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Every $v\in V$ is a finitely supported sequence, so for a finite set of indices $S$,Thus $E$ is a spanning set. If $F$ contained $v\notin E$, the displayed expression would give a nontrivial relationamong the distinct vectors of the finite subset $\{v\}\cup \{e^{(i)}:i\in S\}$ of $F$: the coefficient of $v$ is $1$. This contradicts the linear independence of $F$. It also covers $v=0$, whose singleton is not linearly independent. Hence $F=E$. In particular, $E$ is both a basis and a maximal linearly independent subset of $V$.

Take any finite selection of distinct indices $i_1,\dots ,i_r$ and a relationThe coordinate at $i_j$ on the left is exactly ${\alpha }_j$, while the zero vector has coordinate zero. Hence ${\alpha }_j=0$ for each $j$. Every finite subset of $E$ is therefore linearly independent, which is the definition of linear independence for an infinite set. Thus $E$ is linearly independent in the vector space of finitely supported sequences.

Use the vector triple product identitySince $\mathbf{n}$ is a unit vector, putThus $\mathbf{w}$ is the component of $\mathbf{v}$ in the orthogonal complement of $\mathbf{n}$, and $\mathbf{n}\cdot \mathbf{w}=0$. A cross product is orthogonal to both its factors, so $\mathbf{n}\cdot \mathbf{q}=0$ and $\mathbf{v}\cdot \mathbf{q}=0$. ConsequentlyAll three pairs are orthogonal. This remains true when $\mathbf{v}$ is parallel to $\mathbf{n}$: then $\mathbf{q}$ and $\mathbf{w}$ are zero vectors. Pairwise orthogonality does not assert linear independence in this case.

Three vectors form a basis of three-dimensional space exactly when their scalar triple product is nonzero:It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of $\mathbf{y}$ with $\mathbf{v}\times \mathbf{w}$. The terms containing $\mathbf{v}$ and $\mathbf{w}$ vanish, leaving $\mathbf{y}\cdot (\mathbf{v}\times \mathbf{w})=\alpha \Delta$. Cycling the same argument yieldsThe three cross-product numerators use the reciprocal basis.

Let $\mathbf{d}={\mathbf{a}}_2-{\mathbf{a}}_1$ and $\mathbf{w}={\mathbf{v}}_1\times {\mathbf{v}}_2$. The assumed linear independence gives $\mathbf{w}\ne 0$. Every displacement joining one line to the other isIts dot product with $\mathbf{w}$ is always $\mathbf{d}\cdot \mathbf{w}$. The component in that direction is therefore fixed, so orthogonal decomposition gives $\Vert \mathbf{s}\Vert \ge |\mathbf{d}\cdot \mathbf{w}|/\Vert \mathbf{w}\Vert$.

The two linearly independent vectors ${\mathbf{v}}_1,{\mathbf{v}}_2$ span the plane perpendicular to $\mathbf{w}$. Suitable ${\lambda }_1,{\lambda }_2$ cancel the component of $\mathbf{d}$ in that plane, attaining the lower bound. Thus the distance between skew lines formula, also valid when the lines intersect, gives

Classification and existence. For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$, an ordinary point $x_0$ is one where $p,q$ are analytic. There are two independent analytic power series solutions, with arbitrary initial values $y(x_0),y^{\prime }(x_0)$. A regular singular point is a singular point where $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ are analytic. The Frobenius method then constructs solutions beginning with $(x-x_0{)}^{\sigma }$, where $\sigma$ solves the indicial equation. At least one Frobenius solution exists; two such solutions exist without logarithms when the two exponents differ by a noninteger. Repeated exponents or an integer difference can require a logarithmic second solution.

For this equation,Both are analytic at zero, while $p,q$ themselves have poles there. Thus zero is a regular singular point.

Indicial equation and recurrence. Multiply by $4x$ and insert $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$:The lowest power $x^{\sigma -1}$ has coefficient $[4\sigma (\sigma -1)+2\sigma ]a_0$, givingThe coefficient of $x^{n+\sigma }$ isFor either exponent the common factor is nonzero for $n\ge 0$, so the recurrence relation reduces toNormalizing each leading coefficient to one, the two Frobenius solutions areMore explicitly, $y_2=\sqrt{x}{\sum }_{n\ge 0}{2}^{n}n!x^n/(2n+1)!$. The distinct leading powers establish linear independence, so $y=C_1{y}_1+C_2{y}_2$ on the positive real side of the singular point. The displayed exponential function is the requested closed-form solution; substitution gives $y_1^{\prime }=y_1/2,y_1^{\prime \prime }=y_1/4$, which makes the differential expression zero.

