Basis Transformation

Relevant parts to questions...

• Using ${\mathbf{e}}_i^{\prime }=L_{ij}{\mathbf{e}}_j$ to expand new basis vectors in terms of the old.
• Using $L_{ij}={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}_j$ to read off entries of $L$.
• Using $s^{\prime }=s$ (scalar) and $v_i^{\prime }=L_{ij}{v}_j$ (vector) as the formal definitions of rank-0 and rank-1 objects.

A Basis Transformation expresses one orthonormal basis $\{{\mathbf{e}}_i^{\prime }\}$ in terms of another $\{{\mathbf{e}}_j\}$ via the Coordinate Transformation Matrix $L_{ij}$:

$$\boxed{{\mathbf{e}}_i^{\prime }=L_{ij}{\mathbf{e}}_j}$$

This is the other face of $x_i^{\prime }=L_{ij}{x}_j$: the coordinates and the basis vectors transform with the same matrix (reflecting that the underlying geometric object is unchanged - only the “point of view” has rotated).

Formal Definitions of Scalars and Vectors

The transformation rule ${\mathbf{e}}_i^{\prime }=L_{ij}{\mathbf{e}}_j$ gives us a precise, coordinate-free way to define what “scalar” and “vector” even mean:

• Scalar::a quantity unchanged under coordinate rotation, $s^{\prime }=s$.
• Vector::a quantity whose components transform with exactly one factor of $L$, $v_i^{\prime }=L_{ij}{v}_j$.

Both are special cases of a tensor, which transforms with one factor of $L$ for each free index. Scalars are rank-0 tensors; vectors are rank-1 tensors. See rank and the Tensor Transformation Rule for the generalisation.

Applications

1. Expanding a new basis in the old, using $L$::${\mathbf{e}}_i^{\prime }=L_{i1}{\mathbf{e}}_1+L_{i2}{\mathbf{e}}_2+L_{i3}{\mathbf{e}}_3$.
2. Verifying a quantity is a vector, by checking $v_i^{\prime }=L_{ij}{v}_j$::e.g. the Gradient of a scalar field, or position coordinates themselves.
3. Proving invariance of scalar operations such as the dot product, by transforming and cancelling two $L$s::$(\mathbf{a}\cdot \mathbf{b}{)}^{\prime }=a_i^{\prime }{b}_i^{\prime }=L_{ij}{L}_{ik}{a}_j{b}_k={\delta }_{jk}{a}_j{b}_k=\mathbf{a}\cdot \mathbf{b}$.

Proving $\mathbf{a}\cdot \mathbf{b}$ is a scalar

Under a rotation, $a_i^{\prime }=L_{ij}{a}_j$ and $b_i^{\prime }=L_{ik}{b}_k$. Hence:

$(\mathbf{a}\cdot \mathbf{b}{)}^{\prime }=a_i^{\prime }{b}_i^{\prime }=(L_{ij}{a}_j)(L_{ik}{b}_k)=L_{ij}{L}_{ik}{a}_j{b}_k$.

Using orthogonality $L_{ij}{L}_{ik}={\delta }_{jk}$ collapses the two $L$s into a Kronecker Delta:

$={\delta }_{jk}{a}_j{b}_k=a_k{b}_k=\boxed{\mathbf{a}\cdot \mathbf{b}}$. So $(\mathbf{a}\cdot \mathbf{b}{)}^{\prime }=\mathbf{a}\cdot \mathbf{b}$, confirming it is a scalar.

Basis in, formal tensor-by-rank definition out.