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Suppose $\mathbf{v}=\alpha {\mathbf{e}}_1+\beta {\mathbf{e}}_2$. Applying $M-{\lambda }_{2}I$ would give ${\mathbf{e}}_2=\alpha ({\lambda }_1-{\lambda }_2){\mathbf{e}}_1$, contradicting linear independence of the two eigenvectors. Therefore the generalized eigenvector $\mathbf{v}$ lies outside their span and $({\mathbf{e}}_1,{\mathbf{e}}_2,\mathbf{v})$ is a basis. Write $\mathbf{x}=a{\mathbf{e}}_1+b{\mathbf{e}}_2+d\mathbf{v}$. Since $M\mathbf{v}={\lambda }_2\mathbf{v}+{\mathbf{e}}_2$, the linear system of ordinary differential equations givesSolving these equations gives $a=A_1{e}^{{\lambda }_{1}t}$, $d=A_3{e}^{{\lambda }_{2}t}$ and $b=(A_2+A_{3}t)e^{{\lambda }_{2}t}$. ThusThe length-two Jordan chain solution for the Jordan chain $({\mathbf{e}}_2,\mathbf{v})$ produces the $t{e}^{{\lambda }_{2}t}$ mode. The three arbitrary constants span all initial data, so this is the full general solution.

The three linearly independent eigenvectors still form a basis, so the same coefficient calculation applies to this diagonalizable matrix matrix. The general solution isA repeated eigenvalue alone does not create a polynomial factor in time; that factor is needed only when a nontrivial Jordan block prevents a full eigenvector basis.

The independence of eigenvectors for distinct eigenvalues ensures that such eigenvectors form a linearly independent set. For completeness, apply ${\prod }_{j\ne k}(M-{\lambda }_{j}I)$ to a proposed relation ${\sum }_i{c}_i{\mathbf{e}}_i=0$. It leaves $c_k{\prod }_{j\ne k}({\lambda }_k-{\lambda }_j){\mathbf{e}}_k=0$, so every $c_k=0$. Thus the three eigenvectors form a basis over the relevant field and $\mathbf{x}={\sum }_i{a}_i(t){\mathbf{e}}_i$. Substitution into the linear system of ordinary differential equations gives ${\sum }_i({\dot{a}}_i-{\lambda }_i{a}_i){\mathbf{e}}_i=0$; linear independence implies ${\dot{a}}_i={\lambda }_i{a}_i$. HenceA real matrix may have a complex conjugate pair of eigenvalues. In that case the calculation is over $\mathbb{C}$ and conjugate coefficients yield real solutions; equivalently use real and imaginary parts of the complex modes. No unstated assumption that all three eigenvalues are real is needed.

The Frankl-Wilson theorem. Let $p$ be a prime number, let $L\subseteq {\mathbb{F}}_p$ have $s$ elements, and let $\mathcal{F}\subseteq \mathcal{P}([n])$. Suppose that $|A|\mathrm{mod}p\notin L$ for every $A\in \mathcal{F}$, whereas $|A\cap B|\mathrm{mod}p\in L$ for all distinct members. Thenwith terms beyond $n$ understood as zero.

Work over the finite field ${\mathbb{F}}_p$. For each member define its intersection polynomialReplace every positive power $x_i^d$ by $x_i$ to obtain a multilinear polynomial ${\tilde{f}}_A$ of polynomial degree at most $s$. This Boolean multilinearization preserves its values at every characteristic vector of a set. At the characteristic vector of a set $B$,It is zero if $A\ne B$ and nonzero if $A=B$. Evaluating a proposed linear relation at each $1_B$ therefore establishes linear independence of the ${\tilde{f}}_A$. The vector space of multilinear polynomials of polynomial degree at most $s$ has the basis of monomials ${\prod }_{i\in T}{x}_i$, $|T|\le s$, of size ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the Frankl-Wilson theorem by the polynomial method in combinatorics. The empty product for $s=0$ is $1$, and the argument still applies.

