An explicit reconstruction of a matrix from its minors

Question

$\newcommand{\End}{\operatorname{End}}$ $\newcommand{\GL}{\operatorname{GL}}$ $\newcommand{\Cof}{\operatorname{cof}}$

Let $V$ be a $d$-dimensional real vector space. ($d \ge 4$). Fix an odd integer $2 \le k \le d-2$. Define $H_{>k}=\{ A \in \End(V) \mid \operatorname{rank}(A) > k \}$. Consider the map $$ \psi:H_{>k} \to \End(\bigwedge^k V) \, \,, \, \, \psi(A)=\bigwedge^{k}A. $$

$\psi$ is a smooth injective immersion.

Question: Is there an explicit formula for $\psi^{-1}$?

Comments:

1. Since $\operatorname{rank}(\bigwedge^kA) = \binom {\operatorname{rank}(A)}{k} $, we might start by considering each rank separately.

2. The limitation $k$ is odd is not essential. When $k$ is even, $\psi(A)=\psi(B)$ implies $A=\pm B$ (assuming $A,B \in H_{>k}$), so the inverse is well-defined up to a sign. In fact, the following more general property holds:

Let $V$ be a vector space over an arbitrary field $F$. Then, for $A,B \in H_{>k}$, $\psi(A)=\psi(B)$ iff $A=\lambda B$ when $\lambda^k=1$. So, we can ask the question in this more general setting:

Is there an explicit formula, which given an element in $\text{Image }(\psi)$, produces a source element?

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Part of the problem is that we do not have a closed-form description of $\text{Image }(\psi)$, and it is clear that in general we don't expect $\psi^{-1} $ to have a continuous extension to all $\End(\bigwedge^k V)$ (or even to the space of all endomorphisms of rank bigger than $k$, which is where $\text{Image }(\psi)$ lies, but I am not sure about that.)

I am OK with a formula which uses an inner product and orientation structures. Even though we don't need them in order to define $\psi$, they somehow appear naturally when trying to compute $\psi^{-1}$.

Indeed, in the special case $k=d-1$, $\psi(A) \in \text{GL}(\bigwedge^{d-1}V)$ can be identified with the cofactor matrix of $A$. (The identification is done using the Hodge dual $\star:\bigwedge^{d-1}V \to V$).

Then we have $ \Cof(\Cof A)=(\det A)^{d-2}A$. Now, if $\Cof A=B$, then $\Cof B=(\det A)^{d-2}A$, and $\det(B)=(\det(A))^{d-1}$.

Since $k=d-1$ is odd, we can take the (unique) $ d-1 $-th root, so $(\det(B))^{\frac{1}{d-1}}=\det(A)$.

Thus,

$$ A=(\Cof)^{-1} B=(\det B)^{\frac{2-d}{d-1}} \Cof B $$ gives the formula for $\psi^{-1}$.

Trying to generalize this derivation using the Hodge dual for general $k$ seems to hit a wall.

Answer 1

Assume for simplicity that $kl=d+1$.

Then I claim that $V \otimes \det V$ appears as a summand of $\left( \bigwedge^k V\right)^{\otimes l}$ with multiplicity $l-1$.

The maps $\left( \bigwedge^k V\right)^{\otimes l} \to V \otimes \det V$ are easier to write down - the $i$'th maps sends $$ (v_{1,1} \wedge \dots \wedge v_{1,k}) \otimes \dots \otimes (v_{l,1}\wedge \dots \wedge v_{l,k})$$ to $$\sum_{j=1}^{k} (-1)^{ il+j} v_{i,j} \otimes (v_{1,1} \wedge \dots \wedge v_{1,k} \wedge v_{2,1} \dots \wedge v_{i,j-1} \wedge v_{i,j+1} \wedge \dots \wedge v_{l,k})$$

and we have the relation that the $l$ maps sum to zero. The reverse maps are similar, except that the $i$th map is a sum over all the ways to divide the $kl$ vectors evenly into $l$ boxes, with the first vector in the $i$'th box.

Now if we have an endomorphism of $\wedge^k V$, we can let it act on $\left( \bigwedge^k V \right)^{\otimes l}$, compose with one of these maps on each side, and obtain an endomorphism of $V$. If we apply this to an element of $GL(V)$, we get the individual element back, times its determinant, times some fixed scalar depending on the choice of map. We can divide by the scalar and by the $\frac{1}{ {d-1 \choose k-1}}$ power of the determinant of the endomorphism of $\bigwedge^k V$ to obtain the original matrix exactly.