$\newcommand{\Cof}{\operatorname{Cof}}$
$\newcommand{\Det}{\operatorname{Det}}$
$\newcommand{\Lam}{\operatorname{\Lambda}}$


**Motivation (and the "classic" case):**


I am trying to find a **coordinate-free** criterion to determine when a linear map is an **isometric embedding** of an inner product space in a space of higher dimension. 


Such a criterion exists when the dimensions are equal. Let $V,W$ be oriented inner product spaces, of the **same** dimension $d$. 
Let $\,A: V \to W$ be a linear map. Then one can define invariantly the **cofactor** and the **determinant** of $A$ as follows: $$(1) \Det A := \star\_W^d \circ \bigwedge^d A \circ \star\_V^0 ,$$ $$(2) \Cof A = (-1)^{d-1} \star\_W^{d-1} \circ \bigwedge^{d-1} A \circ \star\_V^1,$$ 


Note that $\Det A \in \mathbb{R}$,$\, \Cof A:V \to W$, (I keep the degree explicit in the notation of the Hodge-dual, i.e $\star\_V^p:\Lam\_p(V) \to \Lam\_{d-p}(V)$). 


The well-known matrix identity $$\Cof X \cdot X^T = \det X \cdot \operatorname{Id}$$ can be invariantly written as $$(\*) \, \Cof A \circ A^T= \Det A \cdot \operatorname{Id}\_W$$


**Immediate observation:** $(\*)$ implies that $A$ is an isometry if and only if $$(3) \Det A \neq 0 \text{ and} \, \Det A \cdot A= \Cof A$$


My question is about the validity of a possible approach of generalizing criterion $(3)$ to the case where $\dim V < \dim W$. 




---


**The case $\dim V < \dim W$:**


*Stage I: generalizing identity $(\*)$.*


Assume $\dim V=d < n=\dim W$. Definitions $(1),(2)$ for the cofactor and the determinant now gives objects of *different types*:


(a) $\Cof A:V \to \bigwedge^{n-d+1}W$, so $\Cof A \circ A^T:W \to \bigwedge^{n-d+1}W$.


(b) $\Det A \in \bigwedge^{n-d}W$.


The natural generalization of identity $(\*)$ should therefore be:


$$(\*\*) \Cof A \circ A^T(w)= \Det A \wedge w \, ,\,\forall w \in W.$$ 
(This is an equality of elements in $\bigwedge^{n-d+1}W $).


*Stage II: generalizing condition $(3)$.*


**I think $A$ will be an isometric embedding if and only if** $$ (3)' \Cof A \, \text{is injective and} \, \Det A \wedge A(v) = \Cof A(v) \, \, \forall v \in V $$ (This is an equality of elements in $\bigwedge^{n-d+1}W $).


Indeed, putting $w=Av$ in $(\*\*)$, one gets that for **any** linear map $A$, $$ \Cof A \circ A^T(Av)= \Det A \wedge Av, $$ so $A$ satisfies condition $(3)'$ if and only if $$ \Cof A \, \text{is injective and} \, \Cof A \circ (A^TA) = \Cof A$$ which holds iff $A^TA=\operatorname{Id}\_V$ 
(which is equivalent to $A$ being an isometry).


**The remaining questions (which are needed for completion of this endeavour) are:**


(1) Does identity $(\*\*)$ hold? (It probably holds "up to sign", i.e both $d,n$ should enter the new definition of the cofactor in a manner which should be determined )


(2) In both cases, part of the conditions were "non-triviality" , in $(3)$ it was $\det A \neq 0$ which is equivalent to the injectivity of $A$. In $(3)'$ it was injectivity of $\Cof A:V \to \bigwedge^{n-d+1}W$. Is it true that injectivity of $A$ implies injectivity of $\Cof A$? (and what about the converse?)



Invariant characterization of isometric embeddings
Here is a proof of $\left( \ast\ast\right) $. As you correctly admitted,
part of the problem is to define the signs that enter into
$\operatorname\*{Cof}A$. Let me actually start from scratch and introduce all
notations that will be needed:


For any $m\in\mathbb{N}$, we shall write $\left[ m\right] $ for the set
$\left\{ 1,2,\ldots,m\right\} $.