Convergence. For either exponent the ratio of successive terms in the coefficient power series isThus both coefficient series have $\boxed{R=\infty }$ by the ratio test. This does not remove the $\sqrt{x}$ branch point of $y_2$: its analytic coefficient factor is entire, while the full Frobenius solution requires a square-root branch away from zero in the complex plane. The differential equation has no other finite singular points to limit continuation of these coefficient series.

Divide the Chebyshev differential equation by $1-x^2$. Its coefficients $-x/(1-x^2)$ and $m^2/(1-x^2)$ are analytic at zero, so $\boxed{x=0\text{ is an ordinary point}.}$ The finite singular points of a differential equation are $x=\pm 1$.

Insert $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$, $a_0\ne 0$. The coefficient of the lowest power $x^{\sigma -2}$ is $\sigma (\sigma -1)a_0$, hence $\sigma =0$ or $1$. The next power gives $\sigma (\sigma +1)a_1=0$, and the remaining powers yieldFor $\sigma =0$, $a_1$ is free andChoosing $a_1=0$ selects the even power series, with $y(0)=a_0,y^{\prime }(0)=0$. For $\sigma =1$, the next-power equation forces $a_1=0$ andThis selects the odd power series, with $y(0)=0,y^{\prime }(0)=a_0$. Their different initial values prove linear independence. Although the Frobenius method produces two possible leading exponents, zero is an ordinary point, and a general $\sigma =0$ series already includes both initial constants.

Use reduction of order by setting $y=xv$. Then $y^{\prime }=v+x{v}^{\prime }$ and $y^{\prime \prime }=2v^{\prime }+x{v}^{\prime \prime }$, so the differential expression simplifies toIn the homogeneous case $x{v}^{\prime }$ is constant. Hence $v=c_1+c_2\ln x$, and a basis of the solution space isTheir Wronskian is $x\ne 0$ on $[1,e]$, proving linear independence. No nonzero homogeneous solution satisfies both endpoint conditions: $y(1)=0$ sets $c_1=0$, and $y(e)=0$ then sets $c_2=0$.

The product rule gives ${\psi }^{\prime }=e^{-x^2/2}(y^{\prime }-xy)$ and ${\psi }^{\prime \prime }=e^{-x^2/2}[y^{\prime \prime }-2x{y}^{\prime }+(x^2-1)y]$. Substituting in the Schrödinger equation cancels the $x^2$ and constant terms, leaving the Hermite differential equationFor $y={\sum }_{n\ge 0}{a}_n{x}^n$, matching coefficients givesThe power series solutions of the Hermite differential equation split into independent even and odd series. With initial data $(y(0),y^{\prime }(0))=(1,0)$ and $(0,1)$, respectively, their first three terms areEach nonterminating parity series has infinite radius of convergence: the ratio of successive terms of that parity has magnitude asymptotic to $2|x{|}^2/n$. Their initial data prove linear independence. If a coefficient shown vanishes at a special parameter, the recurrence supplies the terminating series instead.

A nonzero polynomial solution of degree $N$ requires $\lambda =N$, by its highest-degree term, and the recurrence then terminates the matching parity. Thus all possible values are $\lambda =0,1,2,\dots$. Up to nonzero scalar multiples, the degree-at-most-three solutions are the Hermite polynomials

The Wronskian isSuppose constants $A,B$ satisfy $Af+Bg=0$ throughout the interval. Differentiating gives $A{f}^{\prime }+B{g}^{\prime }=0$. At $x_0$, these form the homogeneous linear systemIts determinant is nonzero by hypothesis, so $A=B=0$. This is exactly linear independence of the two functions on the interval. No differential equation is needed for this implication; the converse for arbitrary differentiable functions would require additional hypotheses.

Use a Frobenius solution $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, with $a_0\ne 0$. The lowest power $x^{\sigma -1}$ has coefficient $2\sigma (2\sigma -1)a_0$, giving the indicial equationFor $n\ge 1$, matching the coefficient of $x^{n+\sigma -1}$ givesFor both candidate exponents the cancelled factor is nonzero in this range, so the recurrence isWith $a_0=1$, the exponent-zero solution isFor $\sigma =1/2$, the coefficients are $a_n=1/(2n+1)!!=2^{n}n!/(2n+1)!$, givingThe ratio test gives infinite radius of convergence for both series factors, since the successive ratios in magnitude are $|x|/(2n+2)$ and $|x|/(2n+3)$. The second full solution still needs a square-root branch in the complex plane; on $x>0$ it is a real solution. The distinct leading behaviors $1$ and $\sqrt{x}$ prove linear independence.

For an independent closed-form check on the structure, put $z=x/2$. The equation becomes $z{Y}^{\prime \prime }+(1/2-z)Y^{\prime }-(1/2)Y=0$, a Kummer differential equation with equal parameters. Its exponential reduction of a confluent hypergeometric equation with equal parameters explains the elementary first solution and the branched companion.

