Are Linear Transformations Always Second Order Tensors? I've been reading a bit about tensors on Wikipedia (so correctness not guaranteed here) and I have a question. The order of a tensor $T$ is defined as $n+m$, where $n$ denotes the number of covariant indices and $m$ the number of contravariant indices. Wikipedia gives linear transformations as examples of second order tensors and scalars as examples of 0 order tensors. Here's what I don't understand:
For the sake of concreteness, let's take a particular example. Consider the vector space $\mathbb{R}^n$ (over $\mathbb{R}$) with the linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$, $T: v \mapsto \alpha v$ for some $\alpha \in \mathbb{R}$. It is my understanding that in any basis we may represent this transformation by the array $\alpha I$, where $I$ is the $n \times n$ identity matrix. Two indices are required to uniquely specify each element of this array, hence $T$ should be a second order tensor, as dictated by Wikipedia. However, what if we simply represented $T$ by the array $(\alpha)$? This is just a scalar, so by Wikipedia this is a 0 order tensor. Does the order of a tensor depend on our representation of it somehow?
Edit: $(\alpha)$ is to be interpreted as a $1 \times 1$ matrix. 

The "type" of a tensor $T$—i.e., of a tensor-valued function, or "tensor field"—determines how $T$ transforms under change of coordinates.


*

*If you view a scalar as a scalar-valued function it behaves like a $0$-tensor.

*If you view a scalar as the operation of scalar multiplication, it behaves like a $2$-tensor.
Strictly speaking those are distinct interpretations, not mere matters of notation. (Matrix notation reflects this distinction, incidentally: You can't (matrix-)multiply a $1 \times 1$ matrix $(\alpha)$ by an $n \times n$ matrix.)
