Let $V$ be an $n$-dimensional vector space. Then in the usual way define $\wedge^2 V$ to be the vector space spanned by the elements $v_1 \wedge v_2$ where $v_1, v_2 \in V$ such that they satisfy the usual properties
$$(cu+dv)\wedge w = c(u\wedge w) + d(v \wedge w)$$ $$u \wedge (cv+dw) = c(u\wedge v) + d(u\wedge w)$$ $$u\wedge v = - v\wedge u$$
Let $M$ be a matrix representing a linear transformation of $V$. Then we can define $\wedge^2M$, as linear transformation of $\wedge^2 V$ by
$$\wedge^2 M (u \wedge v) = (Mu)\wedge(Mv)$$
If $M$ is invertible then so is $\wedge^2 M$ with inverse $\wedge^2(M^{-1})$. My question is, is the converse true? That is, if $\wedge^2 M$ is invertible then does this imply $M$ is invertible?
Further, if we define $\wedge^k V$ as the span of $k$ wedges of elements of $V$, then we can define a linear transformation $\wedge^k M$, in the same way. Is it true in this context that $M$ is invertible if and only if $\wedge^k M$ is invertible?
I know this to be true for when $k = n = \dim(V)$ because then $\wedge^n V$ would be a 1-dimensional vector space and $\wedge^n M$ would just be multiplication by $\det{M}$. So, $\wedge^n M$ is invertible if and only $\det{M} \not = 0$ if and only if $M$ is invertible.
However I feel like the proof on $M$ is invertible if and only if $\det{M}\not = 0$, is the argument outlined above so it would be circular and no good.
You can prove this statement to be true in a metric space; I'm not aware of a proof that applies outside of a metric (or pseudometric) space.
Such a proof relies on a known inversion formula. This could be written in terms of the Hodge dual operator--denoted $\star$--or in terms of a clifford algebra product with some $n$-vector $\epsilon$. I'll use the latter here.
For $B_{(k)} \in \bigwedge^k V$, The inversion formula is
$$\bigwedge^k M^{-1}(B_{(k)}) = \left[\bigwedge^{n-k} M^*(B_{(k)} \epsilon) \right] \epsilon^{-1} [\det M]^{-1}$$
For the case $k=1$, this generates the usual inverse formula in terms of the adjugate matrix.
The product with $\epsilon$ is metrical when $\epsilon$ is an $n$-vector---though in principle, if one chooses the volume form instead, perhaps this could be seen as an overall non-metrical operation.
Regardless, the presence of $\det M$ in the formula means that $\det M$ must be nonzero whenever this inverse exists.