What does $\operatorname{GL}(V) \subset \operatorname{Hom}_k(V,V)$ correspond to in $V^* \otimes V$? Let $V$ be a finite-dimensional vector space over a field $k$.  We have the natural isomorphism $V^* \otimes V \cong \operatorname{Hom}_k(V,V)$ given by linearly extending the map $$h \otimes v \mapsto (w \mapsto h(w)v).$$  What does $\operatorname{GL}(V) \subset \operatorname{Hom}_k(V,V)$ correspond to in $V^* \otimes V$?  Ideally, I would like a nice basis-free description.
 A: Here are a few ways.

*

*There is an algebra structure on $V \otimes V^*$, given on simple tensors by $(v \otimes f) \circ (w \otimes g) = f(w) (v \otimes g)$ and extended bilinearly. Then $\operatorname{GL}(V) \subseteq V \otimes V^*$ corresponds to the group of invertible elements in the algebra. (The unit in this algebra can be written as $\sum_i v_i \otimes v_i^*$, where $\{v_i\}$ is any basis of $V$, and $\{v_i^*\}$ its dual basis).

*As another answer pointed out, the set $\operatorname{GL}(V)$ consists of those tensors which may be written as $\sum_i v_i \otimes f_i$, where $\{v_i\}$ is any basis of $V$, and $\{f_i\}$ is any basis of $V^*$. This is because whenever this happens we can manufacture an inverse: pick $w_i \in V$ such that $f_i(w_i) = 1$ and $f_j(w_i) = 0$ for all $j \neq i$ (we can do this since the $f_i$ form a basis). Then the inverse of $\sum_i v_i \otimes f_i$ is $\sum_i w_i \otimes v_i^*$. The condition for the $\{v_i\}$ and $\{f_i\}$ being bases is also necessary: if the $\{v_i\}$ are not a basis then the image is not the whole of $V$, and if the $\{f_i\}$ are not a basis then the kernel is not zero.

*If a tensor $t = \sum_i v_i \otimes f_i$ has $\dim V$ terms, then you can form the matrix $[t]_{i, j} = f_i(v_j)$ and take the determinant of that matrix, which will be the determinant of the endomorphism. (The $\{v_i\}$ and $\{f_i\}$ here do not need to be a basis for this to work: whenever they are not, the determinant will be zero). If you take a smaller set than $\dim V$ terms, you can still take a determinant of the smaller matrix, and it will correspond to a matrix minor. (I still count this as relatively coordinate-free because the matrix is not defined with respect to a basis, it is just an array recording how the $v_i$ pair with the $f_i$).

*There is a polynomial map $\Lambda^n \colon \operatorname{End}(V) \to \operatorname{End}(\Lambda^n V) \cong k$, where $n = \dim V$. The nonvanishing of this polynomial map is precisely the set of invertible elements, so we just need to figure out how to write this map down for $V \otimes V^*$, to get a polynomial map $\Lambda^n \colon V \otimes V^* \to \Lambda^n V \otimes \Lambda^n V^*$. If you take the time to figure this out, you end up essentially at the previous determinant-of-a-matrix characterisation.

A: First, any sum
$$
\sum u_i \otimes v_i
$$
where $u_i \in V^*$, can be expressed as such a sum where the $v_i$ are linearly independent. And then that can be expressed as a sum where both the $u_i$ and $v_i$ are independent. So let's suppose it's in that form.
Then

*

*there need to be at least $n$ terms (else there's some vector simultaneously in the kernels of all the $u_i$), hence exactly $n$ terms (by independence of the $v_i$).


*I think that's it.
For let's look at
$$
T = \sum u_i \otimes v_i
$$
applied to some nonzero vector $w$ by applying each $u_i$ to $w$. (I think of this, in my head, as "multiply by the column vector $w$, because for me, the $u_i$ are on the right (they eat column vectors) and the $v_i$ on the left, but if that makes no sense to you, that's OK).
Suppose $T(w) = 0$. Then letting $c_i = u_i(w)$, we have
$$
\sum_i c_i v_i = 0
$$
which makes all the $c_i$ be zero because the $v$s are independent.
So: If $T(w) = 0$ (with $w \ne 0$), then $w$ is in the kernel of all the $u_i$s. I believe that this makes the $u_i$ dependent (because there are $n$ of them in the $n$-dimensional space $V^*$). I can't quite construct the proof right off the top of my head, but perhaps you, who've been thinking about dual spaces, etc., can do so.
Oh...wait. It's easy. Look at $(u_1, u_2) : V^* \to F \times F: v \mapsto (u_1(v), u_2(v))$. The kernel of that has dimension at least $n-2$, because each term has nullity $n-1$ (simple dimension-counting argument). And if the nullity is $n-1$ instead of $n-2$, then $u_1$ and $u_2$ are dependent. Adding in subsequent terms similarly to look at maps $(u_1, \ldots, u_k): V^* \to F^k$ gives similar bounds, and shows that when we reach $k = n$, either the nullity is zero OR the $u_i$s are dependent.
So ... we've shown that $T(w) = 0$ for any nonzero $w$ would imply the $u_i$ are dependent. Since they're independent, we've shown this:
If $v_1, \ldots, v_n$ independent in $V$, and $u_i, \ldots, u_n$ independent in $V*$, and $T= \sum u_i \otimes v_i$, then $Tw = 0$ only if $w = 0$, i.e, $T$ is in $GL(n)$.
