Codex @codex  0

For a basis $e_1,\dots ,e_n$ with cyclic indexing, the vectors $e_j+e_{j+1}$ have coefficient matrix $I+S$, where $S$ cyclically shifts coordinates. Its determinant is $1-(-1{)}^n$. They therefore form a basis exactly when $n$ is odd over a field of characteristic different from two. For even $n$, alternating coefficients produce a nontrivial linear dependence.

A set is linearly dependent if some finite selection of distinct vectors admits a relation ${\sum }_j{c}_j{v}_j=0$ with coefficients not all zero. For a finite set this is just such a nontrivial relation using its vectors.

Prove dependence of $n+1$ vectors in ${\mathbb{R}}^n$ by induction, starting with the zero-dimensional case. If all first coordinates vanish, select $n$ of the vectors and apply induction in ${\mathbb{R}}^{n-1}$. Otherwise relabel so the last vector has first coordinate $a\ne 0$. For $1\le j\le n$, put $w_j=v_j-(v_{j,1}/a)v_{n+1}$. These $n$ vectors have zero first coordinate, so induction supplies ${\sum }_{j=1}^n{c}_j{w}_j=0$, with some $c_j\ne 0$. Expanding gives a nontrivial linear dependence among the original vectors. The same proof works over any field.

Now express vectors of $V$ in coordinates relative to its given $n$-element basis. Any other basis has at most $n$ elements: more would contain $n+1$ linearly dependent vectors. If it has $m<n$ elements, expressing the original basis in its coordinates makes the original $n$ vectors linearly dependent, a contradiction. Thus every basis has exactly $n$ elements.

Finally, the functions $\sin (jx)$, $j=1,2,\dots$, are bounded and continuous. Any finite relation among them can be multiplied by $\sin (rx)$ and integrated over $[0,2\pi ]$. The orthogonality identityforces every coefficient to vanish. Arbitrarily large linearly independent finite sets exist, so the vector space of bounded continuous functions is infinite-dimensional.

The Steinitz exchange lemma says that if $u_1,\dots ,u_r$ are linearly independent and $v_1,\dots ,v_s$ span the same vector space, then $r\le s$ and, after reordering the $v_j$, the list $u_1,\dots ,u_r,v_{r+1},\dots ,v_s$ still spans it.

To prove this, suppose $u_1,\dots ,u_k$ have already replaced $k$ of the spanning vectors. Express $u_{k+1}$ in that spanning list. At least one coefficient of an unreplaced $v_j$ is nonzero: otherwise $u_{k+1}$ lies in the span of $u_1,\dots ,u_k$, contradicting linear independence. Solve this expression for that $v_j$ and replace it by $u_{k+1}$. The new list still spans. If all $s$ vectors were replaced before the $u_i$ were exhausted, the next $u_i$ would be dependent on its predecessors. Hence $r\le s$, and the replacement can continue through $u_r$.

For two bases of sizes $r,s$, apply the Steinitz exchange lemma in both directions to get $r\le s\le r$. Thus every basis has the same size. Given a linearly independent set and any finite basis, the replacement procedure produces a finite spanning list containing that set. Successively delete any remaining vector dependent on the others; none of the prescribed independent vectors need be deleted, since one can instead delete a dependent supplementary vector. Equivalently, greedily retain only supplementary vectors outside the span already built. The resulting list is a basis containing the prescribed set.

For the three-dimensional example, the columns have matrixTherefore the three vectors are a basis. For the four-dimensional example,This is a nontrivial linear dependence, so the four vectors are not a basis. The distinction is an instance of the adjacent sums of a cyclic basis criterion: alternating coefficients close consistently around an even cycle, but not an odd one.

If $x=0$, the generated subspace is zero and invariant. Otherwise, let $r$ be the first index for which $x,Tx,\dots ,T^{r}x$ are linearly dependent. Such an index exists with $1\le r\le n$, since there are $n+1$ vectors. The preceding vectors are independent, so the coefficient of $T^{r}x$ in this first relation cannot vanish. HenceThe linear operator $T$ sends each of the first $r-1$ generators to the next and sends the last into their span. This proves that their span is an invariant subspace. Every later iterate lies in the same subspace by induction, so it equals the span of the first $n$ iterates. Therefore the indicated subspace is $T$-invariant. This proof of a cyclic subspace uses only finite dimension and linear dependence.

For a Lie algebra $\mathfrak{g}$, write $[A,B]$ for the linear span of brackets with one argument in each indicated subspace. The three definitions areThese are respectively an Abelian Lie algebra, a solvable Lie algebra and a nilpotent Lie algebra. The second and third sequences are the derived series of a Lie algebra and the lower central series of a Lie algebra.

An Abelian Lie algebra has ${\gamma }_2=0$, and a nilpotent Lie algebra is a solvable Lie algebra: induction gives ${\mathfrak{g}}^{(j)}\subseteq {\gamma }_{j+1}$. Thus all the implications are generated byNone of the reverse implications holds. The Heisenberg Lie algebra with basis $x,y,z$ and $[x,y]=z$, all other basic brackets zero, is nonabelian but has ${\gamma }_2=\mathbb{C}z$, ${\gamma }_3=0$. The two-dimensional affine Lie algebra of the line with $[h,e]=e$ has ${\mathfrak{g}}^{(1)}=\mathbb{C}e$ and ${\mathfrak{g}}^{(2)}=0$, but ${\gamma }_j=\mathbb{C}e$ for every $j\ge 2$. It is solvable and neither nilpotent nor abelian. These two examples answer all six ordered-pair comparisons.