Let $d$ and $n$ be two nonnegative integers. Let $A\in\mathbf{k}^{n\times d}$
be an $n\times d$-matrix over a commutative ring $\mathbf{k}$. Regard $A$ as a
$\mathbf{k}$-linear map $\mathbf{k}^{d}\rightarrow\mathbf{k}^{n}$.


Let $\left( f\_{1},f\_{2},\ldots,f\_{d}\right) $ be the standard basis of the
$\mathbf{k}$-module $\mathbf{k}^{d}$. (Thus, $f\_{i}$ is the vector whose
$i$-th entry is $1$ and whose all other entries are $0$.)


Let $\left( e\_{1},e\_{2},\ldots,e\_{n}\right) $ be the standard basis of the
$\mathbf{k}$-module $\mathbf{k}^{n}$. For every subset $I=\left\{ i\_{1}
<i\_{2}<\cdots<i\_{k}\right\} $ of $\left[ n\right] $, we let $e\_{I}$ be the
vector $e\_{i\_{1}}\wedge e\_{i\_{2}}\wedge\cdots\wedge e\_{i\_{k}}\in\wedge
^{k}\left( \mathbf{k}^{n}\right) $.


If $U\subseteq\left[ n\right] $ and $V\subseteq\left[ d\right] $, then we
let $A\_{U,V}$ denote the submatrix of $A$ obtained by removing all rows except
for those in $U$ and all columns except for those in $V$. (In more rigorous
terms: We let $A\_{U,V}$ be the matrix $\left( a\_{u\_{i},v\_{j}}\right) \_{1\leq
i\leq\left\vert U\right\vert ,\ 1\leq j\leq\left\vert V\right\vert }$, where
$A$ is written in the form $\left( a\_{i,j}\right) \_{1\leq i\leq n,\ 1\leq
j\leq d}$, and where the sets $U$ and $V$ are written in the forms $U=\left\{
u\_{1}<u\_{2}<\cdots<u\_{\left\vert U\right\vert }\right\} $ and $V=\left\{
v\_{1}<v\_{2}<\cdots<v\_{\left\vert V\right\vert }\right\} $.)


If $S$ is a finite set of integers, then $\sum S$ shall mean the sum of the
elements of $S$.


Regard the $d\times n$-matrix $A^{T}\in\mathbf{k}^{d\times n}$ as a
$\mathbf{k}$-linear map $\mathbf{k}^{n}\rightarrow\mathbf{k}^{d}$.


Define an element $\operatorname\*{Det}A\in\wedge^{n-d}\left( \mathbf{k}
^{n}\right) $ by $\operatorname\*{Det}A=\sum\_{\substack{I\subseteq\left[
n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum
I-\left\vert I\right\vert }\det\left( A\_{\left[ n\right] \setminus
I,\left[ d\right] }\right) e\_{I}$.


Define a $\mathbf{k}$-linear map $\operatorname\*{Cof}A:\mathbf{k}
^{d}\rightarrow\wedge^{n-d+1}\left( \mathbf{k}^{n}\right) $ by setting


$\left( \operatorname\*{Cof}A\right) \left( f\_{i}\right) =\sum
\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1}}\left( -1\right) ^{i+\sum J}\det\left( A\_{\left[ n\right]
\setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) e\_{J}$.


**Theorem 1.** For every $w\in\mathbf{k}^{n}$, we have $\left( \left(
\operatorname\*{Cof}A\right) \circ A^{T}\right) \left( w\right) =\left(
\operatorname\*{Det}A\right) \wedge w$ in $\wedge\left( \mathbf{k}
^{n}\right) $.


*Proof of Theorem 1.* Let $w\in\mathbf{k}^{n}$. We must prove the equality
$\left( \left( \operatorname\*{Cof}A\right) \circ A^{T}\right) \left(
w\right) =\left( \operatorname\*{Det}A\right) \wedge w$. This equality is
$\mathbf{k}$-linear in $w$; thus, we can WLOG assume that $w$ is an element of
the basis $\left( e\_{1},e\_{2},\ldots,e\_{n}\right) $ of $\mathbf{k}^{n}$. In
other words, we can WLOG assume that $w=e\_{k}$ for some $k\in\left[ n\right]
$. Assume this, and consider this $k$.


