I'm trying to calculate an integral over the generalized Poincare upper half plane, then I find that I need to show the following identity: 



> 
> Let $X=(X\_{i,j})\in\mathrm{GL}(n,\mathbb R)(n\geq 3)$ and $r$ be a positive integer such that $2\leq r\leq n-1$. For any $1\leq \ell\_1<\ell\_2<\cdots< \ell\_r\leq n$, we define
>  \begin{eqnarray\*}
> \det(\ell\_1,\ell\_2,\cdots,\ell\_r):=\det\left\{
> \left(
>  \begin{array}{cccc}
>  X\_{n-r+1,\ell\_1} & X\_{n-r+1,\ell\_2} & \cdots & X\_{n-r+1,\ell\_r} \\
>  X\_{n-r+2,\ell\_1} & X\_{n-r+2,\ell\_2} & \cdots & X\_{n-r+2,\ell\_r} \\
>  \vdots & \vdots & \cdots & \vdots \\
>  X\_{n,\ell\_1} & X\_{n,\ell\_2} & \cdots & X\_{n,\ell\_r} \\
>  \end{array}
> \right)\right\}.
> \end{eqnarray\*}
>  Let 
>  \begin{eqnarray\*}
> A&=&\left(\sum\limits\_{1\leq \ell\_1<\cdots<\ell\_{r-1}\leq n-1}[\det(\ell\_1,\cdots,\ell\_{r-1})]^2\right)\times\left(\sum\limits\_{1\leq \ell\_1<\cdots<\ell\_{r-1}\leq n-1}[\det(\ell\_1,\cdots,\ell\_{r-1},n)]^2\right)\\
> B&=&\left(\sum\limits\_{1\leq \ell\_1<\cdots<\ell\_{r}\leq n-1}[\det(\ell\_1,\cdots,\ell\_{r})]^2\right)\cdot\left(\sum\limits\_{1\leq \ell\_1<\cdots<\ell\_{r-2}\leq n-1}[\det(\ell\_1,\cdots,\ell\_{r-2},n)]^2\right)\\
> C&=& \left(\sum\limits\_{1\leq \ell\_1<\cdots<\ell\_{r-1}\leq n-1}\det(\ell\_1,\cdots,\ell\_{r-1})\cdot \det(\ell\_1,\cdots,\ell\_{r-1},n)\right)^2.
> \end{eqnarray\*}
>  Show that $A-B=C$.
> 
> 
> 


The cases $n=3,4$ can be verified by Mathematical.


The case $r=2$ with any $n\geq 3$ can be verified easily by a direct calculation, but how to prove this identity for general $r$ ? Any comments and suggestions are welcomed!



A Problem on Linear Algebra
The following proof is due to Yeping ZHANG:


Let $V$ be a $n$ dimensional Euclidean vector space. Let $(e\_1,\cdots,e\_n)$ be an orthogonal basis.


For any $k=1,\cdots,n$, let $\Lambda^k V$ the $k$-th exterior product of $V$. We equip $\Lambda^k V$ with a metric ${\lVert \cdot \rVert}\_{\Lambda^k V}$ such that $(e\_{l\_1}\wedge\cdots\wedge e\_{l\_k})\_{1\leq l\_1 < \cdots < l\_k\leq n}$ is an orthogonal basis. Let ${\langle \cdot,\cdot \rangle}\_{\Lambda^k V}$ be the associated inner product.


Let $v\_1,\cdots,v\_k\in V$, whose coodinates with respect to $(e\_1,\cdots,e\_n)$ are given by
\begin{equation}
v\_i = \sum\_{j=1}^n t\_{ij}e\_j \;,
\end{equation}
where $i = 1,\cdots,k$. Let $T=(t\_{ij})\_{1\leq i \leq k, 1\leq j \leq n}$ be the associated matrix. For $1\leq l\_1 < \cdots < l\_k\leq n$, set $T(l\_1,\cdots,l\_k)=(t\_{ij})\_{1\leq i \leq k, j = l\_1,\cdots l\_k}$. Then
\begin{equation}
v\_1 \wedge\cdots\wedge v\_k = \sum\_{1\leq l\_1 < \cdots < l\_k\leq n} \, \det \, T(l\_1,\cdots,l\_k) \, e\_{l\_1}\wedge\cdots\wedge e\_{l\_k} \;,
\end{equation}
in particular
\begin{equation}
{\lVert v\_1 \wedge\cdots\wedge v\_k \rVert}\_{\Lambda^k V}^2 = \sum\_{1\leq l\_1 < \cdots < l\_k\leq n} [ \, \det \, T(l\_1,\cdots,l\_k) \, ]^2 \;.
\end{equation}


With the above observation, we can interpret the question as the follows.