The constant-intersection family bound, directly. Write $\mathcal{F}=\{F_1,\dots ,F_q\}$ and assume $n\ge 1$. The cases $q\le 1$ are immediate. If $\lambda =0$, the nonempty members are pairwise disjoint and there can also be the empty set, givingFor $\lambda >0$, every member has size at least $\lambda$. If every member has size strictly greater than $\lambda$, form the real matrix $X$ whose rows are their characteristic vectors of sets. Its Gram matrix satisfiesFor any nonzero real vector $z$,Thus the rows are linearly independent and $q\le n$. If a member $F_1$ has size $\lambda$, it lies inside every other member. The sets $F_i\setminus F_1$, $i>1$, are nonempty and pairwise disjoint, so $q-1\le n-\lambda$. This also givesThese bounds are sharp: the singletons together with $\varnothing$ attain $n+1$ at $\lambda =0$, while all complements of singletons attain $n$ at $\lambda =n-2>0$ when $n\ge 3$.

The constant t-wise intersection dichotomy. In its substantive form the last argument requires $m\ge t$. The printed statement does not explicitly impose this. Without it the $t$-fold condition can be vacuous and the claimed dichotomy is false: take $t=3$, $m=2$, $n=1$, $A_1=\varnothing$, $A_2=\{1\}$ and $\lambda =1$. There is no $3$-tuple to test, no common $1$-set, and $k=0$, so $m=2>k+t-2=1$. We therefore prove the intended assertion for $m\ge t$ and record this necessary qualification.

If $\lambda =0$, the first alternative always holds with $S=\varnothing$. Suppose $\lambda >0$. Choose $t-2$ indices attaining the minimum intersection size $k$, and writeFor each of the $q=m-t+2\ge 2$ remaining indices put $B_j=K\cap A_j$. The hypothesis gives $|B_j\cap B_{\ell }|=\lambda$ for distinct remaining indices. This reduces the constant t-wise intersection dichotomy to a constant-intersection family bound on the $k$-element set $K$. The two possibilities are treated below.

A nilpotent endomorphism $x$ satisfies $x^N=0$ for some positive integer $N$. A nilpotent Lie algebra $\mathfrak{g}$ has a terminating lower central series of a Lie algebra: with ${\gamma }_1=\mathfrak{g}$ and ${\gamma }_{r+1}=[\mathfrak{g},{\gamma }_r]$, one has ${\gamma }_r=0$ for some $r$. These are different conditions: the first concerns a particular linear map, whereas the second concerns repeated Lie brackets.

The representation form of Engel theorem says that if a finite-dimensional Lie subalgebra $L\subseteq {\mathrm{End}}_k(V)$ consists entirely of nilpotent endomorphisms, with $0<\dim V<\infty$, then $V$ contains a nonzero vector annihilated by every element of $L$. Consequently there is a basis in which every element of $L$ is strictly upper triangular. No assumption on the characteristic of a field or algebraic closure is required.

We prove the common-vector assertion by induction on $\dim L$, simultaneously for every nonzero finite-dimensional representation space. The zero algebra is immediate. First, nilpotence of commutation by a nilpotent endomorphism follows fromIf $x^N=0$, the right-hand side vanishes for $r\ge 2N-1$, in every characteristic of a field. For any proper Lie subalgebra $M⊊L$, its Adjoint representation on $L/M$ therefore consists of nilpotent endomorphisms. Its image has dimension at most $\dim M<\dim L$, so the inductive assertion gives a nonzero class $y+M$ annihilated by $M$. Equivalently $y\notin M$ and $[M,y]\subseteq M$. Thus the Engel normalizer lemma gives $M⊊N_L(M)$.

Choose a maximal proper Lie subalgebra $M$. Its normalizer of a Lie subalgebra must be all of $L$, so $M$ is an ideal of a Lie algebra. Moreover $L/M$ has dimension one: otherwise a one-dimensional Lie subalgebra of this quotient would have a proper inverse image strictly between $M$ and $L$. By induction the space $W=\{v\in V:Mv=0\}$ is nonzero. It is invariant under $L$, since $m(xw)=x(mw)+[m,x]w=0$. Write $L=M+kx$. The restriction of the nilpotent endomorphism $x$ to $W$ has a nonzero kernel, and any vector in that kernel is annihilated by all of $L$. This completes the induction. Apply the assertion repeatedly to $V/kv$ to obtain a complete invariant flag $0=V_0\subset V_1\subset \cdots \subset V_d=V$ with $L{V}_j\subseteq V_{j-1}$. A basis adapted to that flag makes every matrix strictly upper triangular.

