The dimension formula for a sum of subspaces and the zero intersection implyBut $U+U^{\perp }$ is a vector subspace of the $n$-dimensional space $V$, so its dimension is at most $n$. Equality follows, and the sum is all of $V$. Since the intersection is zero, it is a direct sum:The argument remains valid for a degenerate ambient symmetric bilinear form; it uses the stated intersection hypothesis rather than assuming nondegeneracy everywhere.

Under the bidual evaluation map, a vector $y$ lies in $(\mathrm{im}\Phi {|}_U{)}^{\circ }$ exactly whenThat is precisely membership of the orthogonal complement for a bilinear form. ThereforeThe dimension formula in part (ii) givesso $\boxed{\dim U+\dim U^{\perp }\ge n}$. More precisely, the rank-nullity theorem givesThis is the orthogonal complement dimension for a degenerate bilinear form. The possible excess comes from vectors annihilated by the form against all of $V$.

The annihilator of a vector subspace $W\le V^{\ast }$ isIf $\dim W=k$, extend a basis of $W$ to one of $V^{\ast }$. In the resulting dual basis of $V^{\ast \ast }$, the last $n-k$ coordinate functionals span $W^{\circ }$. Hence $\boxed{\dim W+\dim W^{\circ }=n}$.

The bidual evaluation map $j:V\to V^{\ast \ast }$ is $j(x)(\varphi )=\varphi (x)$. It is a linear isomorphism in finite dimension. Its inverse image of the annihilator isThis identification uses evaluation only; no choice of inner product is involved.

The rank-nullity theorem states that for a linear map $A:X\to Y$ with finite-dimensional domain,Equivalently, nullity plus rank equals the dimension of the domain. One proof extends a basis of $\ker A$ to a basis of $X$; the images of the added vectors form a basis of $\mathrm{im}A$.

The fixed points of reciprocal root pairing are $1$ and $-1$, so part (iii) alone does not establish their parity. Use the polynomial identity itself.

Write $P(x)=(x-1{)}^{r}A(x)$, where $A(1)\ne 0$ and allow $r=0$ if one is not a root. After substituting in $P(x)=x^{2m}P(1/x)$ and canceling $(x-1{)}^r$ away from $x=1$, obtainLetting $x\to 1$ yields $A(1)=(-1{)}^{r}A(1)$, so $r$ is even.

Similarly write $P(x)=(x+1{)}^{s}B(x)$ with $B(-1)\ne 0$. This time cancellation gives $B(x)=x^{2m-s}B(1/x)$. Letting $x\to -1$ yields $B(-1)=(-1{)}^{2m-s}B(-1)$, so $s$ is even. ThusThe conclusion includes multiplicity zero. It is a consequence of the reciprocal identity in even degree, which is stronger at these fixed points than just pairing the roots.

Let $P={\chi }_M$, and suppose $P(z)=(z-\lambda {)}^{d}R(z)$ with $R(\lambda )\ne 0$. The reciprocal identity gives, for $x\ne 0$,Near $x={\lambda }^{-1}$ the factor $x^{2m-d}R(1/x)$ is nonzero. Also $1-\lambda x=-\lambda (x-{\lambda }^{-1})$ has a simple zero there. HenceThis proves equality of the algebraic multiplicities, not merely that the inverse is another eigenvalue. Invertibility excludes zero eigenvalues throughout.

For $\lambda \ne 0$, factor the matrix inside the determinant:The second line uses $\det M=1$; the third uses the even matrix size; the last uses part (i). ConsequentlyIn other words, the characteristic polynomial equals its reciprocal polynomial.

Rearrange $M^{T}JM=J$ asThus $M^{-1}$ is similar to $M^T$. Similar matrices have the same characteristic polynomial by part (a), and transposition also preserves it:Therefore $\boxed{{\chi }_{M^{-1}}={\chi }_M}$. This is the underlying algebraic source of the reciprocal spectrum of a symplectic matrix.

