Mixing
Derivate of the cofactor and the determinant
$begingroup$
$newcommand{Cof}{operatorname{Cof}}$
$newcommand{Det}{operatorname{Det}}$
$newcommand{id}{operatorname{Id}}$
$newcommand{tr}{operatorname{Tr}}$
Let $A(t)$ be a smooth path in $M_d(mathbb{R})$, $A(0)=A,dot A(0)=B$.
By differentiation the identity $(Cof A)^T circ A=Det A cdot id$, one gets
$$(*) , , big(d(Cof)_A(B)big)^T circ A + (Cof A)^T circ B = tr (Cof A)^T B) cdot id= langle Cof A , Brangle cdot id$$
(The derivative of the determinant is known as Jacobi's formula).
From this, at least in the case when $A$ is invertible we can deduce that
$$ big(d(Cof)_A(B)big)^T = big(langle Cof A , Brangle cdot id -(Cof A)^T circ B big)A^{-1},$$ hence
$$ d(Cof)_A(B) = (A^{T})^{-1}big(langle Cof A , Brangle cdot id - B^T circ Cof A big) $$
Questions:
Does equation $(*)$ uniquely determine $d(Cof)_A(B)$ even when $A$ is singular? Is there a closed formula for $d(Cof)_A(B)$ in this case?
Is there a more "direct way" to calculate $d(Cof)_A(B)$? (without relying on Jacobi's formula)
Remark: As a corollary from $(*)$ we get $ tr bigg( big(d(Cof)_A(B)big)^T circ A bigg) = (d-1) tr (Cof A)^T B) $
This is interesting since the cofactor essentially measures the action of the linear map $A$ on $d-1$ dimensional parallelepiped, so maybe there is a "geometric" way to see this immediately.
linear-algebra geometry determinant matrix-calculus exterior-algebra
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
$endgroup$
add a comment |
$begingroup$
$newcommand{Cof}{operatorname{Cof}}$
$newcommand{Det}{operatorname{Det}}$
$newcommand{id}{operatorname{Id}}$
$newcommand{tr}{operatorname{Tr}}$
Let $A(t)$ be a smooth path in $M_d(mathbb{R})$, $A(0)=A,dot A(0)=B$.
By differentiation the identity $(Cof A)^T circ A=Det A cdot id$, one gets
$$(*) , , big(d(Cof)_A(B)big)^T circ A + (Cof A)^T circ B = tr (Cof A)^T B) cdot id= langle Cof A , Brangle cdot id$$
(The derivative of the determinant is known as Jacobi's formula).
From this, at least in the case when $A$ is invertible we can deduce that
$$ big(d(Cof)_A(B)big)^T = big(langle Cof A , Brangle cdot id -(Cof A)^T circ B big)A^{-1},$$ hence
$$ d(Cof)_A(B) = (A^{T})^{-1}big(langle Cof A , Brangle cdot id - B^T circ Cof A big) $$
Questions:
Does equation $(*)$ uniquely determine $d(Cof)_A(B)$ even when $A$ is singular? Is there a closed formula for $d(Cof)_A(B)$ in this case?
Is there a more "direct way" to calculate $d(Cof)_A(B)$? (without relying on Jacobi's formula)
Remark: As a corollary from $(*)$ we get $ tr bigg( big(d(Cof)_A(B)big)^T circ A bigg) = (d-1) tr (Cof A)^T B) $
This is interesting since the cofactor essentially measures the action of the linear map $A$ on $d-1$ dimensional parallelepiped, so maybe there is a "geometric" way to see this immediately.
linear-algebra geometry determinant matrix-calculus exterior-algebra
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
$endgroup$
add a comment |
$begingroup$
$newcommand{Cof}{operatorname{Cof}}$
$newcommand{Det}{operatorname{Det}}$
$newcommand{id}{operatorname{Id}}$
$newcommand{tr}{operatorname{Tr}}$
Let $A(t)$ be a smooth path in $M_d(mathbb{R})$, $A(0)=A,dot A(0)=B$.
By differentiation the identity $(Cof A)^T circ A=Det A cdot id$, one gets
$$(*) , , big(d(Cof)_A(B)big)^T circ A + (Cof A)^T circ B = tr (Cof A)^T B) cdot id= langle Cof A , Brangle cdot id$$
(The derivative of the determinant is known as Jacobi's formula).
From this, at least in the case when $A$ is invertible we can deduce that
$$ big(d(Cof)_A(B)big)^T = big(langle Cof A , Brangle cdot id -(Cof A)^T circ B big)A^{-1},$$ hence
$$ d(Cof)_A(B) = (A^{T})^{-1}big(langle Cof A , Brangle cdot id - B^T circ Cof A big) $$
Questions:
Does equation $(*)$ uniquely determine $d(Cof)_A(B)$ even when $A$ is singular? Is there a closed formula for $d(Cof)_A(B)$ in this case?
Is there a more "direct way" to calculate $d(Cof)_A(B)$? (without relying on Jacobi's formula)
Remark: As a corollary from $(*)$ we get $ tr bigg( big(d(Cof)_A(B)big)^T circ A bigg) = (d-1) tr (Cof A)^T B) $
This is interesting since the cofactor essentially measures the action of the linear map $A$ on $d-1$ dimensional parallelepiped, so maybe there is a "geometric" way to see this immediately.
linear-algebra geometry determinant matrix-calculus exterior-algebra
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
$endgroup$
$newcommand{Cof}{operatorname{Cof}}$
$newcommand{Det}{operatorname{Det}}$
$newcommand{id}{operatorname{Id}}$
$newcommand{tr}{operatorname{Tr}}$
Let $A(t)$ be a smooth path in $M_d(mathbb{R})$, $A(0)=A,dot A(0)=B$.