A nilpotent abstract Lie algebra need not act by nilpotent matrices. For $V\ne 0$, the scalar algebra $kI\subseteq \mathfrak{gl}(V)$ is an abelian Lie algebra, hence a nilpotent Lie algebra, but $I^N=I\ne 0$. No change of basis makes $I$ strictly upper triangular. The failed implication is precisely the distinction described in nilpotent Lie algebras need not act nilpotently.

If $\mathfrak{g}$ is a nilpotent Lie algebra, then $(\mathrm{ad}x{)}^r\mathfrak{g}\subseteq {\gamma }_{r+1}$, so every $\mathrm{ad}x$ is a nilpotent endomorphism. Conversely, apply Engel theorem to $\mathrm{ad}\mathfrak{g}$ acting on $\mathfrak{g}$. Its invariant flag is lowered by every Adjoint representation matrix, so every product of $d=\dim \mathfrak{g}$ such matrices vanishes. Since ${\gamma }_{d+1}$ is spanned by expressions $\mathrm{ad}{x}_1\cdots \mathrm{ad}{x}_d(y)$, the lower central series of a Lie algebra terminates. ThereforeFinally suppose $k$ is an algebraically closed field. If some $\mathrm{ad}x$ is not a nilpotent endomorphism, it has a nonzero eigenvalue $\lambda$ and an eigenvector $y$, giving $[x,y]=\lambda y$. The vectors $x,y$ are linearly independent and span a nonabelian two-dimensional Lie subalgebra. Conversely, a Lie subalgebra of a nilpotent Lie algebra is nilpotent because its lower central series of a Lie algebra lies termwise in the ambient series. A nonabelian two-dimensional Lie algebra has a basis $u,v$ with $[u,v]=v$: choose $v$ spanning the nonzero derived algebra and rescale a complementary vector. Its Adjoint representation has the nonzero eigenvalue $1$, so it is not nilpotent. This proves the two-dimensional subalgebra criterion for Lie algebra nilpotence:

Define the exterior square over the arbitrary field $k$ by ${\bigwedge }^{2}V=(V\otimes V)/\mathrm{span}\{v\otimes v:v\in V\}$. Write the image of $u\otimes v$ as $u\wedge v$. Then $v\wedge v=0$ and $u\wedge v=-v\wedge u$; the basis is $v_i\wedge v_j$ with $i<j$, in characteristic two as well. The exterior-power Lie algebra representation isThe tensor product of Lie algebra representations descends to this quotient: $x\cdot (v\otimes v)=xv\otimes v+v\otimes xv$ is a linear combination of square tensors, namely $(v+xv)\otimes (v+xv)-v\otimes v-xv\otimes xv$. On the wedge basis its coefficients follow directly from the original representing matrices. Expanding both actions shows $[{\rho }_{{\wedge }^2}(x),{\rho }_{{\wedge }^2}(y)]={\rho }_{{\wedge }^2}([x,y])$.

The inclusions of $V$ and $W$ into their direct sum define the mapA combined basis shows that this sends a basis to the pure-$V$, pure-$W$ and mixed wedge basis vectors, with no overlap and no omission. The defining action on each wedge proves equivariance. Hence the exterior square of a direct sum givesNo division by $2$ is involved, so the proof remains valid in characteristic two.

For the complex special linear Lie algebra ${\mathfrak{sl}}_3$, write ${\Gamma }_{a,b}$ for the irreducible highest-weight representation with Dynkin labels $(a,b)$, and $V={\Gamma }_{1,0}$. First the sl3 decomposition of the symmetric-square dual tensor product isIndeed contraction $c((uv)\otimes f)=f(u)v+f(v)u$ is a surjective Lie algebra representation homomorphism onto $V$. Its 15-dimensional kernel contains the highest-weight vector $e_1^2\otimes e_3^{\ast }$ of weight $(2,1)$. The Weyl dimension formula gives $\dim {\Gamma }_{2,1}=15$, and the Weyl complete reducibility theorem identifies the kernel and splits the map. With $B={\Gamma }_{2,1}$, the direct-sum identity reduces the requested calculation to ${\bigwedge }^{2}B$, $B\otimes V$ and ${\bigwedge }^{2}V$.