Set
\begin{equation}
x\_i = \sum\_{j=1}^n X\_{ij}e\_j \;,
\end{equation}
where $i=1,\cdots,n$. Let
\begin{align}
x\_{n-r+2} \wedge\cdots\wedge x\_n = \alpha + \beta \wedge e\_n \;,\nonumber\\
x\_{n-r+1} \wedge\cdots\wedge x\_n = \gamma + \delta \wedge e\_n \;,
\end{align}
with $\alpha,\beta,\gamma,\delta$ generated by $e\_1,\cdots,e\_{n-1}$.
We have
\begin{align}
A & = {\lVert \alpha \rVert}\_{\Lambda^{r-1} V}^2 \cdot {\lVert \delta \rVert}\_{\Lambda^{r-1} V}^2 \;,\nonumber\\
B & = {\lVert \gamma \rVert}\_{\Lambda^r V}^2 \cdot {\lVert \beta \rVert}\_{\Lambda^{r-2} V}^2 \;,\nonumber\\
C & = {\langle \alpha , \delta \rangle}\_{\Lambda^{r-1} V}^2\;.
\end{align}


We have
\begin{equation}
x\_i = y\_i + X\_{i,n}e\_n \;,
\end{equation}
where $y\_i = X\_{i,1}e\_1 + \cdots X\_{i,n-1}e\_{n-1}$. Then
\begin{align}
\alpha & = y\_{n-r+2}\wedge\cdots\wedge y\_n \;,\nonumber\\
\beta & = \sum\_{l=n-r+2}^n (-1)^{n-l} X\_{l,n} y\_{n-r+2}\wedge\cdots\wedge y\_{l-1}\wedge y\_{l+1}\wedge\cdots\wedge y\_n \;,\nonumber\\
\gamma & = y\_{n-r+1}\wedge\cdots\wedge y\_n \;,\nonumber\\
\delta & = \sum\_{l=n-r+1}^n (-1)^{n-l} X\_{l,n} y\_{n-r+1}\wedge\cdots\wedge y\_{l-1}\wedge y\_{l+1}\wedge\cdots\wedge y\_n \;.
\end{align}


Case $1$. If one of $y\_{n-r+2},\cdots,y\_n$ is zero, we have
\begin{equation}
\alpha = \gamma = 0 \;,
\end{equation}
trivially, we get $A=B+C$.


Case 2. If $y\_{n-r+1}=0$, we have
\begin{align}
\gamma & = 0 \;,\nonumber\\
\delta & = (-1)^{r-1}X\_{n-r+1,n}\alpha \;.
\end{align}
Thus
\begin{align}
A & = X\_{n-r+1,n}^2 {\lVert \alpha \rVert}\_{\Lambda^{r-1} V}^4 \;,\nonumber\\
B & = 0 \;,\nonumber\\
C & = X\_{n-r+1,n}^2 {\lVert \alpha \rVert}\_{\Lambda^{r-1} V}^4 \;.
\end{align}
We find $A=B+C$.


Case 3. We suppose that none of $y\_{n-r+1},\cdots,y\_n$ is zero.


Without changing $\alpha,\beta,\gamma,\delta$, we can replace $x\_i$ by $x\_i - tx\_j$ for any $j>i$, any $t\in\mathbb{R}$. Thus we can suppose that
\begin{equation}
\langle y\_i,y\_j \rangle = 0 \;,
\end{equation}
for any $i \neq j$. Since $A,B,C$ are homogenious functions of $y\_{n-r+1},\cdots,y\_n$, we can suppose that $\lVert y\_i \rVert = 1$.


With the above assumptions, we have
\begin{align}
& {\lVert \alpha \rVert}\_{\Lambda^{r-1} V}^2 = 1 \;,\nonumber\\
& {\lVert \beta \rVert}\_{\Lambda^{r-2} V}^2 = \sum\_{l=n-r+2}^n X\_{l,n}^2 \;,\nonumber\\
& {\lVert \gamma \rVert}\_{\Lambda^{r} V}^2 = 1 \;,\nonumber\\
& {\lVert \delta \rVert}\_{\Lambda^{r-1} V}^2 = \sum\_{l=n-r+1}^n X\_{l,n}^2 \;,\nonumber\\
& {\langle \alpha , \delta \rangle}\_{\Lambda^{r-1} V}^2 = X\_{n-r+1,n}^2 \;.
\end{align}
Hence $A = B + C$.


