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Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.

A binary linear-feedback shift register of length $L$ stores $L$ bits. At each clock it outputs one bit, shifts the state, and inserts a fixed linear combination over ${\mathbb{F}}_2$ of the stored bits. Equivalently, its output obeys a recurrence

The Berlekamp-Massey algorithm processes the observed symbols from left to right. It maintains the shortest current connection polynomial $C(D)$, its degree $L$, and a saved polynomial from the most recent degree increase. At step $n$ it computes the discrepancyIf $d_n=0$, the current recurrence already predicts the new symbol. If $d_n\ne 0$, a shifted multiple of the saved polynomial is added to $C$ to cancel the discrepancy; when $2L\le n$, the algorithm also replaces $L$ by $n+1-L$ and saves the old polynomial.

For the observed binary sequence, the nonzero-discrepancy updates areAll discrepancies through $n=11$ then vanish. Thus the shortest recurrence isor, equivalently, $s_{n+4}=s_{n+1}+s_n$. The connection polynomial is $1+D^3+D^4$, and with the reciprocal convention the feedback polynomial is

A basis of a vector space $V$ is a family of vectors that is both linearly independent and a spanning set of $V$.

Write the given finite basis as $\mathcal{B}=\{v_1,\dots ,v_n\}$. First, any other basis ${\mathcal{B}}^{\prime }$ must also be finite. Each $v_i$ is a finite linear combination of elements of ${\mathcal{B}}^{\prime }$, so the union $S$ of the finitely many elements of ${\mathcal{B}}^{\prime }$ occurring in these $n$ expressions is finite. Since $S$ spans every $v_i$, it spans $V$. If some $w\in {\mathcal{B}}^{\prime }\setminus S$ existed, then $w$ would lie in the span of $S$, contradicting the linear independence of ${\mathcal{B}}^{\prime }$. Hence ${\mathcal{B}}^{\prime }=S$; write $|{\mathcal{B}}^{\prime }|=m$.

It remains to prove the elementary Steinitz exchange lemma directly. If independent vectors $w_1,\dots ,w_r$ lie in the span of $v_1,\dots ,v_n$, express $w_1$ in terms of the $v_i$. Some coefficient is nonzero, so the corresponding $v_i$ can be solved for in terms of $w_1$ and the other $v_i$; replacing it by $w_1$ preserves the span. Inductively, after replacing $k$ of the $v_i$ by $w_1,\dots ,w_k$, the expression for $w_{k+1}$ must have a nonzero coefficient on one of the unreplaced $v_i$, since otherwise $w_{k+1}$ would be a linear combination of $w_1,\dots ,w_k$. That $v_i$ can again be replaced. There are only $n$ original vectors to replace, so $r\le n$.

Apply this argument first to the independent family ${\mathcal{B}}^{\prime }$ and the spanning family $\mathcal{B}$ to obtain $m\le n$, and then with the two bases interchanged to obtain $n\le m$. Therefore

The total angular momentum operator isUsing the cyclic orbital angular momentum commutation relations,while $[L_z,L_z^2]=0$; hence $\boxed{[L_z,{\mathbf{L}}^2]=0}$. In Cartesian coordinates,Because $L_z(x+iy)=\hslash (x+iy)$ and it annihilates $z$ and every radial function,Replacing $i$ by $-i$ gives eigenvalue $-m\hslash$.

The six-dimensional space splits into the five trace-free quadratic spherical harmonics with $\ell =2$ and one radial state with $\ell =0$. A simultaneous eigenbasis for $L_z$ and ${\mathbf{L}}^2$ iswhere the eigenvalue $\ell (\ell +1){\hslash }^2$ of ${\mathbf{L}}^2$ was used. Every displayed state is a complex linear combination of the original six energy eigenstates, so it is still an energy eigenstate at the same level.

Let $r=dim(U\cap W)$. Choose a basis $e_1,\dots ,e_r$ of the intersection of vector subspaces, extend it to a basis $e_1,\dots ,e_r,u_{r+1},\dots ,u_m$ of $U$, and independently extend it to a basis $e_1,\dots ,e_r,w_{r+1},\dots ,w_n$ of $W$. The combined listspans the sum of vector subspaces $U+W$. It is also linearly independent: a vanishing linear combination says that a combination of the $u_i$ belongs to $U\cap W$, and its expression in the chosen basis of $U$ forces all its coefficients to vanish; the coefficients of the $w_i$ then vanish in the same way. This proves the dimension formula for a sum of subspaces

Parametrize the two linear subspaces asPutting a vector of $U$ in $W$ first forces its fourth coordinate to vanish; the remaining defining equation is then automatic. Thusand the dimension formula gives $\dim (U+W)=2+2-1=3$.