For completeness, these decompositions can be checked entirely by formal characters. Let $x_1{x}_2{x}_3=1$, $h_1=x_1+x_2+x_3$, $h_1^{\ast }=x_1^{-1}+x_2^{-1}+x_3^{-1}$ and $h_2={\sum }_i{x}_i^2+{\sum }_{i<j}{x}_i{x}_j$. Then ${\chi }_B=h_2{h}_1^{\ast }-h_1$. The Weyl character formula takes the determinant formSubstitute ${\chi }_B$ into ${\chi }_{{\wedge }^{2}B}(x)=\frac{1}{2}({\chi }_B(x{)}^2-{\chi }_B(x_1^2,x_2^2,x_3^2))$ and collect the determinant characters. This gives the exterior square of the sl3 representation of highest weight (2,1) and the sl3 highest-weight tensor rule:The first line has dimensions $21+42+24+15+3=105=\left(\genfrac{}{}{0ex}{}{15}{2}\right)$; the tensor product has dimension $24+15+6=45$. Equivalently, enumerate weights with the sl3 interlacing character formula and subtract characters from the highest weight downwards. FinallyIts dimension is $21+42+48+30+6+6=153=\left(\genfrac{}{}{0ex}{}{18}{2}\right)$. The exterior square here is taken of the entire 18-dimensional tensor product; taking it only of $S^{2}V$ would be a different representation.

Use the Holstein–Primakoff transformation on the physical occupation number subspace $0\le n\le 2S$. With the canonical commutation relation $[a,a^{†}]=1$, the ordered square root givesBoth endpoints are respected: $S^{+}|0⟩=0$ and $S^{-}|2S⟩=0$. Consequently,Since these Fock states form a basis of the physical spin space, $[S^{+},S^{-}]=2S^z$ there. Similarly $[S^z,S^{\pm }]=\pm S^{\pm }$. The Holstein–Primakoff occupation constraint is essential: unrestricted bosonic occupation would not represent a spin-$S$ Hilbert space.

The symmetric power $S^{n}V$ is the quotient of $V^{\otimes n}$ identifying tensors which differ by permutation of the factors. The exterior power ${\Lambda }^{n}V$ is the alternating quotient, with a tensor having repeated equal factors set to zero. Both quotients inherit the given group action. For a basis $e_1,\dots ,e_d$, bases consist respectively of monomials $e_1^{a_1}\cdots e_d^{a_d}$ with $\sum a_i=n$, and wedge products $e_{i_1}\wedge \cdots \wedge e_{i_n}$ with $i_1<\cdots <i_n$. Thuswith the second dimension zero for $n>d$.

A finite-order group element is diagonalizable over $\mathbb{C}$, because its minimal polynomial divides $X^m-1$, which has no repeated roots. Choose an eigenbasis with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$. The induced eigenvalues, including multiplicities, areThe zeroth powers are the one-dimensional trivial representation, with the empty product equal to $1$.

A matrix is a diagonalizable matrix if there is an invertible matrix $P$ for which $P^{-1}MP$ is diagonal. Its columns then form a basis of eigenvectors, and the diagonal entries are their corresponding eigenvalues.

Here the given eigenvectors form an orthonormal basis. Put them into the columns of $Q$:Define $\Lambda =Q^{T}MQ$. Its entries areThis proves the claimed diagonalization of a matrix, rather than merely guessing it:The trace follows directly from this factorization:because each column of $Q$ has unit length. Trace is the sum of the eigenvalues, with multiplicity.

For the particular matrix, $(M^2{)}_{ij}={\sum }_k{M}_{ik}{M}_{kj}$. If $i=j$, every $k\ne i$ contributes one and $k=i$ contributes zero. If $i\ne j$, precisely $k=i$ and $k=j$ contribute zero; the other $n-2$ terms contribute one. ThusApplying this identity to a nonzero eigenvector givesEvery eigenvalue is therefore $n-1$ or $-1$.

