- Considerations on Basis Transformation Matrix A as a (1,1) Tensor

1. Relationship with Linear Operator U #

The basis transformation matrix $A$, i.e., $e_i = A e_{i’} = A_i^{j’} e_{j’}$, appears not to be a tensor at first glance since it uses both unprimed and primed indices, suggesting it’s not a notation with respect to a single basis.

However, $A$ can effectively be considered a tensor with respect to the unprimed basis.

Let’s consider a linear operator $U$ that transforms the basis as $e_i = U e_i’$. Since a linear operator is a $(1,1)$ tensor, it can be expressed as $U_i^j = U(e_i, e^j)$.

To understand the relationship between $U$ and $A$, we can expand using the relation $e_i = A_i^{k’} e_{k’}$:

$$ U_i^j = U(e_i, e^j) = e^j (U e_i) = e^j (U A_i^{k’} e_{k’}) $$

$$ = A_i^{k’} e^j (U e_{k’}) = \sum_k A_i^{k’} e^j (e_k) $$

$$ = \sum_k A_i^{k’} \delta_k^j = A_i^{j’} $$

Thus, we can see that $U_i^j = A_i^{j’}$ holds. Effectively, $A$ can be considered a $(1,1)$ tensor $U$ in the unprimed basis. In other words, by replacing $A_i^{j’}$ with $A_i^j$ to construct a matrix, we can formulate the basis transformation matrix $A$ in the unprimed basis and express it as $e_i = A e’_i$.

2. Additional Remarks #

In a previous post, we demonstrated that using the Schouten convention (with primes on indices), tensor index notation, and Einstein’s summation convention, we could intuitively construct the basis transformation matrix. Here, we’ve shown that this can be converted back to matrix notation without using the Schouten convention, i.e., it can be expressed as a matrix notation in a single basis.