Calculating determinant of this positive semidefinite matrix is positive or not.
I am dealing with the following problem:
Suppose we have the vector $v=(1,x_1,cdots ,x_n,x_1x_2,cdots ,x_n^2,cdots ,x_n^k)$. Elements of $v$ are monomials up to degree $k$.
I know that for any choice of $x_iin mathbb R^n$ the matrix $v^Tv$ is positive semidefinite. Question is do the following matrix has positive determinant?
$A=sum v^Tv$
Such that the summation is over all vectors $v$ with exactly $d$ of $x_i$ are $1$ and the other are $0$. And $din {1,cdots ,n-1}$.
I think the determinant is positive, could any one prove it or give a counter example?
linear-algebra
asked Nov 27 at 0:08
GhD
795310
add a comment |
I am dealing with the following problem:
Suppose we have the vector $v=(1,x_1,cdots ,x_n,x_1x_2,cdots ,x_n^2,cdots ,x_n^k)$. Elements of $v$ are monomials up to degree $k$.
I know that for any choice of $x_iin mathbb R^n$ the matrix $v^Tv$ is positive semidefinite. Question is do the following matrix has positive determinant?
$A=sum v^Tv$
Such that the summation is over all vectors $v$ with exactly $d$ of $x_i$ are $1$ and the other are $0$. And $din {1,cdots ,n-1}$.
I think the determinant is positive, could any one prove it or give a counter example?
linear-algebra
asked Nov 27 at 0:08
GhD
795310
This cannot be true independently of the values of $d$ and $k$. You have only $dbinom{n}{d}$ many vectors $v$; thus, the rank of $A$ is $leq dbinom{n}{d}$. If this rank is $leq$ to the dimension of your vector space, then $A$ cannot have positive determinant.
– darij grinberg
Nov 27 at 2:21
add a comment |
I am dealing with the following problem:
Suppose we have the vector $v=(1,x_1,cdots ,x_n,x_1x_2,cdots ,x_n^2,cdots ,x_n^k)$. Elements of $v$ are monomials up to degree $k$.
I know that for any choice of $x_iin mathbb R^n$ the matrix $v^Tv$ is positive semidefinite. Question is do the following matrix has positive determinant?
$A=sum v^Tv$
Such that the summation is over all vectors $v$ with exactly $d$ of $x_i$ are $1$ and the other are $0$. And $din {1,cdots ,n-1}$.
I think the determinant is positive, could any one prove it or give a counter example?
linear-algebra
asked Nov 27 at 0:08
GhD
795310
I am dealing with the following problem:
Suppose we have the vector $v=(1,x_1,cdots ,x_n,x_1x_2,cdots ,x_n^2,cdots ,x_n^k)$. Elements of $v$ are monomials up to degree $k$.
I know that for any choice of $x_iin mathbb R^n$ the matrix $v^Tv$ is positive semidefinite. Question is do the following matrix has positive determinant?
$A=sum v^Tv$
Such that the summation is over all vectors $v$ with exactly $d$ of $x_i$ are $1$ and the other are $0$. And $din {1,cdots ,n-1}$.
I think the determinant is positive, could any one prove it or give a counter example?
linear-algebra
linear-algebra
asked Nov 27 at 0:08
GhD
795310
asked Nov 27 at 0:08
GhD
795310
edited Nov 27 at 0:26
asked Nov 27 at 0:08
GhD
795310
asked Nov 27 at 0:08
GhD
795310
asked Nov 27 at 0:08
GhD
795310
795310
This cannot be true independently of the values of $d$ and $k$. You have only $dbinom{n}{d}$ many vectors $v$; thus, the rank of $A$ is $leq dbinom{n}{d}$. If this rank is $leq$ to the dimension of your vector space, then $A$ cannot have positive determinant.
– darij grinberg
Nov 27 at 2:21
add a comment |
This cannot be true independently of the values of $d$ and $k$. You have only $dbinom{n}{d}$ many vectors $v$; thus, the rank of $A$ is $leq dbinom{n}{d}$. If this rank is $leq$ to the dimension of your vector space, then $A$ cannot have positive determinant.
– darij grinberg
Nov 27 at 2:21
This cannot be true independently of the values of $d$ and $k$. You have only $dbinom{n}{d}$ many vectors $v$; thus, the rank of $A$ is $leq dbinom{n}{d}$. If this rank is $leq$ to the dimension of your vector space, then $A$ cannot have positive determinant.
– darij grinberg
Nov 27 at 2:21
This cannot be true independently of the values of $d$ and $k$. You have only $dbinom{n}{d}$ many vectors $v$; thus, the rank of $A$ is $leq dbinom{n}{d}$. If this rank is $leq$ to the dimension of your vector space, then $A$ cannot have positive determinant.
– darij grinberg
Nov 27 at 2:21
add a comment |
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This cannot be true independently of the values of $d$ and $k$. You have only $dbinom{n}{d}$ many vectors $v$; thus, the rank of $A$ is $leq dbinom{n}{d}$. If this rank is $leq$ to the dimension of your vector space, then $A$ cannot have positive determinant.
– darij grinberg
Nov 27 at 2:21