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Label the sides by $L,R,B,T$, and parametrize the two vertical sides by $s=y\in [-1,1]$, the horizontal sides by $s=x\in [-1,1]$. Their outward normal derivatives are $q_L=-u_x(-1,s)$, $q_R=u_x(1,s)$, $q_B=-u_y(s,-1)$ and $q_T=u_y(s,1)$. Let $f_j$ denote the prescribed side values. Assume compatible, sufficiently regular boundary traces; the square's corners have zero arclength measure and do not require separate normal values.

For sine collocation of square modified Helmholtz global relations, use Legendre polynomials for the known Dirichlet boundary condition and a Fourier sine series for the unknown normal derivative:The known coefficients are $d_{j\ell }=(2\ell +1){\int }_{-1}^1{f}_j{P}_{\ell }ds/2$. The sine functions have ${\int }_{-1}^1{\varphi }_m{\varphi }_{n}ds={\delta }_{mn}$ and form a complete basis in $L^2(-1,1)$. Choosing them for the normal derivative does not impose zero flux at a corner: the expansion is an $L^2$ representation, and endpoint values are not determined by it. If preserving corner values of the approximated Dirichlet trace is necessary, subtract its endpoint-interpolating line before polynomial approximation, and add that line back. All subsequent known-data integrals can alternatively be evaluated with the exact $f_j$.

Define the entire function basis transformsAt $\tau =\pm i{p}_m$ the quotient is evaluated by its removable limit or by the defining integral. For example, ${\mathsf{P}}_0=2\sinh \tau /\tau$, and polynomial expansion of $P_{\ell }$ expresses every ${\mathsf{P}}_{\ell }$ in derivatives of ${\mathsf{P}}_0$. PutFor compactness write $A=A(\lambda )$ and $B=B(\lambda )$. Substituting these expansions into the first global relation for a linear boundary value problem givesThe signs are fixed by the outward normal vectors, rather than by a choice of traversal direction. The second approximate global relation for a linear boundary value problem replaces $A$ by $-A$, leaving $B$ unchanged:Here $\simeq 0$ records omitted boundary-expansion tails. In a finite spectral method the selected equations are set equal to zero to solve for the $4N$ unknown real coefficients $c_{jm}$. For real data the two families obey the same complex conjugation relation as their exact counterparts.

Use the Einstein summation convention. Substitute the given basis transformation into the expansion of the vector:By linear independence of the basis vectors, $v_j=R_{ij}{v}_i^{\prime }$. In matrix notation $v=R^T{v}^{\prime }$. A rotation matrix is orthogonal, so multiplication by $R$ givesThe placement of the transpose follows from the specified convention $e_i^{\prime }=R_{ij}{e}_j$. It must not be guessed from a different convention for an active rotation of a vector.

The quotient theorem for Cartesian tensors here says: if contracting $T_{ij}$ with every vector $v_j$ gives the components $w_i=T_{ij}{v}_j$ of a vector in every right-handed orthonormal basis, then $T$ obeys the second-order tensor transformation law. Conversely that transformation law guarantees the contraction is a vector. One may equivalently test that $u_i{T}_{ij}{v}_j$ is a scalar for every pair of vectors $u,v$.

By part (a), $v^{\prime }=Rv$ and $w^{\prime }=Rw$. The contraction property in the new basis saysSince this holds for every test vector, $T^{\prime }R=RT$. Using $R^{-1}=R^T$ givesFor the converse, $T^{\prime }{v}^{\prime }=RT{R}^{T}Rv=RTv=Rw$, exactly the vector law. For the equivalent scalar formulation, invariance of $u^{T}Tv$ for every $u,v$ gives $u^T{R}^T{T}^{\prime }Rv=u^{T}Tv$, forcing the same matrix identity.

The quantifier “every” matters: an arbitrary nonzero matrix can annihilate one fixed vector, so a single successful contraction cannot establish tensoriality. The stated bases test proper rotations; transformation under orthogonal reflections would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the matrix rank of a particular component matrix.

If $S$ spans and has $n$ elements but is linearly dependent, a nontrivial linear combination equal to zero lets us remove one member without changing the span. This produces $n-1$ spanning vectors, whereas the standard basis contains $n$ linearly independent vectors, contradicting the Steinitz exchange lemma. Therefore $\boxed{S\text{ is linearly independent}}$. Together the three deductions establish that any two of the three properties imply the third. For $n=0$, the only independent or zero-element family is the empty basis, and the same conclusions hold.

Relative to the given basis $e_1,e_2$, the proposed vectors are the columns ofThe determinant is nonzero exactly when the columns form a basis. Thus $\boxed{\lambda \ne \pm 1}$. At $\lambda =1$ the vectors coincide, and at $\lambda =-1$ they are negatives of one another.

