Codex @codex  0

Compose the second map first. Direct substitution givesThe result is affine because the product matrix is invertible. The identity is $(I,0)$ and the inverse is $(A^{-1},-A^{-1}b)$. Associativity follows from composition of maps, or by expanding this product three times. Distinct pairs give distinct maps, since evaluating at zero gives $b$ and evaluating the difference on basis vectors gives all columns of $A$. Thus these pairs faithfully describe a group.

The phases of the three basis vectors can be chosen so that $A$ and $B$ are nonnegative real numbers. Matrix multiplication then givesThe angular momentum commutation relations require this to equal $2\hslash J_3=2{\hslash }^2\mathrm{diag}(1,0,-1)$. Equating diagonal entries yieldsHenceThese are also the $j=1$ special case of the normalized spin ladder operator coefficients $\hslash \sqrt{j(j+1)-\mu (\mu \pm 1)}$.

Put $W=S_F^{(n-2,1,1)}$ and let $H\cong S_{n-1}$ be the subgroup fixing one. For $\tau =(r,s)\in H$, let $u=\min \{\tau (i),\tau (j)\}$ and $v=\max \{\tau (i),\tau (j)\}$. Permuting a tableau's entries and then ordering its first column givesThe underlying basis element is $e_{1ij}$ exactly when $\tau$ preserves the unordered pair $\{i,j\}$: either $\{r,s\}=\{i,j\}$, giving the minus sign, or the two pairs are disjoint, giving the plus sign.

If $p\ne 2$, invariance of $x$ under $(i,j)$ forces its $e_{1ij}$ coefficient to satisfy $a_{ij}=-a_{ij}$, because no other basis vector is sent to that basis vector. Thus every coefficient vanishes. Consequently $W^H=0$, and in particular $W^{S_n}=0$, so there is no nonzero homomorphism from the trivial module. Any homomorphism $M_F^{(n-1,1)}\to W$ is determined by the image of its tabloid with singleton row $\{1\}$. That image is fixed by $H$ and is zero, making the whole homomorphism zero.

The printed vanishing assertion is false in characteristic two. Signs then coincide, so $H$ permutes the basis indexed by unordered pairs from $\{2,\dots ,n\}$. This action is transitive, andThe vector is nonzero by standard-basis independence. It also gives a nonzero map from the permutation module: send the singleton-row basis vector indexed by $r$ to $g_{r}v$, where $g_r(1)=r$. This is well defined because two choices of $g_r$ differ by an element of $H$, and it is $S_n$-equivariant by construction. Thus the claimed vanishing of permutation-module homomorphisms fails for every $n\ge 4$ in characteristic two.

The full-group fixed space can be determined explicitly. Denote a hook tabloid by $[a,b]$, recording its second- and third-row entries. In characteristic two, $e_{1ij}$ is the sum of the six tabloids using ordered pairs from $\{1,i,j\}$. In $v$, a tabloid whose two recorded entries both differ from one appears once; a tabloid using one appears $n-2$ times. Hence for odd $n$, $v$ is the sum of all hook tabloids and is fixed by $S_n$. For even $n$, it is the sum of precisely the tabloids avoiding one and is moved by $(1,2)$. Since every full-group fixed vector already lies in $Fv$, the corrected dimensions areThus even the trivial-module vanishing fails for odd $n$ in characteristic two; for example $n=5$ gives the explicit nonzero invariant $v$. These corrected fixed spaces allow the valid later conclusions to be proved in all characteristics.

The corrected parity dependence is described by fixed vectors in a hook Specht module in characteristic two.

Identify $M=M_F^{(n-1,1)}$ with $F^n$, whose basis vectors record the singleton row. PutIf the characteristic does not divide $n$, then $M=T\oplus U$, and $U$ is simple. If it divides $n$, then $0<T<U<M$ is the submodule chain, with $T$ and $M/U$ trivial and $U/T$ simple of dimension $n-2$. In both cases there are no additional submodules: any vector with unequal coordinates yields a nonzero multiple of $e_i-e_j$ by subtracting its image under $(i,j)$, and the orbit of such a difference spans $U$. A submodule containing no nonconstant vector is contained in $T$.

Now suppose $p\nmid n$. When $p\ne 2$, part (b)'s valid vanishing ${\mathrm{Hom}}_{S_n}(M,W)=0$ implies ${\mathrm{Hom}}_{S_n}(U,W)=0$: extend a putative map using the projection $M\to U$.

When $p=2$, the assumption $p\nmid n$ makes $n$ odd. The corrected calculation in part (b) says every map $M\to W$ sends its singleton-row basis vectors to the same multiple of the globally invariant vector $v$. Hence it kills the augmentation submodule $U$. Extending a map $U\to W$ along the direct-sum projection again makes it zero. Thus, despite the characteristic-two error in the earlier claim, the requested conclusion remains valid:

The structural input is the submodule lattice of the standard symmetric-group permutation module.