The linear functionalvanishes on each displayed generator of $U$ and $W$. Its kernel is three-dimensional because $\ell \ne 0$, so the inclusion $U+W\subseteq \ker \ell$ between two three-dimensional subspaces is equality.

Using the states from part (a),The Born rule therefore givesAfter the equal-sign outcomes the state is respectively $|\uparrow \uparrow ⟩$ or $|\downarrow \downarrow ⟩$. These are triplet states, hence eigenstates of ${\mathbf{S}}^2$ with eigenvalueThe opposite-sign product states are each a nontrivial linear combination of the singlet and the $m=0$ triplet, so they are not eigenstates of ${\mathbf{S}}^2$.

For a finite set $S=\{s_1,\dots ,s_k\}$ of vectors in a real vector space, its span is the set of all linear combinations:The span of the empty set is $\{0\}$. The set $S$ is linearly independent if ${\sum }_i{a}_i{s}_i=0$ implies $a_i=0$ for every $i$. It is a spanning set for $V$ if $\mathrm{Sp}(S)=V$. A basis is both a linearly independent set and a spanning set.

For basis extension, start with the given linearly independent set. If it does not span $V$, choose $v\notin \mathrm{Sp}(S)$ and adjoin it. The enlarged set remains linearly independent: a relation $av+{\sum }_i{a}_i{s}_i=0$ with $a\ne 0$ would express $v$ as a linear combination of $S$, a contradiction; if $a=0$, independence of $S$ forces all other coefficients to be zero.

Repeat whenever the current span is proper. Each step increases the set's size by one, while the allowed bound says its size cannot exceed $\dim V$. Thus the process terminates with a linearly independent set spanning $V$, hence a basis containing the original set. This also covers an initially empty set and the zero-dimensional vector space.

For a linear map $T:V\to V$, its kernel of a linear map and image of a linear map areIts nullity of a linear map is $n(T)=\dim \ker T$, and its rank of a linear map is $r(T)=\dim \mathrm{im}T$.

Choose a basis $u_1,\dots ,u_k$ of $\ker T$ and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_m$ of $V$. The vectors $T(v_1),\dots ,T(v_m)$ span $\mathrm{im}T$, because applying $T$ to an arbitrary linear combination of the full basis eliminates the $u_i$ terms. They are also linearly independent: if ${\sum }_j{b}_{j}T(v_j)=0$, then ${\sum }_j{b}_j{v}_j\in \ker T$, so ${\sum }_j{b}_j{v}_j={\sum }_i{a}_i{u}_i$ for some coefficients. Independence of the extended basis forces all $a_i,b_j$ to vanish.

Consequently $r(T)=m$, $n(T)=k$ and $\dim V=k+m$. This proves the rank-nullity theorem:

The rank of a linear map and nullity of a linear map areThe rank-nullity theorem states thatTo prove it, choose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and use basis extension to obtain a basisof $V$. The vectors $T(v_1),\dots ,T(v_{n-k})$ span the image of a linear map: applying $T$ to an arbitrary linear combination of the displayed basis removes all the $u_i$ terms. They are also linearly independent. Indeed,so ${\sum }_j{a}_j{v}_j={\sum }_i{b}_i{u}_i$ for some scalars $b_i$. The linear independence of the extended basis forces every $a_j$ and $b_i$ to vanish. Consequently $r(T)=n-k$ and $n(T)=k$, proving the identity. The same argument covers an empty basis when one of these dimensions is zero.

Every elementary column operation preserves the linear span of the columns: each new column is a linear combination of the old columns, and the inverse operation supplies the reverse inclusion. Thus the column space of $A^{\prime \prime }$ equals that of $A$, which is precisely the image of a linear map $T_{\alpha }$. Consequently (B) is true for every $\alpha \in \mathbb{R}$.

The Steinitz exchange lemma states that if $u_1,\dots ,u_r$ are linearly independent and $w_1,\dots ,w_s$ span a vector space $V$, then $r\le s$ and some $r$ of the $w_j$ can be replaced by $u_1,\dots ,u_r$ while preserving the spanning set property.

Here is the exchange argument. Suppose that after $k$ replacements, a spanning set iswhere the remaining $w_j$ have been relabelled if necessary. Express $u_{k+1}$ as a linear combination of this family. At least one remaining $w_j$ has a nonzero coefficient; otherwise $u_{k+1}$ belongs to the span of $u_1,\dots ,u_k$, contradicting linear independence. Relabel this entry $w_{k+1}$. Solving for $w_{k+1}$ shows that it lies in the span ofThe new family therefore still spans $V$. Starting at $k=0$ repeats this step until all $r$ independent vectors have been inserted. If $r>s$, after $s$ steps the next $u_{s+1}$ would lie in the span of $u_1,\dots ,u_s$, a contradiction. Thus $r\le s$, and the construction proves the replacement assertion as well.

