonormal basis is equivalent to preserving every inner product. In the converse, a single chosen orthonormal basis suffices.

Use $f,g$ from i. Their inner products with $x^2$ areThus $h=x^2-2/3$. Its values at $-1,0,1$ are $1/3,-2/3,1/3$, soBy ii the three polynomials are orthogonal vectors, and normalizing $h$ givesThis orthonormal set is a basis of the three-dimensional vector space $V$. It is the result of the Gram-Schmidt process applied to $1,x,x^2$.

For any orthonormal basis $f,g$ of $U$, bilinearity givesTogether with $⟨f,g⟩=0$, these identities prove pairwise orthogonality. The construction subtracts the orthogonal projection onto a finite-dimensional subspace $U$ from $x^2$, leaving the orthogonal projection residual.

The relevant inner products areHence $1$ and $x$ are orthogonal vectors. Dividing by their norms gives the orthonormal basisof their linear span $U$.

Using the convention ${\chi }_T(x)=\det (xI-A)$, the characteristic polynomial isSince $x^2+1>0$ for every $x\in \mathbb{R}$, the only eigenvalue over the real numbers is $1$. Solving $(A-I)v=0$ givesThus the only eigenspace over the real numbers is the axis of the rotation.

Choose the rotation taking $e_1$ to $e_2$, $e_2$ to $-e_1$ and $e_3$ to itself. The columns of the matrix of a linear map are these images of the standard basis, soThis fixes the orientation used throughout this question.

A system of linear equations $T_{\alpha }x=b$ is solvable exactly when $b\in \mathrm{im}{T}_{\alpha }$. If $x_0$ is one solution, thenso its solution set is $x_0+\ker T_{\alpha }$, a translate of the kernel of a linear map.

For $\alpha \ne 2$, the matrix is an invertible matrix, andEvery $b$ is in the image of a linear map and has exactly one preimage. For $\alpha =2$, its first two columns span the image of a linear map, while the corresponding homogeneous linear system gives the kernel of a linear map:The prescribed $b=(-1,\beta ,1{)}^T$ satisfies the displayed condition precisely when $\beta =0$. If $\beta \ne 0$, it lies outside the image of a linear map and there is no solution. If $\beta =0$, it lies in the image of a linear map, and the nonzero kernel of a linear map supplies infinitely many solutions.

The necessary and sufficient condition isFor necessity, if $ST={\mathrm{id}}_V$ and $v\in \ker T$, then $v=S(Tv)=S(0)=0$.

For sufficiency, let $e_1,e_2$ be a basis of $V$ and suppose the kernel of a linear map is trivial. The vectors $T{e}_1,T{e}_2$ are linearly independent, since $aT{e}_1+bT{e}_2=0$ implies $a{e}_1+b{e}_2\in \ker T$. Extend them to a basis $T{e}_1,T{e}_2,w_3$ of $W$. Define a linear map $S$ byThen $ST$ agrees with the identity operator on a basis, hence on all of $V$. Thus a left inverse exists precisely when $T$ is injective; this is the linear left inverse criterion.

If $TS={\mathrm{id}}_W$, every $w\in W$ would satisfy $w=T(Sw)$, so $T$ would be a surjective linear map. But its image of a linear map has dimension at most $\dim V=2$, whereas $\dim W=3$. Therefore no such $S$ exists. This is the obstruction to a right inverse: $T$ cannot reach every target vector.

Here $\dim V=2$. By the rank-nullity theorem, $r(T)+n(T)=2$, and a dimension is a nonnegative integer. All three possibilities occur:The images of a linear map in these examples have dimensions $0,1,2$, respectively, while the corresponding kernel of a linear map is ${\mathbb{R}}^2$, $\{(0,y):y\in \mathbb{R}\}$ and $\{0\}$. The identity above excludes all other pairs.

For $S>b$, maximize the coefficient $H(b)=b^{\gamma }(K-b{)}^n$ of $S^{-\gamma }$. Its logarithmic derivative isIt changes sign once, at $b=\gamma K/(\gamma +n)$. The perpetual power put option therefore has candidate valueThe same boundary follows from smooth fit, $-\gamma (K-b{)}^n/b=-n(K-b{)}^{n-1}$. To establish optimality among all stopping times, rather than only fixed barriers, we verify the candidate. Let $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$ be the infinitesimal generator. The maximum property of $H$ gives $V_n\ge g_n$ everywhere, and $(\mathcal{L}-r)V_n=0$ for $S>b$. For $S\le b$, set $y=S/(K-S)\le \gamma /n$. ThenThus the discounted candidate is a nonnegative supermartingale. The smooth fit removes the derivative-jump local time of a semimartingale term in the Itô formula. The optional stopping theorem and localization bound every exercise policy's discounted expected value by $V_n$. Immediate exercise attains it below $b$; the first hitting time of $b$ attains it above $b$. The terminal remainder vanishes since $0\le e^{-ru}{V}_n(S_u)\le K^n{e}^{-ru}\to 0$. This proves the displayed value.

For $n=1$ it reduces to the ordinary perpetual put option. Larger $n$ lowers the optimal trigger, because the more strongly curved financial payoff makes further downward moves more valuable. If $r=0$, under the risk-neutral measure the stock tends to zero almost surely, so the supremum is $K^n$, approached as the exercise time tends to infinity and not attained at a finite time when $S>0$. If $r<0$, the values at deterministic exercise times grow without bound; the perpetual value is infinite. These cases do not admit the positive-rate formula above.

Assume $r>0$, put $b=S^{\star }\in (0,K)$ and let ${\tau }_b=\inf \{u\ge 0:S_u\le b\}$. In the continuation region $S>b$, the time-independent Black-Scholes equation is a Cauchy-Euler equation. Substituting $V=S^m$ givesSet $\gamma =2r/{\sigma }^2>0$. The two powers are $S$ and $S^{-\gamma }$. The boundary condition $V(S)\to 0$ as $S\to \infty$ removes the growing solution; value matching at $b$ sets the coefficient. Hence the fixed-trigger value isFor $0\le S\le b$ the policy exercises immediately, giving $V_b(S)=(K-S{)}^n$. The quoted continuation expression alone does not apply below the exercise boundary. Equivalently, the discounted boundary-hitting representation gives ${\mathbb{E}}_S[e^{-r{\tau }_b};{\tau }_b<\infty ]=(S/b{)}^{-\gamma }$, which multiplies the exercise financial payoff.

Write $\tau =T-t>0$ and $g_n(s)=(K-s{)}_{+}^n$. The European power put formula isHere $Y>0$ has a lognormal distribution. Since $g_n$ is decreasing and is a convex function, its discounted expected value is also decreasing and convex as a function of $S$. For $S>0$ it is smooth and strictly positive: the normal distribution assigns positive probability to $SY<K$ whatever the current positive spot price. Its endpoints areThe dominated convergence theorem justifies the second limit, because $0\le (K-SY{)}_{+}^n\le K^n$. For a quantitative sketch, the binomial theorem and truncated lognormal moment givewhere $\Phi$ is the standard normal distribution function. This finite sum is useful for drawing the curve; its positive expectation form makes its qualitative shape clearer.

The solid curves use $K=1$, $r=0.05$, $\sigma =0.30$ and $\tau =1$; the dashed curves show their terminal financial payoffs for comparison. Each price remains positive at $S=K$ and tends smoothly towards zero beyond it. At small $S$, discounting gives a price below the undiscounted financial payoff; European exercise is permitted only at expiry.