# A Problem on Linear Algebra

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!

• this seems to be a moment problem in disguise? Commented Nov 9, 2013 at 14:49
• I've seen some determinant identities that I'd classify as "lost to time" ... you could track down references 11 and 14 in math.ubc.ca/~gerg/papers/downloads/AAIMHNIE.pdf and scan through them for inspiration. Commented Nov 9, 2013 at 21:02
• what does det$(l_1,..,l_k,n)$ mean? the question is about the last $n$, a number. Commented Nov 11, 2013 at 14:41
• All the $\ell_j$ are numbers too.... $\det(\ell_1,\dots,\ell_k,n)$ means the same thing as $\det(\ell_1,\dots,\ell_k,\ell_{k+1})$ where $\ell_{k+1}$ happens to equal $n$. Commented Nov 11, 2013 at 17:59
• My friend Yeping ZHANG had already proved this result last night using Wedge product. I will email to you latter. Thank you very much for the answers. Commented Nov 12, 2013 at 1:00

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 $$v_i = \sum_{j=1}^n t_{ij}e_j \;,$$ 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 $$v_1 \wedge\cdots\wedge v_k = \sum_{1\leq l_1 < \cdots < l_k\leq n} \, \det \, T(l_1,\cdots,l_k) \, e_{l_1}\wedge\cdots\wedge e_{l_k} \;,$$ in particular $${\lVert v_1 \wedge\cdots\wedge v_k \rVert}_{\Lambda^k V}^2 = \sum_{1\leq l_1 < \cdots < l_k\leq n} [ \, \det \, T(l_1,\cdots,l_k) \, ]^2 \;.$$

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

Set $$x_i = \sum_{j=1}^n X_{ij}e_j \;,$$ 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 $$x_i = y_i + X_{i,n}e_n \;,$$ 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 $$\alpha = \gamma = 0 \;,$$ 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 $$\langle y_i,y_j \rangle = 0 \;,$$ 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 $A = B + C$.