Choose a basis and let $A$ be the matrix of $T$. Use the monic conventionUnder a change of basis, $A$ becomes $B=P^{-1}AP$ for an invertible matrix $P$. Thenso multiplicativity of the determinant gives $\det (xI-B)=\det (xI-A)$. Therefore the characteristic polynomial is independent of the chosen basis and is well-defined for the linear transformation itself.

The Specht module $S^{(n-1,1)}$ is the augmentation subspace of the natural permutation module $F^n$, with the standard tabloid bilinear form. Its orthogonal complement in $F^n$ is the constant-vector line. That line lies in the augmentation subspace exactly when the field characteristic divides $n$, giving the displayed radical. Equivalently, the basis $v_i-v_n$ has Gram matrix $I_{n-1}+J_{n-1}$. The James submodule theorem makes every nonzero radical quotient simple, with dimension $n-2$ when $p\mid n$ and $n-1$ otherwise. The case $n=2,p=2$ gives the zero quotient rather than a nonzero simple module.

For the Legendre form of an elliptic curve $y^2=x(x-1)(x-\lambda )$, a change in the $x$ coordinate that permutes its branch values lifts with either of two opposite signs in the $y$ coordinate. These lifts differ by the involution $(x,y)\mapsto (x,-y)$. At $\lambda =e^{\pi i/3}$ the parameter transformation $x\mapsto 1-x/\lambda$ has order three. Its lift fixing $y$ has order three, whereas its lift negating $y$ has order six, since its cube is the nontrivial involution. Thus the order must be computed for a specified lift; both imply the same hexagonal shape of the associated period lattice.

Over the complex numbers, the multiplicity of $S^{\mu }$ in the Young permutation module $M^{\lambda }$ equals $K_{\mu \lambda }$. Its character is ${\mathrm{Ind}}_{S_{\lambda }}^{S_n}1$, so Frobenius reciprocity expresses this multiplicity as the average of the restricted irreducible character over the Young subgroup. For $\lambda =(2,1^{n-2})$ that subgroup contains only the identity and one transposition, giving $({\chi }_{\mu }(1)+{\chi }_{\mu }((12)))/2$. For $\lambda =(1^n)$ it is trivial, giving the character degree. This supplies a calculation from ordinary character values without enumerating semistandard tableaux. An indexing convention writing the multiplicity as $m_{\lambda ,\mu }$ reverses the conventional Kostka indices.

Let $\Phi$ be a nonempty set of unit vectors in a real Euclidean plane, closed under the reflections $s_v(u)=u-2⟨u,v⟩v$ and satisfying $2⟨u,v⟩\in \mathbb{Z}$ for every pair. Then it consists of an antipodal pair, a square, or a regular hexagon. Relative to any fixed member, the inner products can only be $0,\pm 1/2,\pm 1$, so only eight angles are initially possible. Reflection gives every negative vector. If a perpendicular pair occurs, it and its negatives exhaust the set, since any extra sixty-degree direction would have forbidden inner product $\pm \sqrt{3}/2$ with one of them. Otherwise any noncollinear pair gives, after negation, a sixty-degree angle; reflection supplies the third direction and negation supplies the other three hexagon vertices. If no such pair occurs, only the antipodal pair remains. Each configuration directly satisfies the conditions.

These are the equal-length cases of the classification of rank-two root systems, with the antipodal case having rank one. The square is the orthogonal sum of two A1 root systems, and the hexagon is an A2 root system. The root system is considered in the span of its vectors when the ambient plane is not spanned.

In an orthonormal coordinate system on ${\mathbb{R}}^4$, these forty-eight vectors form the reduced crystallographic root system of type $F_4$. There are twenty-four long roots of squared length two, and twenty-four short roots of squared length one. Signed coordinate permutations preserve the set. Reflections in coordinate roots are sign changes, and reflections in long roots are signed permutations. For the remaining short roots it suffices, by symmetry, to use $h=(1/2,1/2,1/2,1/2)$; its reflection is $v\mapsto v-({\sum }_i{v}_i)(1,1,1,1)/2$. It sends coordinate roots to half-coordinate roots, long roots to long roots, and half-coordinate roots to half-coordinate or coordinate roots. Thus all required reflections preserve the set. The Cartan integrality condition follows directly from its integral or half-integral coordinate pairings.

One base of a root system is $e_2-e_3,e_3-e_4,e_4,(e_1-e_2-e_3-e_4)/2$. Its two long roots followed by two short roots give the F4 Dynkin diagram, with the double bond between the middle nodes and the arrow toward the short root.

For a finite-dimensional complex sl2 Lie algebra module $V$, the number of simple summands, counted with multiplicity, is $\dim V_0+\dim V_1$. Indeed, the simple module of highest weight $m\ge 0$ has weights $m,m-2,\dots ,-m$, each once. Exactly one of zero and one appears: zero when $m$ is even, one when $m$ is odd. The classification of finite-dimensional sl2 representations decomposes $V$ into these modules, and dimensions of weight spaces add over the direct sum.

Suppose the Cartan matrix has diagonal two, entries minus one between adjacent indices, and all other entries zero. Normalize the simple-root squared lengths to two. Every root is a Weyl group translate of a simple root, so all roots have squared length two. If a positive root has coefficients $c_1,\dots ,c_l$ in the simple-root basis, let $c_0=c_{l+1}=0$. ThenThe nonnegative integer coefficients force exactly one upward jump of one and one downward jump of one. Thus the coefficients are one on a consecutive interval and zero elsewhere. Conversely, an interval root can be extended by the next simple root: its Cartan pairing with that root is minus one, so the corresponding simple reflection adds it. Starting from a simple root constructs every interval and proves the displayed description. This also identifies the root system as type $A_l$ without needing to enumerate reflection words individually.

Let $n\ge 4$ and let the coefficient field have characteristic two. The subgroup fixing one permutes the standard hook basis $e_{1ij}$ indexed by unordered pairs from $\{2,\dots ,n\}$. Its fixed space is the line generated by $v={\sum }_{2\le i<j\le n}{e}_{1ij}$. Record a hook tabloid by its second- and third-row entries. In $v$, every recorded ordered pair avoiding one occurs once, and every pair involving one occurs $n-2$ times: a polytabloid is the sum of the six ordered pairs from its three column entries. For odd $n$, $v$ is therefore the sum of all tabloids and is fixed by the full symmetric group. For even $n$, it is the sum of the tabloids avoiding one and is moved by $(1,2)$. No other full-group fixed vector is possible because the point-stabilizer fixed space is one-dimensional. In characteristic different from two, swapping $i,j$ negates the $e_{1ij}$ coefficient and forces it to vanish, so even the point-stabilizer fixed space is zero.