Assume that $V$ is a $n$ -dimensional inner product space. Basis $(e_1,...,e_n)$, $(f_1,...,f_n)$ are said to be dual if $\langle e_i, f_j\rangle=0$ for $i\neq j$, $i,j=1,...,n$ and $\langle e_i, f_i\rangle=1$ for $i=1,...,n$.

The Gram determinant of vectors $x_1,...,x_r\in V$ is the determinant $G(x_1,...,x_r)=\det[\langle x_i,x_j\rangle]$.

For dual basis $(e_1,...,e_n)$, $(f_1,...,ef_n)$ we have: $f_i=\sum_{j=1}^n g^{i j}e_j$ for $i=1,...,n$, where $g^{ij}$ are elements of the inverse matrix to $[g_{ij}]$ with $g_{ij}=\langle e_i,e_j\rangle$.

Is it true for fixed $p\in \{1,...,n-1 \}$ the formula $$ \frac{G(e_1,..,e_p)}{G(f_{p+1},...,f_{n})}=G(e_1,...,e_n) $$ and how to prove it.

Assume that $V$ is a $n$ -dimensional inner product space. Basis $(e_1,...,e_n)$, $(f_1,...,f_n)$ are said to be dual if $\langle e_i, f_j\rangle=0$ for $i\neq j$, $i,j=1,...,n$ and $\langle e_i, f_i\rangle=1$ for $i=1,...,n$.

The Gram determinant of vectors $x_1,...,x_r\in V$ is the determinant $G(x_1,...,x_r)=\det[\langle x_i,x_j\rangle]$.

For dual basis $(e_1,...,e_n)$, $(f_1,...,f_n)$ we have: $f_i=\sum_{j=1}^n g^{i j}e_j$ for $i=1,...,n$, where $g^{ij}$ are elements of the inverse matrix to $[g_{ij}]$ with $g_{ij}=\langle e_i,e_j\rangle$.

Is it true for fixed $p\in \{1,...,n-1 \}$ the formula $$ \frac{G(e_1,..,e_p)}{G(f_{p+1},...,f_{n})}=G(e_1,...,e_n) $$ and how to prove it.

Assume that $V$ is a $n$ -dimensional inner product space. Basis $(e_1,...,e_n)$, $(f_1,...,f_n)$ are said to be dual if $ <e_i, f_j>=0$ for $i\neq j$, $i,j=1,...,n$ and $<e_i, f_i>=1$ for $i=1,...,n$.

The Gram determinant of vectors $x_1,...,x_r\in V$ is the determinant $G(x_1,...,x_r)=det[<x_i,x_j>]$.

For dual basis $(e_1,...,e_n)$, $(f_1,...,ef_n)$ we have: $f_i=\sum_{j=1}^n g^{i j}e_j$ for $i=1,...,n$, where $g^{ij}$ are elements of the inverse matrix to $[g_{ij}]$ with $g_{ij}=<e_i,e_j>$.

Is it true for fixed $p\in \{1,...,n-1 \}$ the formula $$ \frac{G(e_1,..,e_p)}{G(f_{p+1},...,f_{n})}=G(e_1,...,e_n) $$ and how to prove it.

Assume that $V$ is a $n$ -dimensional inner product space. Basis $(e_1,...,e_n)$, $(f_1,...,f_n)$ are said to be dual if $\langle e_i, f_j\rangle=0$ for $i\neq j$, $i,j=1,...,n$ and $\langle e_i, f_i\rangle=1$ for $i=1,...,n$.

The Gram determinant of vectors $x_1,...,x_r\in V$ is the determinant $G(x_1,...,x_r)=\det[\langle x_i,x_j\rangle]$.

For dual basis $(e_1,...,e_n)$, $(f_1,...,ef_n)$ we have: $f_i=\sum_{j=1}^n g^{i j}e_j$ for $i=1,...,n$, where $g^{ij}$ are elements of the inverse matrix to $[g_{ij}]$ with $g_{ij}=\langle e_i,e_j\rangle$.

Is it true for fixed $p\in \{1,...,n-1 \}$ the formula $$ \frac{G(e_1,..,e_p)}{G(f_{p+1},...,f_{n})}=G(e_1,...,e_n) $$ and how to prove it.