Are the linear Lie groups matrices, tensors, or both?
In some ways, this is a question about notation.
In my experience, I have only seen the classical Lie groups — such as $GL(n,\mathbb{R})$, $SL(n,\mathbb{R})$, $O(n,\mathbb{R})$, $SO(n,\mathbb{R})$, $Sp(2n,\mathbb{R})$, etc. — introduced and defined in terms of (real) matrices (I'm sticking with $\mathbb{R}$ as the field just to keep things simpler).
My question: can a 2nd order tensor be a member of one of these groups? Or is it only the matrix representation of a tensor that can be a member of one of these groups? If the answer is the latter, then are there corresponding tensor groups for tensors whose matrix representation (in any basis) is a member of one of the classic <PERSON> groups?
Example: The general linear group, $GL(n,\mathbb{R})$, is defined for matrices simply as the group of all non-singular real square matrices: $$ GL(n,\mathbb{R}) \;=\; { A\in M_{n,n}(\mathbb{R}) \;|\; \text{det}(A)\neq 0 } $$
where $M_{n,n}(\mathbb{R})$ denotes $n\times n$ real matrices. Now, consider some finite n-dimensional (real) vector space, $V$, with an inner product. 364
If we have some (1,1)-tensor $\mathbf{Q}\in \mathcal{T}^1_1(V)$ with non-zero determinant (i.e., $\mathbf{Q}$ is non-degenerate), then can we say that $\mathbf{Q}\in GL(n,\mathbb{R})$? Or maybe the proper notation would be $\mathbf{Q}\in GL(V)$?
Although I used a (1,1)-tensor for this example, I don't think anything would change if $\mathbf{Q}$ were a (2,0)-tensor or a (0,2)-tensor, would it? (I am assuming we have an inner product and can thus raise and lower indices as we please)
Does anything change for the other classical Lie groups? For instance, if $[Q]$ is the matrix representation (in any basis) of $\mathbf{Q}$ and satisfies $[Q][Q]^{\top}=I$ such that $[Q]\in SO(n,\mathbb{R})$, can we say $\mathbf{Q}\in SO(n,\mathbb{R})$? Or perhaps $\mathbf{Q}\in SO(V)$? (this would mean that $\mathbf{Q}\cdot\mathbf{Q}^{\top}=\delta^i_j\mathbf{e}_i\otimes\boldsymbol{\epsilon}^j=\mathbf{I}$, where, here, the dot denotes a contraction using the inner product/metric, $\mathbf{e}_i$ and $\boldsymbol{\epsilon}^i$ are basis vectors and basis 1-forms for $V$ and $V^*$, and $\mathbf{I}$ is the identity tensor for $V$. )
Another question: same question for the associated Lie algebras. I.e., can a tensor be a member of $\mathfrak{so}(n,\mathbb{R})$ or only a matrix?
Edit: To clarify, I'm using the term tensor such that $T\in \mathcal{T}^p_q(V)$ means $T$ is (p+q)-linear map, $T:(V^)^{\times p} \times V^{\times q} \to \mathbb{R}$. That is, $T= V^{\otimes p}\otimes (V^)^{\otimes q}$. 656