The identity is symplectic because $I^{T}JI=J$. For two symplectic matrices $M,N$,Thus the set is closed under matrix multiplication. Every member is invertible by part (i). Multiplying $M^{T}JM=J$ on the left by $M^{-T}$ and on the right by $M^{-1}$ givesso it is also closed under inverses. Associativity is inherited from the general linear group. These facts prove that the symplectic group is a subgroup of $\mathrm{GL}(2m,\mathbb{R})$.

The given $J$ satisfies $J^2=-I_{2m}$ and is invertible. Taking determinants in the defining identity of a symplectic matrix givesSince $\det (M^T)=\det M$ and $\det J\ne 0$, cancellation yields $(\det M{)}^2=1$. HenceIn particular every symplectic matrix is invertible. This argument establishes the requested alternatives; the later parts explicitly supply the determinant-one hypothesis that they use.

Let $P=b(T)q(T)$ and $Q=a(T)p(T)$. By part (ii), these are complementary projection operators with images $U=\ker (T-\lambda I{)}^{\alpha }$ and $W=\ker (T-\mu I{)}^{\beta }$. Thus $D=\lambda P+\mu Q$ acts as $\lambda I$ on $U$ and $\mu I$ on $W$.

The linear operator $T$ preserves both summands because it commutes with its own polynomials. Therefore $N=T-D$ preserves them too, and its restrictions satisfyPut $k=\max (\alpha ,\beta )$. Both restrictions of $N^k$ vanish, so the direct sum givesHence $T-D$ is a nilpotent linear map. The operator $D$ is diagonalizable, since its two summands are its eigenspaces, and it commutes with $T$. This illustrates the Jordan–Chevalley decomposition in the case of two distinct eigenvalues.

Write $U=\ker p(T)$, $W=\ker q(T)$. The linear operator $E=a(T)p(T)$ vanishes on $U$. On $W$, the identity $E=I-b(T)q(T)$ gives $Ew=w$.

By the direct sum in part (i), every vector has a unique expression $u+w$, andConsequentlyThe complementary projection operator is $I-E=b(T)q(T)$, onto $U$ along $W$. These are polynomial projections for a coprime decomposition. In particular the projection does not depend on which Bézout coefficients are chosen: its action on the two summands uniquely determines it.

Evaluate Bézout's identity for polynomials at $T$:All these linear operators commute because they are polynomials in the same operator. Thus every $v\in V$ can be writtenThe first summand belongs to $\ker p(T)$ and the second to $\ker q(T)$, since applying the corresponding operator introduces the product $p(T)q(T)=0$.

If $v$ lies in both kernels, the same identity gives $v=0+0=0$. Therefore the sum is direct:This is the kernel decomposition for coprime polynomials. It uses both annihilation by $pq$ and the coprimality expressed by the Bézout identity.

Suppose $E^2=E$. Every vector has the decompositionIf $w=Ez$ also belongs to $\ker E$, then $w=E^{2}z=Ew=0$. HenceOn its image of a linear map, $E$ acts as the identity; on its kernel of a linear map, it acts as zero. Concatenate a basis of the image with a basis of the kernel. The resulting matrix is $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}E$.

Conversely, this block matrix squares to itself, so any linear map having that matrix in a basis satisfies $E^2=E$. Thus projection operators are precisely the linear maps with this block form. This is an algebraic projection; no inner product or orthogonality is required.

The minimal polynomial $m_T$ is the monic polynomial of least polynomial degree such that $m_T(T)=0$. Here a polynomial $p(x)={\sum }_j{c}_j{x}^j$ is evaluated at the linear operator by $p(T)={\sum }_j{c}_j{T}^j$, with the constant term acting as $c_{0}I$.

An annihilating polynomial exists because $I,T,T^2,\dots$ lie in the finite-dimensional vector space ${\mathrm{End}}_F(V)$ and are therefore eventually linearly dependent. Uniqueness follows by subtracting two monic annihilators of the same least degree: their difference has smaller degree and still annihilates $T$.

Equivalently,For the forward implication evaluate the factorization at $T$. For the reverse implication divide $p$ by $m_T$ using polynomial division; the remainder annihilates $T$, so minimality makes it zero. This also identifies the annihilating polynomials as the principal ideal generated by $m_T$.