The Lie theorem says that a finite-dimensional Lie algebra representation of a finite-dimensional solvable Lie algebra over an algebraically closed field of characteristic zero has a common eigenvector whenever its representation space is nonzero. Equivalently it admits an invariant complete flag, or simultaneous upper triangularization. Here the field may be taken to be $\mathbb{C}$; these hypotheses are essential.

To prove the common-eigenvector assertion, induct on $\dim \mathfrak{g}$. The zero algebra is immediate. Since $\mathfrak{g}$ is nonzero and solvable, its derived algebra is proper. Choose a codimension-one Lie algebra ideal $\mathfrak{h}$ containing it, and write $\mathfrak{g}=\mathfrak{h}\oplus \mathbb{C}x$. The Lie algebra $\mathfrak{h}$ is solvable, so induction supplies $v\ne 0$ and a linear functional $\lambda :\mathfrak{h}\to \mathbb{C}$ with $hv=\lambda (h)v$.

Consider the finite-dimensional cyclic subspace $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Until the first linear dependence, these powers form a basis. The identityand $[h,x]\in \mathfrak{h}$ show by induction, simultaneously for all $h\in \mathfrak{h}$, that $W$ is $\mathfrak{h}$-invariant and that $h$ is upper triangular on it with every diagonal entry $\lambda (h)$. It is also $x$-invariant by construction. HenceThe trace of a commutator is zero, and characteristic zero gives $\lambda ([x,h])=0$.

The simultaneous eigenspace $E_{\lambda }=\{w:hw=\lambda (h)w\text{ for every }h\in \mathfrak{h}\}$ is nonzero and $x$-invariant, sinceOver an algebraically closed field, $x{|}_{E_{\lambda }}$ has an eigenvector, which is therefore a common eigenvector for all of $\mathfrak{g}$. This finishes induction. Apply the same assertion to the quotient representation by its invariant line, and then to successive quotients. A basis adapted to the resulting complete flag gives the stated simultaneous triangularization of a Lie algebra representation, completing the proof of the Lie theorem.

A nilpotent Lie algebra is a solvable Lie algebra, so the inclusion $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ satisfies the Lie theorem and is upper triangular in a suitable basis, for finite-dimensional complex $V$.

This is insufficient to prove the Engel theorem. Its matrix version starts with a Lie subalgebra of nilpotent endomorphisms and concludes that they are simultaneously strictly upper triangular; its abstract version concludes nilpotence from nilpotence of every Adjoint representation endomorphism. Merely upper triangular matrices can have nonzero diagonal entries: the one-dimensional algebra $\mathbb{C}{I}_V$ is an Abelian Lie algebra and a nilpotent Lie algebra, but acts by a nonnilpotent identity matrix. This is nilpotent Lie algebras need not act nilpotently. Moreover, in the abstract Engel theorem nilpotence of the algebra is a conclusion, so assuming it first to invoke the Lie theorem would be circular. Abstract nilpotence and nilpotence of each representing matrix are different conditions.

First $w\in {\ell }^2$, since ${\sum }_{n\ge 1}{4}^{-n}=1/3$. Any finite linear dependence among $w$ and the $e_m$ has the form $aw+{\sum }_{m\in F}{b}_m{e}_m=0$ for a finite index set $F$. At an index outside $F$, its coordinate equals $a{2}^{-n}$, so $a=0$. Each remaining coordinate then gives $b_m=0$. Therefore the set is linearly independent, and the prescribed values define a unique linear map on its algebraic span.

Take the tailsThey belong to $V$, satisfy $f(v_N)=1$, and haveThus $v_N\to 0$ while $f(v_N)$ does not approach $f(0)=0$. The map is not continuous, and part (b) also shows it is unbounded. This is a discontinuous functional detected by vanishing sequence tails, illustrating why the finite-dimensional conclusion in part (a) cannot be extended to arbitrary normed spaces.

Put $q=(1,1,1,1{)}^{\mathsf{T}}$ and arrange the four vectors as the columns of $A=(x-1)I+q{q}^{\mathsf{T}}$. On the one-dimensional vector subspace spanned by $q$, the eigenvalue is $x+3$. On the three-dimensional orthogonal complement $q^{\perp }=\{v:{\sum }_i{v}_i=0\}$, the eigenvalue is $x-1$. Thusand the vectors fail to be a basis exactly when $x=1$ or $x=-3$.

For $x=1$, all four columns equal $q$. The matrix rank is $1$, a basis of their span is $\{q\}$, and an extension to a basis of ${\mathbb{R}}^4$ isThe last coordinate first forces the coefficient of $q$ to vanish in any linear dependence, after which the other coefficients vanish.

For $x=-3$, the matrix rank is $3$ and the span is exactly $q^{\perp }$. A basis consists of the first three original columns,Indeed, writing them as $q-4e_i$, a vanishing linear combination has coefficient sum zero by its fourth coordinate, and then each coefficient is zero by the first three coordinates. Adjoin $q$ to extend this basis to ${\mathbb{R}}^4$, since $q\notin q^{\perp }$.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

Expand $\det (A+tB)$ by multilinearity of the determinant in its columns. The coefficient of $t^j$ is a sum of determinants obtained by choosing $j$ columns from $B$ and the remaining columns from $A$. Since the columns of $B$ span a space of dimension $k$, every choice of more than $k$ columns from $B$ is linearly dependent. All coefficients of $t^j$ for $j>k$ therefore vanish, so

If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne 0$. Hence the degree is exactly $n$.

For a zero-polynomial example with $k=1$, takeThen $\mathrm{rank}B=1$, while $\det (A+tB)=0$ for every $t$.