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An identity of determinant
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Let $L$ be an $ntimes n$ matrix and $W= operatornamediag(w_1,cdots,w_n)$.
Show that $det(I-WL)= sum_Ssubset[n] det(L[S])(-1)^w^S$. Where $I$ is the $ntimes n$ identity matrix, $L[S]$ is the principal submatrix of $L$ whose rows and columns are indexed by $S$ and $w^S = prod_iin S w_i$.
I am thinking there is some way to apply Cauchy-Binet in this case.
linear-algebra matrices determinant
edited Aug 2 at 18:15
Bernard
110k635102
asked Aug 2 at 18:08
Sudipta Roy
1517
add a comment |Â
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Let $L$ be an $ntimes n$ matrix and $W= operatornamediag(w_1,cdots,w_n)$.
Show that $det(I-WL)= sum_Ssubset[n] det(L[S])(-1)^w^S$. Where $I$ is the $ntimes n$ identity matrix, $L[S]$ is the principal submatrix of $L$ whose rows and columns are indexed by $S$ and $w^S = prod_iin S w_i$.
I am thinking there is some way to apply Cauchy-Binet in this case.
linear-algebra matrices determinant
edited Aug 2 at 18:15
Bernard
110k635102
asked Aug 2 at 18:08
Sudipta Roy
1517
First prove that $detleft(I + Aright) = sumlimits_S subseteq left[nright] detleft(Aleft[Sright]right)$ for any $n times n$-matrix $A$. (This is Proposition 1 in math.stackexchange.com/a/1752326 where I give a reference to a detailed proof.) Now apply it to $A = - WL$ and simplify $detleft(left(WLright)left[Sright]right)$ to $left(-1right)^ w^S detleft(Lleft[Sright]right)$ (because $- WL$ is obtained from $L$ by multiplying each row with one of the $-w_i$).
– darij grinberg
Aug 4 at 16:52
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
Let $L$ be an $ntimes n$ matrix and $W= operatornamediag(w_1,cdots,w_n)$.
Show that $det(I-WL)= sum_Ssubset[n] det(L[S])(-1)^w^S$. Where $I$ is the $ntimes n$ identity matrix, $L[S]$ is the principal submatrix of $L$ whose rows and columns are indexed by $S$ and $w^S = prod_iin S w_i$.
I am thinking there is some way to apply Cauchy-Binet in this case.
linear-algebra matrices determinant
edited Aug 2 at 18:15
Bernard
110k635102
asked Aug 2 at 18:08
Sudipta Roy
1517
Let $L$ be an $ntimes n$ matrix and $W= operatornamediag(w_1,cdots,w_n)$.
Show that $det(I-WL)= sum_Ssubset[n] det(L[S])(-1)^w^S$. Where $I$ is the $ntimes n$ identity matrix, $L[S]$ is the principal submatrix of $L$ whose rows and columns are indexed by $S$ and $w^S = prod_iin S w_i$.
I am thinking there is some way to apply Cauchy-Binet in this case.
linear-algebra matrices determinant
edited Aug 2 at 18:15
Bernard
110k635102
asked Aug 2 at 18:08
Sudipta Roy
1517
edited Aug 2 at 18:15
Bernard
110k635102
edited Aug 2 at 18:15
Bernard
110k635102
edited Aug 2 at 18:15
Bernard
110k635102
110k635102
asked Aug 2 at 18:08
Sudipta Roy
1517
asked Aug 2 at 18:08
Sudipta Roy
1517
asked Aug 2 at 18:08
Sudipta Roy
1517
1517
First prove that $detleft(I + Aright) = sumlimits_S subseteq left[nright] detleft(Aleft[Sright]right)$ for any $n times n$-matrix $A$. (This is Proposition 1 in math.stackexchange.com/a/1752326 where I give a reference to a detailed proof.) Now apply it to $A = - WL$ and simplify $detleft(left(WLright)left[Sright]right)$ to $left(-1right)^ w^S detleft(Lleft[Sright]right)$ (because $- WL$ is obtained from $L$ by multiplying each row with one of the $-w_i$).
– darij grinberg
Aug 4 at 16:52
add a comment |Â
First prove that $detleft(I + Aright) = sumlimits_S subseteq left[nright] detleft(Aleft[Sright]right)$ for any $n times n$-matrix $A$. (This is Proposition 1 in math.stackexchange.com/a/1752326 where I give a reference to a detailed proof.) Now apply it to $A = - WL$ and simplify $detleft(left(WLright)left[Sright]right)$ to $left(-1right)^ w^S detleft(Lleft[Sright]right)$ (because $- WL$ is obtained from $L$ by multiplying each row with one of the $-w_i$).
– darij grinberg
Aug 4 at 16:52
First prove that $detleft(I + Aright) = sumlimits_S subseteq left[nright] detleft(Aleft[Sright]right)$ for any $n times n$-matrix $A$. (This is Proposition 1 in math.stackexchange.com/a/1752326 where I give a reference to a detailed proof.) Now apply it to $A = - WL$ and simplify $detleft(left(WLright)left[Sright]right)$ to $left(-1right)^ w^S detleft(Lleft[Sright]right)$ (because $- WL$ is obtained from $L$ by multiplying each row with one of the $-w_i$).
– darij grinberg
Aug 4 at 16:52
First prove that $detleft(I + Aright) = sumlimits_S subseteq left[nright] detleft(Aleft[Sright]right)$ for any $n times n$-matrix $A$. (This is Proposition 1 in math.stackexchange.com/a/1752326 where I give a reference to a detailed proof.) Now apply it to $A = - WL$ and simplify $detleft(left(WLright)left[Sright]right)$ to $left(-1right)^ w^S detleft(Lleft[Sright]right)$ (because $- WL$ is obtained from $L$ by multiplying each row with one of the $-w_i$).
– darij grinberg
Aug 4 at 16:52
add a comment |Â
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First prove that $detleft(I + Aright) = sumlimits_S subseteq left[nright] detleft(Aleft[Sright]right)$ for any $n times n$-matrix $A$. (This is Proposition 1 in math.stackexchange.com/a/1752326 where I give a reference to a detailed proof.) Now apply it to $A = - WL$ and simplify $detleft(left(WLright)left[Sright]right)$ to $left(-1right)^ w^S detleft(Lleft[Sright]right)$ (because $- WL$ is obtained from $L$ by multiplying each row with one of the $-w_i$).
– darij grinberg
Aug 4 at 16:52