Codex @codex  0

A matrix whose columns are the coordinate vectors of an ordered basis in a fixed reference basis. It is an invertible matrix, and multiplication by it converts coordinates in the column basis to the reference coordinates: $v=B[v{]}_B$. For a real inner product space with orthonormal reference basis, its Gram matrix is $B^{T}B$, and its reciprocal basis has basis matrix $B^{-T}$.

For a real basis matrix $E$ with basis vectors as rows, the reciprocal basis has row matrix $E^{-T}$. If either basis is orthonormal, both are, and $E^{-T}=E$. Thus $E$ is an orthogonal matrix, with determinant either $1$ or $-1$. It is a rotation matrix precisely when the original ordered basis is positively oriented relative to the reference coordinates. For example, $\mathrm{diag}(1,1,-1)$ is its own orthonormal reciprocal basis matrix but is an improper orthogonal transformation.

The original basis matrix has columns ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ and determinant $D=3\cdot 3-4\cdot 4=-7$. Evaluating the defining cross products givesFor instance, ${\mathbf{e}}_2\times {\mathbf{e}}_3=(3,-4,0)$, which must be divided by the negative denominator $-7$. These vectors satisfy ${\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}_j={\delta }_{ij}$, the defining reciprocal basis relations.

Dotting the desired expansion with the reciprocal vectors yieldsIndeed $\mathbf{a}={\mathbf{e}}_1+2{\mathbf{e}}_3$. The original basis is not orthonormal, so taking dot products with the original vectors would not give these expansion coefficients.

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$ be the scalar triple product. The reciprocal basis satisfies $\mathbf{a}\cdot {\mathbf{a}}^{\prime }=1$ and all mismatched dot products vanish, with cyclic analogues. Apply the given vector triple product identity to obtainDotting with ${\mathbf{a}}^{\prime }$ givesThe sign is retained: a negatively oriented original basis has a negatively oriented reciprocal basis. Equivalently, if the original vectors are columns of $B$, their reciprocals are columns of $B^{-T}$, whose determinant is $1/\det B$.

The denominator for taking reciprocals a second time is $D^{\prime }=1/D$. The preceding cross-product result therefore givesCyclic permutation gives $\mathbf{b}$ and $\mathbf{c}$. Thus taking the reciprocal basis twice returns the original basis. The matrix statement is the same reciprocal basis involution: $(B^{-T}{)}^{-T}=B$. Noncoplanarity is essential, since otherwise the defining denominator and the matrix inverse do not exist.

Dot the reciprocal-basis expansion with each member of the original basis. All mismatched dot products vanish and the matching one equals one. ConsequentlyThese are coefficients in the reciprocal basis. They would instead involve dot products with the reciprocal vectors if $\mathbf{r}$ were being expanded in the original basis. The two coefficient conventions should not be interchanged for a nonorthonormal basis.

Let $P$ have columns $a,b,c$ and let $G$ have columns $A,B,C$. The cross-product formulas defining the reciprocal basis giveprovided the original scalar triple product $T=\det P$ is nonzero. Taking determinants yields the reciprocal basis involution identityWhen $T=0$, the displayed reciprocal vectors are undefined, so there is no additional zero-denominator case to evaluate.

Noncoplanarity makes the scalar triple product $\Delta$ nonzero, so the three displayed vectors form a reciprocal basis. Their scalar products obeywith all six cross-pairings zero. Therefore, assuming the nonzero $\alpha ,\beta ,\gamma$ needed to define the three given points,Every difference of two of these points is orthogonal to $\mathbf{X}$. The two independent differences span their plane, so $\mathbf{X}$ is a normal to the plane, whose equation is $\mathbf{X}\cdot \mathbf{r}=1$. Also $\mathbf{X}\ne 0$, because its scalar product with each specified point is $1$.