The two-torsion square-class homomorphism image on $E$ is all of $\{1,-1,5,-5\}$, realized by the identity, $(0,0)$, $(-1,3)$ and multiplication of classes. Its two-isogeny formula companion is $E^{\prime }:Y^2=X^3-10X^2+45X$. Positivity and prime support allow only $1,3,5,15$. Class five is realized by $(0,0)$, whereas class three would require $N^2=3U^4-10U^2{V}^2+15V^4$. Reduction modulo three forces one of coprime $U,V$ to be divisible by three; either choice gives odd valuation one on the right, contradicting a square. Multiplication by class five also excludes fifteen. The two-isogeny rank formula gives $2^r=4\cdot 2/4=2$.

For $C:y^2=x^3+\alpha x^2-\alpha x$, $\alpha \in \mathbb{Z}\setminus \{0,-4\}$, the two-isogeny formula companion is $D:v^2=u(u^2-2\alpha u+\alpha (\alpha +4))$. With the positive-sign standard isogenies, the two preimages of $(1,1)$ under the dual are $(\alpha ,-2\alpha )$ and $(\alpha +4,2(\alpha +4))$. The two-torsion square-class homomorphism on $D$ detects membership in the image of the forward isogeny of elliptic curves by the class of $u$. Thus either preimage lifts precisely when its first coordinate is a rational square. Their being integers turns rational squareness into integer squareness.

For $C:y^2=x(x^2+3x-3)$, the abscissa square class map has image contained in $\{1,-1,3,-3\}$. It contains $1,-3$ from the identity and $(0,0)$. Class three would require a primitive quartic solution $N^2=3U^4+3U^2{V}^2-V^4$. Modulo three this forces $V,N$ divisible by three; modulo nine the right side is then three and the left side zero, a contradiction. Class minus one would multiply by the known class minus three to give three, so is also excluded. This image has size two.

The dual curve is $D:y^2=x(x^2-6x+21)$. Its nonzero real abscissas are positive because $(x-3{)}^2+12>0$, so its square class image is contained in $\{1,3,7,21\}$. The point $(0,0)$ gives 21 and $(3,6)$ gives three, and their product gives seven. Thus this image has size four. The two-isogeny rank formula $2^{r+2}=|\mathrm{im}{\alpha }_C||\mathrm{im}{\alpha }_D|$ gives $2^{r+2}=8$, proving rank one. This combines a local obstruction on one covering with explicit rational witnesses on the other.

For $C:y^2=x(x^2+x-2)$, the two-isogenous curve is $D:v^2=u(u^2-2u+9)$. The first-coordinate square-class image on $C$ is $\{1,-2\}$. The prime-support bound permits only $1,-1,2,-2$; square class two would require a primitive integer solution to $N^2=2M^4+M^2{e}^2-e^4$, whose right side is two modulo three whenever $M,e$ are not both divisible by three. Thus that class is excluded, as is minus one by the subgroup property. The identity and the point $(0,0)$ supply the remaining two classes.

The image on $D$ is $\{1,3\}$: the quadratic factor $(u-1{)}^2+8$ is positive, so only positive square classes dividing nine occur, and $(3,6)$ supplies class three. The square-class index formula for two-isogeny descent consequently gives $2^r=(2\cdot 2)/4=1$, so the rank of an elliptic curve is zero. Alternatively, the dual-isogeny kernel lies in the image of the forward isogeny because nine is a square, giving $[C(\mathbb{Q}):2C(\mathbb{Q})]=4$. The four rational two-torsion points already account for that index, again forcing rank zero.

In characteristic zero, inducing a Specht module from ${\Sigma }_{n-1}$ to ${\Sigma }_n$ gives one copy of each Specht module obtained by adding one box in a valid Young-diagram position. By Frobenius reciprocity, its multiplicity in the induced module equals the multiplicity of $S^{\lambda }$ in the restriction of that larger Specht module. The restriction branching rule for a symmetric group says this multiplicity is one exactly when removing one corner gives $\lambda$, and zero otherwise. Semisimplicity in characteristic zero turns these multiplicities into the displayed direct-sum decomposition.