If $r$ is the multiplicity of $n-1$, the trace formula for a diagonalizable matrix and the zero diagonal giveso $r=1$. Up to ordering, the required diagonal form isFor $n=1$ this means just $(0)$. For $n\ge 2$, the eigendirections are also transparent: the all-ones vector has eigenvalue $n-1$, and the $(n-1)$-dimensional subspace whose coordinate sum is zero has eigenvalue $-1$, since $(M\mathbf{x}{)}_i={\sum }_j{x}_j-x_i$. The spectrum consists of one collective mode and $n-1$ zero-sum modes.

The characteristic polynomial of the given matrix is $(t-1)(t-2{)}^2$. Solving the two eigenvalue equations givesFor example, multiplying the matrix by these three vectors gives respectively $1$, $2$ and $2$ times the original vector. The eigenvectors form a basis: the matrix with these columns has nonzero determinant $-1$. Thus an eigenbasis and its diagonal matrix are

Since $x{p}_n-p_{n+1}$ has degree of a polynomial at most $n$, expand it in the basis $p_0,\dots ,p_n$. Multiplication by $x$ is self-adjoint for this weighted inner product: $⟨xp,q⟩=⟨p,xq⟩$. For $k\le n-2$, the polynomial $x{p}_k$ has degree at most $n-1$, soOnly the $p_n$ and $p_{n-1}$ coefficients remain. Taking their inner products gives the three-term recurrenceFor the second formula, $x{p}_{n-1}=p_n+$ a polynomial of lower degree, hence $⟨x{p}_n,p_{n-1}⟩=⟨p_n,x{p}_{n-1}⟩=⟨p_n,p_n⟩$. At $n=0$, take $p_{-1}=0$, ${\beta }_0=0$ and the same formula for ${\alpha }_0$. Positivity of the weight ensures that all the squared norms in the denominators are nonzero.

To prove that trace is the unique normalized cyclic linear functional, let $E_{ij}$ be the matrix units. For $i\ne j$, $E_{ii}{E}_{ij}=E_{ij}$ whereas $E_{ij}{E}_{ii}=0$, so the assumed identity for the linear functional gives $f(E_{ij})=0$. Also, $E_{ij}{E}_{ji}=E_{ii}$ and $E_{ji}{E}_{ij}=E_{jj}$, giving $f(E_{ii})=f(E_{jj})$. Write this common value as $c$. Since the matrix units form a basis,For $n=1$ the same conclusion follows immediately from the one-dimensional basis $E_{11}$. Finally, $f(I)=nc$; if this equals $n$, then $c=1$ over either specified field. The normalization makes $f$ exactly the trace.

Group the matrix entries by columns. Each column of $BA$ is $B$ times the corresponding column of $A$. In the column-ordered basis of matrix units, the linear map $L_B$ consequently has a block diagonal matrix with $n$ copies of $B$. The determinant of a block diagonal matrix is the product of its block determinants, givingThis argument does not require $B$ to be an invertible matrix.

If $A$ is a Hermitian matrix, then $(BA{B}^{\ast }{)}^{\ast }=BA{B}^{\ast }$, so $M_B$ preserves $W$. Addition and multiplication by real scalars commute with this operation, making $T_B$ a real linear map.

A real basis of $W$ consists of $E_{ii}$ and, for $i<j$, $E_{ij}+E_{ji}$ and $i(E_{ij}-E_{ji})$. It has $n^2$ elements and is also a complex basis of all matrices. Indeed every matrix is uniquely $H+iK$ with $H,K$ Hermitian matrices, by takingReal linear independence of the displayed basis therefore implies complex linear independence. Since its images under $M_B$ are Hermitian matrices, the matrix of $M_B$ in this complex basis has exactly the same real entries as the matrix of $T_B$ in the real basis. Their determinants are equal. There is no further squaring when passing to this real subspace:This is the determinant of Hermitian congruence, including the case of singular $B$.