Can all rank two tensors be represented as matrices?

Question

I’m curious about wether every Tensor of second order can be represented as a matrix. The only possibilities are (to my understanding, correct me if i’m wrong): $$\mathcal{T}^{(2;0)}= \mathbf{V\otimes V}$$ $$\mathcal{T}^{(1;1)}= \mathbf{V\otimes V^*}$$ $$\mathcal{T}^{(0;2)}= \mathbf{V^* \otimes V^*}$$ Can all these be represented as $n\times n$ matrices? If so, i would like to know if there is a difference between all these matrices.

Answer 1

At risk of reading too much into the question:

If $V$ is a finite-dimensional real vector space, then upon fixing a basis of $V$ (which determines a unique dual basis of $V^{*}$) every real-valued $2$-tensor $T$ is uniquely determined by a doubly-indexed set of real coefficients, namely the values of $T$ on appropriate basis and/or dual basis vectors. Since any such collection may be written as a matrix, "yes (in this sense)."

That said, a $2$-tensor is not just a set of components in a specific basis, it's a bilinear function, or a set of components in an arbitrary basis. A matrix alone does not capture that. Concretely, a matrix of components alone does not tell us how the components transform under change of basis. That transformation law is an essential datum.

Tangential anecdote: Some years back there was a question on Math.SE about the intersection form of a particular simply-connected $4$-manifold, and an apparent contradiction between using two particular bases for two-dimensonal homology. The resolution was that an intersection form, contrary to OP's understandable matrix-based habit, does not transform by similarity like a linear transformation (tensor in $V^{*} \otimes V$), but like a quadratic form (tensor in $V^{*} \otimes V^{*}$).