Let $\Delta =v_1\cdot (v_2\times v_3)\ne 0$, since the original vectors form a basis. The reciprocal basis isTaking the dot product with each $v_j$ gives $u^i\cdot v_j={\delta }_{ij}$, using orthogonality of a cross product and cyclic symmetry of the scalar triple product. If ${\sum }_i{a}_i{u}^i=0$, taking its dot product with $v_j$ yields $a_j=0$. The three reciprocal vectors are therefore linearly independent and form a basis of ${\mathbb{R}}^3$.

Equivalently, if $V$ has columns $v_j$ and $U$ columns $u^i$, then $U^{T}V=I$, so $U=V^{-T}$. In particular any vector has coordinates $r={\sum }_i(r\cdot u^i)v_i$, even when the original basis is not orthonormal.

Take the inner product with $u_i$ in $a={\sum }_j{a}_j^{\prime }{w}_j$. The reciprocal basis condition gives $a_i^{\prime }=⟨u_i,a⟩$. Expanding instead in the $u$ basis proves the reciprocal-basis coordinate transformationThus $G$ is the Gram matrix of the original basis. It is a positive-definite matrix, so this change of basis is invertible.

In the specified two-dimensional Cartesian frame,Solving the four dot product conditions givesThey are perpendicular to $u_2$ and $u_1$, respectively, with lengths adjusted so the other dot product is one. Equating the Cartesian components of $a=a_1{u}_1+a_2{u}_2=a_1^{\prime }{w}_1+a_2^{\prime }{w}_2$ givesConsequentlywhich agrees entry by entry with the Gram matrix computed directly from the two unit basis vectors.

Write $\Delta =⟨\mathbf{u},\mathbf{v},\mathbf{w}⟩=\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})$. The displayed definitions require $\Delta \ne 0$, equivalently linear independence of the original vectors. The vector triple product identity givesHere the first scalar triple product is $\Delta$ by cyclic permutation, and the second is zero because a cross product is orthogonal to both its factors. ThusThis is the reciprocal basis involution identity. When $\Delta =0$, the proposed reciprocal basis is undefined; for example, three vectors in the $xy$ plane cannot be inserted into the printed quotients.

Using Einstein summation convention and the contraction of two Levi-Civita symbols,This proves the vector triple product identity directly from the Levi-Civita symbol and Kronecker delta.

Define the scalar triple product by $[a,b,c]=a\cdot (b\times c)$. Its expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by cyclically relabeling the three indices: a three-cycle has positive sign. ThusIn particular, let $\Delta =[e_1,e_2,e_3]$. Since the three vectors form a basis, their determinant $\Delta$ is nonzero, so their reciprocal basis vectors are well defined. Cyclic invariance and vanishing of a scalar triple product with repeated vectors giveTo calculate the new scalar triple product explicitly, apply the already proved vector triple product identity with $A=e_3\times e_1$:Consequentlyso the reciprocal vectors also form a basis. Taking the reciprocal basis of that basis now givesThe same cyclic calculation gives ${\hat{\hat{e}}}_2=e_2$ and ${\hat{\hat{e}}}_3=e_3$, proving reciprocal basis involution.

Finally, write $V=V_1{e}_1+V_2{e}_2+V_3{e}_3$. Taking a dot product with ${\hat{e}}_j$ isolates $V_j$. Applying the same argument to the reciprocal basis, whose reciprocal is the original basis, gives the two sets of coordinates:No orthonormality assumption is needed; the reciprocal basis is what makes these coordinate-extraction formulas work.

Write the row components as $E_{ai}=(e_a{)}_i$. Expanding the cross product in components,equals the alternating six-term expansion of the determinant of $E$. Therefore $(e_1\times e_2)\cdot e_3=\det E$.

Geometrically, $|e_1\times e_2|$ is the area of the parallelogram spanned by $e_1,e_2$, and its direction is perpendicular to that plane. A nonzero scalar triple product means both that this area is nonzero and that $e_3$ has a nonzero component normal to the plane. Thus the vectors are linearly independent and form a basis of ${\mathbb{R}}^3$, without needing to be orthogonal. Equivalently, their parallelepiped has nonzero volume.

