Codex @codex  0

Keep the same origin and the same vector $p=(2,1{)}^T$, but use the nonperpendicular basis vectors $b_1=(1,0{)}^T$ and $b_2=(1,1{)}^T$. They form a basis because the column matrix $B$ has nonzero determinant:The reconstruction is $p=b_1+b_2$. Its coordinates are read along lines parallel to the other axis, producing a parallelogram. These lines need not be perpendicular to the axis being measured. In contrast, the dot products are $p\cdot b_1=2$ and $p\cdot b_2=3$; treating those as coordinates would reconstruct $2b_1+3b_2=(5,3{)}^T$, a different vector.

The dual basis supplies the right measuring rules for any basis.^[ref] Here these coordinate functionals are $b_1^{\ast }(x,y)=x-y$ and $b_2^{\ast }(x,y)=y$, the rows of $B^{-1}$. They give $b_1^{\ast }(p)=b_2^{\ast }(p)=1$. Equivalently, their representing vectors $d_1=(1,-1{)}^T$, $d_2=(0,1{)}^T$ satisfy $d_i\cdot b_j={\delta }_{ij}$ and extract the coordinates as $d_i\cdot p$. These are Riesz dual basis in an inner product space representatives, rather than the original basis vectors.

The Gram matrix records how the nonperpendicular directions interact:For any real basis represented by a square invertible matrix $B$, writing $p=Bc$ gives these identities immediately. One can recover the coordinates either as $c=B^{-1}p$ or by solving $Gc=B^{T}p$. In this example, if $c=(a,b{)}^T$, the squared norm is $a^2+2ab+2b^2$, not $a^2+b^2$. Oblique coordinates remain useful when natural directions are not perpendicular, but the Gram matrix must accompany them when computing lengths or dot products.

A coordinate functional extracts one coefficient in a specified finite basis. It is an element of that basis's dual basis, rather than a coordinate rule independent of the basis. For the standard basis of $F^n$, the $i$th coordinate functional sends $(a_1,\dots ,a_n)$ to $a_i$.

A coordinate of a vector in a chosen basis is one entry of its coordinate vector. The corresponding coordinate functional in the dual basis extracts it. For an orthonormal basis, it can instead be read as an inner product with the basis vector, as in orthonormal eigenbasis. For a nonorthogonal basis, ordinary dot products need not give the coordinates.

Let $F$ be the common field. The dual map is $S^{\prime }(f)=f\circ S$ for $f\in W^{\prime }$, where $W^{\prime }$ and $V^{\prime }$ are the dual spaces.

First assume $S^{\prime }$ is injective. If $S$ were not surjective, its image of a linear map $\mathrm{im}S$ would be a proper vector subspace of $W$. Choose a basis $w_1,\dots ,w_r$ of $\mathrm{im}S$ and use basis extension to obtain a basis $w_1,\dots ,w_m$ of $W$, with $r<m$. There is a linear functional $f\in W^{\prime }$ withIt is nonzero and vanishes on $\mathrm{im}S$, so $S^{\prime }(f)=f\circ S=0$, contradicting injectivity. Equivalently, the annihilator of a vector subspace $\mathrm{im}S$ would give a nonzero kernel for $S^{\prime }$. Therefore

Now assume $S^{\prime }$ is surjective. If $0\ne v\in \ker S$, basis extension gives a linear functional $h\in V^{\prime }$ with $h(v)=1$: include $v$ in a basis and take its coordinate functional. By surjectivity, $h=S^{\prime }(f)$ for some $f\in W^{\prime }$. Buta contradiction. Hence $\ker S=0$, which proves

Put $b(v,w)=\phi (w)(v)$. Each $\phi (w)$ belongs to the dual space $V^{\prime }$, so it is a linear functional. For real numbers $a,b_0$ and vectors $v_1,v_2,w$, this givesThe linearity of $\phi$ gives linearity in the second argument as well:Thus $b$ is a bilinear form.

Its right radical of a bilinear form is exactly $\ker \phi$, becauseIf $b$ is a nondegenerate bilinear form, this kernel is zero. Since $V$ is a finite-dimensional vector space and $\dim V^{\prime }=\dim V$, the rank-nullity theorem makes $\phi$ surjective.

Conversely, if $\phi$ is surjective, the same dimension equality makes it injective, so the right radical is zero. If $v$ lies in the left radical, then $f(v)=0$ for every $f\in V^{\prime }$, because every $f$ is some $\phi (w)$. A nonzero $v$ can be included in a basis and detected by its coordinate functional, so this forces $v=0$. Both radicals vanish, proving

Writeon $V=W\times W^{\prime }$, where $W^{\prime }$ is the dual space. The dual pairing is linear separately in the vector and the linear functional. For example, for real numbers $a,b$,Thus $\omega$ is a bilinear form. It is also an alternating bilinear form, since

To prove that this is a nondegenerate bilinear form, suppose $(v,f)$ pairs to zero with every $(w,g)$. Setting $g=0$ gives $f(w)=0$ for every $w\in W$, hence $f=0$. Setting $w=0$ gives $g(v)=0$ for every $g\in W^{\prime }$. If $v\ne 0$, basis extension supplies a coordinate functional with $g(v)=1$, a contradiction. Thus $v=0$. The left radical is zero, and the displayed change of sign implies the right radical is zero as well. Hence $\omega$ is a nondegenerate bilinear form, making $V$ a symplectic vector space.

For the required example take the coordinate subspaceUsing the definition of the bilinear orthogonal complement in part (b),ThereforeThis is a Lagrangian subspace. Its dimension is $\dim W$, exactly half of $\dim V=2\dim W$, so part (b)(i) is satisfied.

There is no contradiction with part (b)(ii): that part asks whether a direct sum is guaranteed, and the answer is already no. Moreover, when $W\ne 0$, the present alternating bilinear form is not a symmetric bilinear form. Choose $v\ne 0$ and $f\in W^{\prime }$ with $f(v)=1$; thenIt therefore does not satisfy the symmetry hypothesis of part (b)(ii). If $W=0$, then $V=U=0$, and both $U=U^{\perp }$ and $V=U\oplus U^{\perp }$ hold trivially.

If $(e_1,\dots ,e_m)$ and $(f_1,\dots ,f_n)$ are bases of finite-dimensional vector spaces $V,W$ over $F$, then the $mn$ products $e_i\otimes f_j$ form a basis of $V{\otimes }_{F}W$. There is one independent coordinate for every pair of indices.^[ref]

For a proof outline, bilinearity expands each pure tensor asso these products form a spanning set. For linear independence, take the coordinate functionals $e^k,f^{\ell }$ from the two dual bases. The bilinear map $(v,w)\mapsto e^k(v)f^{\ell }(w)$ induces a linear functional ${\lambda }_{k\ell }$ on the tensor product, satisfying ${\lambda }_{k\ell }(e_i\otimes f_j)={\delta }_{ki}{\delta }_{\ell j}$. Applying it to a proposed zero linear combination selects just $c_{k\ell }$, forcing every coefficient to vanish. Thus dimensions multiply, rather than add. This finite-dimensional count includes a zero factor, whose tensor product has dimension zero.