The homogeneous equation is a Cauchy-Euler equation. With the power-law ansatz $y=x^r$ on $x>0$, its indicial equation isThus $r=(1-\alpha )/2$ is a repeated indicial root. The two independent solutions are $x^r$ and $x^r\log x$, givingTheir Wronskian, with $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, is $x^{2r-1}=x^{-\alpha }\ne 0$, verifying linear independence.

Expand the quadratic form of this Gram matrix:For $\mathbf{c}\ne 0$, linear independence of the real vectors implies ${\sum }_i{c}_i{\mathbf{x}}_i\ne 0$, so its squared Euclidean norm is strictly positive. Thusproving the positive definiteness of a Gram matrix in the independent-vector case.

For an exponential trial $y=e^{rx}$, the characteristic equation isIts distinct complex roots are $r=-a\pm ib$. Thus the general real solution of the homogeneous second-order linear differential equation isThe two displayed solutions have nonzero Wronskian $b{e}^{-2ax}$ and hence linear independence. The same formula permits complex solutions when $C_1,C_2$ are complex.

At zero the indicial equation is $\nu (\nu -1)+\nu ={\nu }^2=0$, with repeated indicial root zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:Substitute $y_1={\sum }_{n\ge 0}{b}_n{x}^n$ and equate coefficients. With $b_j=0$ for negative indices, the linear recurrence relation isEvery leading coefficient on the left is nonzero. Hence $b_0\ne 0$ is free and all later coefficients are determined; the Frobenius method guarantees local convergence up to the next singularity. In particular,For a second solution, use reduction of order. Writing $y_2=y_{1}v$ gives $W=y_1^2{v}^{\prime }$, so part (c) yieldsThe additive constant in the integral only adds a multiple of $y_1$. As $x\downarrow 0$ within $(0,m)$, its integrand isThereforeThe nonzero Wronskian proves linear independence, and the logarithm is unavoidable for this second local solution.

The homogeneous linear differential equation can be written in divergence form:One integration gives $u^{\prime }=C{x}^{-n-1}$. Therefore the two cases areIn each case the displayed functions have linear independence throughout the interval. Imposing both homogeneous Dirichlet boundary conditions forces $A=B=0$: for $n\ne 0$, $2^n\ne 1$, and for $n=0$, $\log 2\ne 0$. This also establishes uniqueness for the inhomogeneous boundary value problem.

Set $t=y-1$. Then $y^2-1=t(t+2)$, and a Frobenius solution has the formIn the differential equation, the lowest power is $t^{\sigma -1}$, whose coefficient isThe indicial equation and its indicial roots are consequentlyTheir difference is not an integer, so the Frobenius method produces two independent series without a logarithmic term:Choose a branch of $(y-1{)}^{1/2}$, or take $y>1$ for real solutions. A general local solution is $C_1{x}_1+C_2{x}_2$ by linear independence and the superposition principle. The analytic factors converge at least for $|y-1|<2$, up to the next singular point at $y=-1$.

For each fixed finite tuple length $n$, use the finite conjunctionIt asserts that no nontrivial scalar combination vanishes, exactly linear independence. It is an $L$-formula because the field is finite and every scalar multiplication is named. Scalar coefficients are indices of function symbols, not variables ranging over a separate field sort. There is no single finite formula with arbitrarily many free variables; the assertion is a formula for each prescribed finite $n$.

Embed the two off-diagonal Pauli matrices in each of the $(u,d)$, $(u,s)$ and $(d,s)$ two-dimensional subspaces. Together with two independent embedded diagonal Pauli matrices, this produces the Gell-Mann matrices:To construct the last matrix explicitly, the $(u,s)$ diagonal embedding is $D_{us}=\mathrm{diag}(1,0,-1)$ and the $(d,s)$ embedding is $D_{ds}=\mathrm{diag}(0,1,-1)$. Their difference is ${\lambda }_3$, and their sum divided by $\sqrt{3}$ is ${\lambda }_8$. There are only two independent diagonal generators.

Each matrix is Hermitian and traceless. Their inner products satisfy $\mathrm{tr}({\lambda }_a{\lambda }_b)=2{\delta }_{ab}$, proving linear independence. With $T_a={\lambda }_a/2$, arbitrary transformations near the identity have $U=\exp (i{\theta }_a{T}_a)$. The eight matrices span the generators; they are not a vector-space basis of group elements themselves.

There are $p^n$ choices for the translation. Choose the columns of an invertible matrix successively: the first is any nonzero column, the second avoids its span, and at stage $j+1$ one avoids a $j$-dimensional span of size $p^j$. ConsequentlyThe successive linear independence conditions are exactly the criterion that the columns form a basis.