= Cartesian and oblique coordinate reconstruction
{c}
{title2=$c=B^{-1}p,\quad B^Tp=Gc$}
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>:
$$
B=[\,b_1\ b_2\,]=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad B^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix},\qquad [p]_{\mathcal B}=B^{-1}p=\begin{pmatrix}1\\1\end{pmatrix}.
$$
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>.
\Image[/cartesian-oblique-basis-reconstruction.png]
{title=The same <vector> $p=(2,1)^T$ reconstructed in an oblique <basis>: <coordinates> $(1,1)^T$ differ from <dot products> $(2,3)^T$}
The <dual basis> supplies the right measuring rules for any <basis>.<Linear Algebra Done Right by Sheldon Axler (4th edition)>[§3F, Definition 3.112 and Theorem 3.114, printed p. 106: dual basis and coefficient extraction]{ref} Here these <coordinate functionals> are $b_1^*(x,y)=x-y$ and $b_2^*(x,y)=y$, the rows of $B^{-1}$. They give $b_1^*(p)=b_2^*(p)=1$. 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:
$$
G=B^TB=\begin{pmatrix}1&1\\1&2\end{pmatrix},\qquad B^Tp=G[p]_{\mathcal B}=\begin{pmatrix}2\\3\end{pmatrix},\qquad \|p\|^2=[p]_{\mathcal B}^TG[p]_{\mathcal B}=5.
$$
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 $Gc=B^Tp$. 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>.
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