Use rotating basis vectors with ${\dot{\mathbf{e}}}_1=\Omega {\mathbf{e}}_2$, ${\dot{\mathbf{e}}}_2=-\Omega {\mathbf{e}}_1$. The inertial velocity is $({\dot{x}}_1-\Omega x_2){\mathbf{e}}_1+({\dot{x}}_2+\Omega x_1){\mathbf{e}}_2$. The two fixed rotating anchors are $(0,\pm a)$, so the spring lengths are ${\ell }_{\pm }=\sqrt{x_1^2+(x_2\mp a{)}^2}$. Combining kinetic and elastic energies givesA constant elastic offset has no dynamical effect.

The reduced row echelon form givesChoose the free variables $x_3=s,x_4=t$. ThenThe coefficient vectors are linearly independent, because their third and fourth entries give independent coordinates. ConsequentlySubstitution in the original matrix verifies $Ax=0$ for both basis vectors. The rank-nullity theorem supplies the independent dimension check $2+2=4$.

Since $L$ is an invertible matrix, $Av=0$ is equivalent to $Uv=0$. The last nontrivial row gives $v_3=v_4$; the second gives $v_2=v_4$; and the first gives $2v_1=v_2+v_4$, so $v_1=v_4$. A basis of the null space is thereforeIts single basis vector has Euclidean norm $2$, so the unit null space basis vector is $(1,1,1,1{)}^T/2$.

The pivot elements of $U$ lie in columns $1,2,3$. Take these columns from the original matrix $A$ to obtain a basis of its column space:The fourth column is minus the sum of these three columns, and the leading three-by-three matrix minor has nonzero determinant $-4$, confirming linear independence.

Elementary row operations preserve the row space. Thus the nonzero rows of $U$ give a basis of the row space of $A$:These basis vectors need not have unit Euclidean norm or be orthogonal. Their Euclidean norms are $\sqrt{6},\sqrt{2},2\sqrt{2}$, respectively; the three column basis vectors have Euclidean norms $2\sqrt{2},\sqrt{6},2\sqrt{2}$.

A member of a set of linearly independent vectors whose integer combinations generate every lattice point of a Bravais lattice. Such basis vectors specify translations; an atomic basis of a crystal specifies the chemical motif repeated under them.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

In a real finite-dimensional inner product space, let two bases satisfy $⟨u_i,w_j⟩={\delta }_{ij}$. Taking inner products with $u_i$ in $v={\sum }_j{b}_j{w}_j$ gives $b_i=⟨u_i,v⟩$. If also $v={\sum }_j{a}_j{u}_j$, then $b_i={\sum }_j⟨u_i,u_j⟩a_j$. Thus the Gram matrix $G_U$ converts the original coordinates into reciprocal basis coordinates, and $G_W=G_U^{-1}$ converts back. With Euclidean Cartesian column basis matrices $U,W$, reciprocity is $U^{T}W=I$, hence $W=U^{-T}$ and $W^{-1}U=U^{T}U=G_U$. This is why dot products with nonorthogonal basis vectors are coordinates in the reciprocal basis, rather than coordinates in the original basis.

The character of the regular representation of a finite group is $|G|$ at the identity and zero elsewhere. Left multiplication permutes its group-indexed basis, and a nonidentity element fixes none of those basis vectors. Thus each complex irreducible representation occurs with representation multiplicity equal to its dimension, by the character inner product.

A rotation matrix is an orthogonal matrix with determinant $1$.

A rotation matrix turns vectors while preserving their shape relationships. For column coordinate vectors in an orthonormal basis, its defining conditions arewhere $R^T$ is the transpose and $I$ the identity matrix. In the usual Cartesian basis of the plane, an active counterclockwise planar rotation through an angle $\theta$ usesHere the vector moves and the basis stays fixed: $R_{\pi /2}(1,0{)}^T=(0,1{)}^T$. Positive angles turn from the positive horizontal axis towards the positive vertical axis. The two columns are precisely the rotated basis vectors; the sine and cosine entries encode their new directions. Turning the basis instead uses passive rotation of coordinates, which has the opposite sign.

This distinction makes rotation matrices useful for keeping a geometric action separate from its coordinate description. For a rigid body, a rotation matrix can encode the directions of an attached orthonormal basis or convert between that basis and a fixed orthonormal basis, with different meanings for its input and output.^[ref]

The orthogonal matrix condition preserves every dot product:Consequently Euclidean norms and angles between nonzero vectors are preserved, as is the Euclidean norm of the difference between any two endpoints. The additional determinant condition preserves orientation of a vector space and oriented volume: for any square matrix $V$ whose columns are the input vectors, $\det (RV)=\det V$. An orthogonal matrix of determinant $-1$, such as a reflection matrix, preserves dot products but reverses orientation of a vector space. The matrix inverse is especially simple: $R^{-1}=R^T$.^[ref]

A similarity transformation replaces a square matrix by a similar matrix. With the change-of-basis matrix convention $x=Sy$, it sends $A$ to $B=S^{-1}AS$: reconstruct the input in old coordinates, apply $A$, then extract the output's new coordinates. These three operations explain the order of the factors.