Write the matrix $A$ in the form $A=\left( a\_{i,j}\right) \_{1\leq i\leq
n,\ 1\leq j\leq d}$. Thus, $A^{T}=\left( a\_{j,i}\right) \_{1\leq i\leq
d,\ 1\leq j\leq n}$.


From $w=e\_{k}$, we obtain $A^{T}\left( w\right) =A^{T}\left( e\_{k}\right)
=\sum\_{i\in\left[ d\right] }a\_{k,i}f\_{i}$ (since $A^{T}=\left(
a\_{j,i}\right) \_{1\leq i\leq d,\ 1\leq j\leq n}$). Now,


$\left( \left( \operatorname\*{Cof}A\right) \circ A^{T}\right) \left(
w\right) $


$=\left( \operatorname\*{Cof}A\right) \underbrace{\left( A^{T}\left(
w\right) \right) }\_{=\sum\_{i\in\left[ d\right] }a\_{k,i}f\_{i}}=\left(
\operatorname\*{Cof}A\right) \left( \sum\_{i\in\left[ d\right] }a\_{k,i}
f\_{i}\right) $


$=\sum\_{i\in\left[ d\right] }a\_{k,i}\underbrace{\left( \operatorname\*{Cof}
A\right) \left( f\_{i}\right) }\_{\substack{=\sum\_{\substack{J\subseteq
\left[ n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right)
^{i+\sum J}\det\left( A\_{\left[ n\right] \setminus J,\left[ d\right]
\setminus\left\{ i\right\} }\right) e\_{J}\\\text{(by the definition of
}\operatorname\*{Cof}A\text{)}}}$


$=\sum\_{i\in\left[ d\right] }a\_{k,i}\sum\_{\substack{J\subseteq\left[
n\right] ;\\\left\vert J\right\vert =n-d+1}}\left( -1\right) ^{i+\sum
J}\det\left( A\_{\left[ n\right] \setminus J,\left[ d\right]
\setminus\left\{ i\right\} }\right) e\_{J}$


$=\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1}}\left( \sum\_{i\in\left[ d\right] }\left( -1\right) ^{i+\sum
J}a\_{k,i}\det\left( A\_{\left[ n\right] \setminus J,\left[ d\right]
\setminus\left\{ i\right\} }\right) \right) e\_{J}$


$=\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1}}\left( -1\right) ^{\sum J}\left( \sum\_{i\in\left[ d\right]
}\left( -1\right) ^{i}a\_{k,i}\det\left( A\_{\left[ n\right] \setminus
J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right) e\_{J}$


**(1)** $=\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert
J\right\vert =n-d+1;\\k\in J}}\left( -1\right) ^{\sum J}\left( \sum
\_{i\in\left[ d\right] }\left( -1\right) ^{i}a\_{k,i}\det\left( A\_{\left[
n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right)
\right) e\_{J}$


$+\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1;\\k\notin J}}\left( -1\right) ^{\sum J}\left( \sum\_{i\in\left[
d\right] }\left( -1\right) ^{i}a\_{k,i}\det\left( A\_{\left[ n\right]
\setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right) \right)
e\_{J}$


(here we have split the sum into two subsums, according to whether $k\in J$ or
$k\notin J$).


If $J$ is a finite set of integers, and if $k$ is an integer, then $\#\left(
J<k\right) $ shall denote the number of all elements of $J$ that are smaller
than $k$. Similarly, $\#\left( J>k\right) $ shall denote the number of all
elements of $J$ that are greater than $k$.


But every $\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $
satisfying $k\in J$ satisfies


**(2)** $\sum\_{i\in\left[ d\right] }\left( -1\right) ^{i+\#\left( \left[
n\right] \setminus J<k\right) +1}a\_{k,i}\det\left( A\_{\left[ n\right]
\setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right)
=\det\left( A\_{\left( \left[ n\right] \setminus J\right) \cup\left\{
k\right\} ,\left[ d\right] }\right) $


(by Laplace expansion, applied to the matrix $A\_{\left( \left[ n\right]
\setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }$, with
respect to the $\#\left( \left[ n\right] \setminus J<k\right) +1$-th row
of this matrix (which used to be the $k$-th row of $A$)). Hence, every
$\left( n-d+1\right) $-element subset $J$ of $\left[ n\right] $ satisfying
$k\in J$ satisfies


**(3)** $\sum\_{i\in\left[ d\right] }\left( -1\right) ^{i}a\_{k,i}
\det\left( A\_{\left[ n\right] \setminus J,\left[ d\right] \setminus
\left\{ i\right\} }\right) =\left( -1\right) ^{\#\left( \left[
n\right] \setminus J<k\right) +1}\det\left( A\_{\left( \left[ n\right]
\setminus J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) $


(this follows by multiplying both sides of **(2)** with $\left( -1\right)
^{\#\left( \left[ n\right] \setminus J<k\right) +1}$).