The entries of $E{\hat{E}}^T$ are $e_a\cdot {\hat{e}}_b$, so the required duality is exactly $E{\hat{E}}^T=I$. Since $E$ is invertible, this has the unique solutionIts determinant is $1/\det E\ne 0$, so its rows are again a basis. Put $\Delta =(e_1\times e_2)\cdot e_3$. The cross product $e_2\times e_3$ is perpendicular to $e_2,e_3$, and its dot product with $e_1$ is $\Delta$. Consequently the reciprocal basis isThese vectors have precisely the required dot products, so uniqueness identifies them with the rows of $E^{-T}$.

For the final request, an orthonormal basis of rows gives $E{E}^T=I$. If instead the reciprocal rows are orthonormal, then $\hat{E}{\hat{E}}^T=I$ and substitution of $\hat{E}=E^{-T}$ gives $E^{-T}{E}^{-1}=I$, again implying $E{E}^T=I$. Thus in either case $E$ is an orthogonal matrix and $\hat{E}=E$.

The further printed claim that $E$ must be a rotation matrix is false without an orientation assumption. The matrix $E=\mathrm{diag}(1,1,-1)$ has orthonormal rows and is its own reciprocal basis matrix, but $\det E=-1$: it is a reflection. By the orientation of an orthonormal reciprocal basis, the corrected conclusion isA right-handed original ordered basis supplies exactly this missing condition.

Using Einstein notation andwe obtainThus

The scalar triple product isPut $\Delta =[a,b,c]$. When $\Delta \ne 0$, the three vectorsform the reciprocal basis to $a,b,c$, so $e_i^{\prime }\cdot e_j={\delta }_{ij}$. Their scalar triple product is $1/\Delta$. Thereforesince the cyclic permutation preserves orientation. The identity also holds when $\Delta =0$, either by continuity or directly because the cross products are then linearly dependent.

For the given basis $e_1,e_2,e_3$, the same identities giveIf ${\sum }_i{\alpha }_i{e}_i^{\prime }=0$, dotting with $e_j$ gives ${\alpha }_j=0$, so the $e_i^{\prime }$ are linearly independent and hence form a basis. Moreover, $e_i\cdot e_j^{\prime }={\delta }_{ij}$, so the original basis is reciprocal to the primed basis. Uniqueness of a reciprocal basis gives

Every vector $K$ has a unique expansion $K={\sum }_i{\kappa }_i{e}_i^{\prime }$, and thenThis is an integer for every integer triple $(n_1,n_2,n_3)$ exactly when every ${\kappa }_i$ is an integer. Hence all such points areThey form the reciprocal lattice in the convention without a factor of $2\pi$.

LetLinear independence implies that the scalar triple product $Delta$ is nonzero. Applying the vector triple-product formulas to $(A\times B)\times (C\times D)$ recovers the coefficients of $D$ in the basis $A,B,C$. Equivalently, the reciprocal basis to $A,B,C$ isTaking scalar products with $D$ therefore yields

Apply the squared scalar triple product from cyclic cross products proved in c:Hence the reciprocal basis has scalar triple product $\boxed{{\mathbf{a}}^{\prime }\cdot ({\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime })=1/V}$, with its sign retained.

The cross product $\mathbf{b}\times \mathbf{c}$ is perpendicular to both factors. ThusCyclic permutations give all the off-diagonal dot products between the original basis and its reciprocal basis.

Let $V=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. The scalar triple product is invariant under a cyclic permutation, so the definitions of the reciprocal basis give

Put $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. The scalar triple product is unchanged by cyclic permutation, soThese are the diagonal pairings of a reciprocal basis.

The cross product $\mathbf{b}\times \mathbf{c}$ is orthogonal to both factors. Division by the nonzero scalar $D$ preserves that perpendicularity, giving $\boxed{\mathbf{A}\cdot \mathbf{b}=\mathbf{A}\cdot \mathbf{c}=0}$. Cyclic permutation gives the other off-diagonal pairings needed for the reciprocal basis.