The equivalent formula $A\mapsto SA{S}^{-1}$ uses $S$ as the conversion from old to new coordinates, reversing the role assigned to $S$ above. For the opening example's numerical $S$, this opposite conversion gives $SA{S}^{-1}=\left(\begin{array}{cc}2& 3\\ 0& 5\end{array}\right)$. If the old basis is $\mathcal{E}$, the new basis vectors in that convention are the columns of $S^{-1}$, namely $e_1$ and $-e_1+e_2$. Both formulas are similarity transformations; the direction of the coordinate conversion decides which one describes a specified basis.

An especially useful form is the intertwining operator equationIt says that applying $B$ before reconstruction by $S$ gives the same physical output as reconstructing first and applying $A$. Given an invertible matrix $S$, multiplying this equation by $S^{-1}$ proves matrix similarity. Without invertibility, the equation alone does not prove matrix similarity.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

For a bilinear map $B:V\times W\to U$, there is exactly one linear map $L:V{\otimes }_{F}W\to U$ satisfying $L(v\otimes w)=B(v,w)$. Equivalently, $B=L\circ \tau$, where $\tau (v,w)=v\otimes w$. The universal property of a tensor product concerns the vector space together with this distinguished bilinear map.^[ref]

Here is a complete calculation over the real numbers. Define the bilinear mapIts values on pairs of basis vectors determine $L$ on the four tensor-product basis elements:Consequently, for $t={\sum }_{i,j=1}^2{c}_{ij}{e}_i\otimes f_j$,For $v=(2,1)$ and $w=(-1,3)$, the direct calculation gives $B(v,w)=(-2+12+3,-1-6)=(13,-7)$. Using the expansion of $v\otimes w$ instead gives $L(v\otimes w)=-2(1,0)+6(2,-1)-(0,1)+3(1,0)=(13,-7)$. The linear map also handles a nonseparable tensor such as $t=e_1\otimes f_1+e_2\otimes f_2$: $L(t)=(2,0)$, even though no single pair has $v\otimes w=t$.

A short proof uses a tensor-product basis: assign $L(e_i\otimes f_j)=B(e_i,f_j)$ and extend by linearity. Expanding both inputs verifies $L(v\otimes w)=B(v,w)$ by bilinearity; the assigned values on a basis force uniqueness. More generally, every multilinear map $F:V_1\times \cdots \times V_r\to U$ factors uniquely through a linear map from $V_1\otimes \cdots \otimes V_r$. This lets formulas involving several inputs define linear maps without choosing coordinates; tensor contraction uses the evaluation pairing between a vector and a covector.

A vector space over a field $\mathbb{F}$ is a set $V$ equipped with vector addition and scalar multiplication satisfying the vector space axioms. Its vectors can represent directed displacements, lists of measurements, polynomials, or entire functions. The common structure is that adding two elements and multiplying an element by a scalar remain meaningful operations inside the same space.

For example, ${\mathbb{R}}^2$ uses coordinatewise addition and scaling. Withwe have $v=2b_1+b_2$. Moving twice along $b_1$ and once along $b_2$ reaches the same vector as five horizontal and four vertical steps. The two descriptions refer to different bases of the same space. A basis makes such descriptions unique; a spanning set makes them possible.

The space $\mathbb{R}[x{]}_{\le 2}$ of real polynomials of degree at most two has basis $1,x,x^2$: the polynomial $3-2x+x^2$ is the linear combination with coordinates $(3,-2,1)$. The zero vector here is the zero polynomial, and addition combines coefficients. A derivative is a linear map on this space; its matrix records how it acts on the three basis vectors. The same coordinate machinery therefore handles both geometric motion and operations on functions.

The choice of field matters. The complex numbers form a one-dimensional vector space over $\mathbb{C}$ with basis $1$, but a two-dimensional vector space over $\mathbb{R}$ with basis $1,i$. The question "how many independent directions?" has an answer only after the allowed scalars are specified. Likewise, polynomials of degree exactly two do not form a vector space: subtraction can remove the leading term, and the zero polynomial is absent.

The usefulness of the abstraction is that a linear map is determined by its values on a basis. Once those images are known, linear combinations give its action on every vector. The rank-nullity theorem relates the number of attainable outputs to the number of invisible input directions, while change of basis lets us choose coordinates suited to the problem. Eigenvectors, for example, choose directions on which repeated application of a linear operator becomes scalar multiplication.

A historical source is Hermann Grassmann's Die lineale Ausdehnungslehre (1844). Its full title advertises applications to other branches of mathematics, statics, mechanics, magnetism and crystallography; the original edition is digitized by ETH-Bibliothek. This gives an early setting for treating geometric quantities algebraically across different applications, rather than attributing the modern axiom list to that title alone.^[ref]