Furthermore, every $\left( n-d+1\right) $-element subset $J$ of $\left[
n\right] $ satisfying $k\notin J$ satisfies


**(4)** $\sum\_{i\in\left[ d\right] }\left( -1\right) ^{i}a\_{k,i}
\det\left( A\_{\left[ n\right] \setminus J,\left[ d\right] \setminus
\left\{ i\right\} }\right) =0$


(by Laplace expansion, applied to the matrix $A\_{\left[ n\right] \setminus
J,\left[ d\right] }$ with an extra copy of the $k$-th row of $A$ added to
its top; the result is $0$ because the resulting matrix has this $k$-th row
appear twice inside it). Now, **(1)** becomes


$\left( \left( \operatorname\*{Cof}A\right) \circ A^{T}\right) \left(
w\right) $


$=\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1;\\k\in J}}\left( -1\right) ^{\sum J}\underbrace{\left( \sum
\_{i\in\left[ d\right] }\left( -1\right) ^{i}a\_{k,i}\det\left( A\_{\left[
n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right)
\right) }\_{\substack{=\left( -1\right) ^{\#\left( \left[ n\right]
\setminus J<k\right) +1}\det\left( A\_{\left( \left[ n\right] \setminus
J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) \\\text{(by
(3))}}}e\_{J}$


$+\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1;\\k\notin J}}\left( -1\right) ^{\sum J}\underbrace{\left( \sum
\_{i\in\left[ d\right] }\left( -1\right) ^{i}a\_{k,i}\det\left( A\_{\left[
n\right] \setminus J,\left[ d\right] \setminus\left\{ i\right\} }\right)
\right) }\_{\substack{=0\\\text{(by (4))}}}e\_{J}$


$=\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1;\\k\in J}}\underbrace{\left( -1\right) ^{\sum J}\left( -1\right)
^{\#\left( \left[ n\right] \setminus J<k\right) +1}}\_{=\left( -1\right)
^{\sum J+\#\left( \left[ n\right] \setminus J<k\right) +1}}\det\left(
A\_{\left( \left[ n\right] \setminus J\right) \cup\left\{ k\right\}
,\left[ d\right] }\right) e\_{J}$


$=\sum\_{\substack{J\subseteq\left[ n\right] ;\\\left\vert J\right\vert
=n-d+1;\\k\in J}}\left( -1\right) ^{\sum J+\#\left( \left[ n\right]
\setminus J<k\right) +1}\det\left( A\_{\left( \left[ n\right] \setminus
J\right) \cup\left\{ k\right\} ,\left[ d\right] }\right) e\_{J}$


$=\sum\_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\}
\right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{
k\right\} \right) <k\right) +1}\underbrace{\det\left( A\_{\left( \left[
n\right] \setminus\left( I\cup\left\{ k\right\} \right) \right)
\cup\left\{ k\right\} ,\left[ d\right] }\right) }\_{\substack{=\det\left(
A\_{\left[ n\right] \setminus I,\left[ d\right] }\right) \\\text{(since
}\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right)
\right) \cup\left\{ k\right\} =\left[ n\right] \setminus I\text{)}
}}\underbrace{e\_{I\cup\left\{ k\right\} }}\_{\substack{=\left( -1\right)
^{\#\left( I>k\right) }e\_{I}\wedge e\_{k}\\\text{(because the sign of the
permutation}\\\text{that takes }e\_{I\cup\left\{ k\right\} }\text{ to }
e\_{I}\wedge e\_{k}\text{ is }\left( -1\right) ^{\#\left( I>k\right)
}\text{)}}}$