The vectors are linearly independent ifThey span $V$ if every $v\in V$ is a linear combination $v={\sum }_{j=1}^n{a}_j{v}_j$ with real coefficients. They form a basis of $V$ if they are linearly independent and span $V$. Equivalently, every $v\in V$ has exactly one such representation: existence is the spanning set condition, and subtracting two representations shows that uniqueness is precisely linear independence.

The preimage isIt contains $0$, and if $u,v$ belong to it and $a,b\in \mathbb{R}$, then $T(au+bv)=aT(u)+bT(v)\in X$, since $X$ is a vector subspace. Thus $T^{-1}(X)$ is a vector subspace. It contains the kernel of $T$ because $0\in X$.

To establish the dimension of the preimage of a vector subspace formula without assuming the theorem to be deduced, choose a basis $k_1,\dots ,k_{n_T}$ of $\ker T$ and a basis $y_1,\dots ,y_d$ of $X\cap \mathrm{im}T$, where $d=\dim (X\cap \mathrm{im}T)$. For each $j$, choose $u_j\in U$ with $T(u_j)=y_j$. These $u_j$ lie in $T^{-1}(X)$.

The family $k_1,\dots ,k_{n_T},u_1,\dots ,u_d$ is linearly independent: applying $T$ to a vanishing linear combination forces all $u_j$ coefficients to vanish by linear independence of the $y_j$, and then all remaining coefficients vanish by linear independence of the $k_i$. It also spans $T^{-1}(X)$: for $u$ in that preimage, write $T(u)={\sum }_j{b}_j{y}_j$. Then $u-{\sum }_j{b}_j{u}_j\in \ker T$, so it is a linear combination of the $k_i$. ConsequentlySet $X=V$. Then $T^{-1}(V)=U$ and $V\cap \mathrm{im}T=\mathrm{im}T$, giving the rank-nullity theorem:

For any matrix $Y$ and compatible invertible matrix $P$, the linear map $r\mapsto rP$ maps $\mathrm{rowspace}(Y)$ onto $\mathrm{rowspace}(YP)$: a linear combination of rows of $Y$ is sent to the same linear combination of rows of $YP$. Its inverse function is $s\mapsto s{P}^{-1}$, so these row spaces have equal dimensions. Hence right matrix multiplication by $P$ also preserves row rank.

If $X=P^{-1}YP$, ii now givesThis proves the required matrix similarity invariant without needing to assume equality of row rank and column rank.

Write $A=(a_{ij})$ as an $m\times r$ matrix and $B$ as an $r\times n$ matrix, with rows $b_1,\dots ,b_r$. The rule for matrix multiplication givesEvery row of $AB$ is therefore a linear combination of rows of $B$. Taking linear spans gives

Over the finite field $F={\mathbb{Z}}_p$, a basis of the hyperplane has $n-1$ vectors by ii. Each vector on the hyperplane has a unique expression as a linear combination of that basis, and each coefficient has $p$ possibilities. ThusEquivalently, choose freely all coordinates except $x_j$, where $a_j\ne 0$, and solve uniquely for $x_j$. For $n=1$ this counts just the zero vector, consistently with $p^0=1$.

Bilinearity of the real inner product gives, for the specified linear combinations,The last equality is the entrywise rule for matrix multiplication, with $x^T$ the transpose of the column vector $x$.

Yes. Consider the following $n^2$ matrices:Each has only row $i$ nonzero and has $i$th diagonal entry $1$, so part (c)(ii) proves that it is a diagonalisable matrix. The collection is a spanning set, because it contains every $E_{ii}$ andThus it spans the matrix unit basis. A spanning set of $n^2$ vectors in an $n^2$-dimensional vector space is a basis. Alternatively, in a vanishing linear combination of $\mathcal{B}$, each off-diagonal entry first forces its own coefficient to be zero, and then each diagonal entry forces the remaining coefficient to be zero.

The vector triple product identity givesIt is the requested linear combination of $b$ and $c$. The notation $\wedge$ in this question denotes the three-dimensional cross product.

For $\lambda =5$, the eigenvalue equation reduces to a one-dimensional eigenspace spanned by $(1,2,-1)$. For $\lambda =-3$, the rows of $A+3I$ are multiples of $(1,2,-3)$, so its eigenspace is the plane $x+2y-3z=0$. One possible set of three independent eigenvectors isEvery eigenvector for $-3$ is a nonzero linear combination of the last two. Their two-dimensional eigenspace shows that the repeated root is not defective: the matrix is diagonalizable.