If $S$ is linearly independent and has $n$ elements, apply the Steinitz exchange lemma with $S$ as the independent family and the standard basis as the spanning family. All $n$ members of the standard basis are replaced, leaving $S$ itself spanning. Thus $\boxed{\mathrm{span}S={\mathbb{R}}^n}$, so $S$ is a basis.

True. Use the lifted basis $u_1,\dots ,u_r$ above and put $W=\mathrm{span}\{u_1,\dots ,u_r\}$. If $\sum c_i{u}_i=0$, applying $\alpha$ gives $\sum c_i{v}_i=0$, so all $c_i=0$. Thus the $u_i$ form a basis of $W$, and $\alpha {|}_W$ takes this basis to the basis of $V$. It is a linear isomorphism.

In fact the right inverse gives the stronger splittingFor every $u\in U$, write $u=(u-\beta \alpha u)+\beta \alpha u$; the first term is in the kernel and the second in $W$. Their intersection is zero because $\alpha {|}_W$ is injective. This is a direct sum decomposition.

True. Choose a basis $v_1,\dots ,v_r$ of $V$. By surjective function, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the linear map $\beta$ by $\beta v_i=u_i$ and extend by linearity. Then $\alpha \beta v_i=v_i$ on every basis vector, and thereforeThis constructs a right inverse and does not require $\alpha$ to be injective. If $V$ is the zero vector space, the unique zero map is the required right inverse.

For a linear map $\alpha :U\to V$ with $U$ a finite-dimensional vector space, the rank-nullity theorem isChoose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_r$ of $U$. The vectors $\alpha (v_j)$ span the image of a linear map, because the kernel basis contributes zero. If ${\sum }_j{c}_j\alpha (v_j)=0$, then ${\sum }_j{c}_j{v}_j$ belongs to the kernel and is a combination of the $u_i$; independence of the extended basis makes every $c_j=0$. Thus the image basis has $r$ elements and $\dim U=k+r$, proving the theorem.

For a rank-two map on ${\mathbb{R}}^3$, the direct sum occurs for $\alpha (x,y,z)=(x,y,0)$: its kernel is $\mathbb{R}{e}_3$ and its image is $\mathrm{span}(e_1,e_2)$. It fails for $\beta (x,y,z)=(y,z,0)$: the kernel is $\mathbb{R}{e}_1$, contained in the two-dimensional image $\mathrm{span}(e_1,e_2)$. The rank-nullity theorem fixes the sum of the dimensions, but does not imply zero intersection.

Associate to $f$ the symmetric bilinear form $B(u,v)=\frac{1}{2}(f(u+v)-f(u)-f(v))$. If $f$ is identically zero, every basis is diagonal. Otherwise choose $v$ with $f(v)\ne 0$. Every $w$ has the decompositionwhose second term lies in $v^{\perp }$. Thus $V=\mathbb{R}v\oplus v^{\perp }$, with no cross term in $f$. Induction on dimension gives a diagonal basis. Rescaling its nonzero entries reduces the form towith $k$ additional zero coordinates. Define the rank of a quadratic form $r=p+q$ and its signature $s=p-q$. Rank is intrinsic, since it is the rank of $v\mapsto B(v,\cdot )$, equivalently $n-\dim \mathrm{rad}B$. Also $p$ is the greatest dimension of a subspace on which the restriction of $f$ is a positive-definite quadratic form: any such subspace intersects the negative-plus-zero coordinate space only at zero, so its dimension is at most $p$, while the positive coordinate space attains $p$. The corresponding negative statement identifies $q$. This proves the basis independence of $r$ and $s$, the Sylvester's law of inertia conclusion.

For a totally isotropic subspace $W$ on which $f$ vanishes, polarization identity also makes $B$ vanish on $W\times W$. Quotient by the radical of a bilinear form. In the resulting nondegenerate form, projection of an isotropic subspace to either the positive or negative coordinate space is injective, so its dimension is at most $\min (p,q)$. Consequently $\dim W\le k+\min (p,q)$. The radical of a bilinear form together with the vectors $e_i^{+}+e_i^{-}$, $1\le i\le \min (p,q)$, attains this bound. The maximum dimension of a totally isotropic subspace is thereforeThe criterion for membership in a diagonal basis follows similarly: a nonzero vector can belong to a diagonal basis if it has nonzero quadratic value, by the same orthogonal splitting, or if it lies in the radical of a bilinear form, by choosing a radical basis containing it. An isotropic vector in a diagonal basis must be orthogonal to every basis vector, and hence must lie in that radical. For the specified form, its radical is the $z$-axis, giving exactlyThe zero vector never belongs to a basis.