(here, we have substituted $I\cup\left\{ k\right\} $ for $J$ in the sum,
because every $\left( n-d+1\right) $-element subset $J$ of $\left[
n\right] $ satisfying $k\in J$ must have the form $I\cup\left\{ k\right\} $
for some $\left( n-d\right) $-element subset $I$ of $\left[ n\right] $
satisfying $k\notin I$)


$=\sum\_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left( -1\right) ^{\sum\left( I\cup\left\{ k\right\}
\right) +\#\left( \left[ n\right] \setminus\left( I\cup\left\{
k\right\} \right) <k\right) +1}\det\left( A\_{\left[ n\right] \setminus
I,\left[ d\right] }\right) \left( -1\right) ^{\#\left( I>k\right)
}e\_{I}\wedge e\_{k}$


$=\sum\_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\underbrace{\left( -1\right) ^{\sum\left( I\cup\left\{
k\right\} \right) +\#\left( \left[ n\right] \setminus\left(
I\cup\left\{ k\right\} \right) <k\right) +1+\#\left( I>k\right) }
}\_{\substack{=\left( -1\right) ^{\sum I-\left\vert I\right\vert
}\\\text{(since }\sum\left( I\cup\left\{ k\right\} \right) +\#\left(
\left[ n\right] \setminus\left( I\cup\left\{ k\right\} \right)
<k\right) +1+\#\left( I>k\right) \\\equiv\underbrace{\sum\left(
I\cup\left\{ k\right\} \right) }\_{=\sum I+k}+\#\left( \left[ n\right]
\setminus\left( I\cup\left\{ k\right\} \right) <k\right)
+1-\underbrace{\#\left( I>k\right) }\_{\substack{=\left\vert I\right\vert
-\#\left( I<k\right) \\\text{(since }k\notin I\text{)}}}\\=\sum
I+k+\#\left( \left[ n\right] \setminus\left( I\cup\left\{ k\right\}
\right) <k\right) +1-\left\vert I\right\vert +\#\left( I<k\right) \\=\sum
I+k+1-\left\vert I\right\vert +\underbrace{\#\left( \left[ n\right]
\setminus\left( I\cup\left\{ k\right\} \right) <k\right) +\#\left(
I<k\right) }\_{=k-1}\\=\sum I+k+1-\left\vert I\right\vert +k-1=\sum
I+2k-\left\vert I\right\vert \\\equiv\sum I-\left\vert I\right\vert
\operatorname{mod}2\text{)}}}\det\left( A\_{\left[ n\right] \setminus
I,\left[ d\right] }\right) e\_{I}\wedge e\_{k}$


$=\sum\_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }
\det\left( A\_{\left[ n\right] \setminus I,\left[ d\right] }\right)
e\_{I}\wedge e\_{k}$.


Comparing this with


$\underbrace{\operatorname\*{Det}A}\_{=\sum\_{\substack{I\subseteq\left[
n\right] ;\\\left\vert I\right\vert =n-d}}\left( -1\right) ^{\sum
I-\left\vert I\right\vert }\det\left( A\_{\left[ n\right] \setminus
I,\left[ d\right] }\right) e\_{I}}\wedge\underbrace{w}\_{=e\_{k}}$


$=\left( \sum\_{\substack{I\subseteq\left[ n\right] ;\\\left\vert
I\right\vert =n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }
\det\left( A\_{\left[ n\right] \setminus I,\left[ d\right] }\right)
e\_{I}\right) \wedge e\_{k}$


$=\sum\_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert
=n-d}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }\det\left(
A\_{\left[ n\right] \setminus I,\left[ d\right] }\right) e\_{I}\wedge
e\_{k}$


$=\sum\_{\substack{I\subseteq\left[ n\right] ;\\\left\vert I\right\vert
=n-d;\\k\notin I}}\left( -1\right) ^{\sum I-\left\vert I\right\vert }
\det\left( A\_{\left[ n\right] \setminus I,\left[ d\right] }\right)
e\_{I}\wedge e\_{k}$


(here, we have removed all addends which satisfy $k\in I$, because all such
addends contain the vanishing factor $e\_{I}\wedge e\_{k}=0$), we should obtain
$\left( \left( \operatorname\*{Cof}A\right) \circ A^{T}\right) \left(
w\right) =\left( \operatorname\*{Det}A\right) \wedge w$. This proves Theorem 1.


