= Cartesian basis
{c}
{title2=$\mathcal E=(e_1,\ldots,e_n)$}
A <Cartesian basis> is an <ordered basis> of <unit vectors> along the mutually perpendicular axes of a <Cartesian coordinate system>. It is an <orthonormal basis> of a real <Euclidean space>: $e_i\cdot e_j=\delta_{ij}$, with $\delta_{ij}$ the <Kronecker delta>. In the plane, the familiar <standard basis> is $e_1=(1,0)^T$, $e_2=(0,1)^T$; the <vector> $p=(2,1)^T$ is $2e_1+e_2$. A rotated pair of perpendicular <unit vectors> is also a <Cartesian basis>, even though its <basis vectors> have different <standard basis> entries.
The advantage is that <coordinates> can be measured separately by <dot products>, and the squared <norm> is the sum of their squares:
$$
p=\sum_i(p\cdot e_i)e_i,\qquad [p]_{\mathcal E}=(p\cdot e_1,\ldots,p\cdot e_n)^T,\qquad \|p\|^2=\sum_i(p\cdot e_i)^2.
$$
These are the real <Euclidean space> instances of the <orthonormal basis> expansion.<Linear Algebra Done Right by Sheldon Axler (4th edition)>[§6B, Definition 6.27, printed p. 199, and Theorem 6.30(a–b), printed p. 200]{ref}
An <orthogonal basis> only requires perpendicular <basis vectors>; their lengths need not be one. For example, $(2,0)^T,(0,3)^T$ are <orthogonal>, but $p=(2,1)^T$ has <coordinates> $(1,1/3)^T$, while its <dot products> with them are $4,3$. For an <orthogonal basis> $(a_i)$, the <coordinate> along $a_i$ is $(p\cdot a_i)/(a_i\cdot a_i)$. Dividing each nonzero <basis vector> by its <norm> makes it a <unit vector> without changing its direction. Unit length alone does not imply <orthogonality>: $(1,0)^T$ and $(1,1)^T/\sqrt2$ are both <unit vectors> but have <dot product> $1/\sqrt2$.
The examples below show why these distinctions matter: rotating a <Cartesian basis> changes the <coordinate vector> while keeping lengths easy to compute; using nonperpendicular <basis vectors> still gives unique <coordinates>, but requires the <dual basis> or a <matrix inverse> to extract them.
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