By differentiation the identity $(Cof A)^T circ A=Det A cdot id$, one gets
$$(*) , , big(d(Cof)_A(B)big)^T circ A + (Cof A)^T circ B = tr (Cof A)^T B) cdot id= langle Cof A , Brangle cdot id$$
(The derivative of the determinant is known as Jacobi's formula).
From this, at least in the case when $A$ is invertible we can deduce that
$$ big(d(Cof)_A(B)big)^T = big(langle Cof A , Brangle cdot id -(Cof A)^T circ B big)A^{-1},$$ hence
$$ d(Cof)_A(B) = (A^{T})^{-1}big(langle Cof A , Brangle cdot id - B^T circ Cof A big) $$
Questions:
Does equation $(*)$ uniquely determine $d(Cof)_A(B)$ even when $A$ is singular? Is there a closed formula for $d(Cof)_A(B)$ in this case?
Is there a more "direct way" to calculate $d(Cof)_A(B)$? (without relying on Jacobi's formula)
Remark: As a corollary from $(*)$ we get $ tr bigg( big(d(Cof)_A(B)big)^T circ A bigg) = (d-1) tr (Cof A)^T B) $
This is interesting since the cofactor essentially measures the action of the linear map $A$ on $d-1$ dimensional parallelepiped, so maybe there is a "geometric" way to see this immediately.
linear-algebra geometry determinant matrix-calculus exterior-algebra
linear-algebra geometry determinant matrix-calculus exterior-algebra
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
asked Feb 9 '17 at 14:56
Asaf ShacharAsaf Shachar
5,79031145
5,79031145
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $Cof_{ij}(A)=(-1)^{i+j}det(Delta_{ij}(A))$, where $Delta_{ij}(A)$ is the submatrix obtained from $A$ by deleting row $i$ and column $j$. Notice that $Delta_{ij}(cdot)$ is a linear transformation. Let $Adj(D)$ be the adjugate matrix of $D$.
Let ${e_1,ldots,e_d}$ be the canonical basis of $mathbb{R}^d$.
Now, $d(Cof)_A(e_me_n^t)=lim_{hrightarrow 0}dfrac{Cof(A+h e_me_n^t)-Cof(A)}{h}$.
$Cof_{ij}(A+h e_me_n^t)=(-1)^{i+j}det(Delta_{ij}(A+h e_me_n^t))=$
$(-1)^{i+j}det(Delta_{ij}(A)+h Delta_{ij}(e_me_n^t))=
(-1)^{i+j}det(Delta_{ij}(A)+h uv^t)=$
By the matrix determinant lemma,
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h (v^tAdj(Delta_{ij}(A))u)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))uv^t)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$
$=Cof_{ij}(A)+h (-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$.
Thus, $d(Cof)_A(e_me_n^t)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t)))_{i,j}$.
Since $B=sum_{m,n=1}^db_{m,n}e_me_n^t$ then $d(Cof)_A(B)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(B)))_{i,j}$.
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
$endgroup$
$begingroup$
Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(Cof)_A(B)$ when $A$ is singular?
$endgroup$
– Asaf Shachar
Feb 13 '17 at 16:40
$begingroup$
$Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question.
$endgroup$
– Daniel
Feb 13 '17 at 16:42
$begingroup$
If $rank(A)leq d-2$ then $Cof(A)=0$ and your formula becomes $big(d(Cof)_A(B)big)^t circ A =0$. I cannot see a way to determine $big(d(Cof)_A(B)big)^t $ from the equality $XA=0$, there are infinitely many solutions.
$endgroup$
– Daniel
Feb 13 '17 at 16:54
$begingroup$
You are right. Thanks again!
$endgroup$
– Asaf Shachar
Feb 13 '17 at 17:00
$begingroup$
You are welcome.
$endgroup$
– Daniel
Feb 13 '17 at 17:03
add a comment |
$begingroup$
As mentioned by Daniel, when $operatorname{rank}(A) le d-2$, $operatorname{Cof}(A)=0$, hence equation $(*)$ becomes
$$ big(d(operatorname{Cof})_A(B)big)^T A=0$$
However, the equation $XA=0$ has infinitely many solutions, as can be seen as follows:
Look at the linear map $T:M_d to M_d$, defined by $T(X)=X^TA$. $T$ is not surjective, since every matrix in its image has rank smaller than $d-1$. Thus, $T$ is not injective.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Let $Cof_{ij}(A)=(-1)^{i+j}det(Delta_{ij}(A))$, where $Delta_{ij}(A)$ is the submatrix obtained from $A$ by deleting row $i$ and column $j$. Notice that $Delta_{ij}(cdot)$ is a linear transformation. Let $Adj(D)$ be the adjugate matrix of $D$.
Let ${e_1,ldots,e_d}$ be the canonical basis of $mathbb{R}^d$.
Now, $d(Cof)_A(e_me_n^t)=lim_{hrightarrow 0}dfrac{Cof(A+h e_me_n^t)-Cof(A)}{h}$.
$Cof_{ij}(A+h e_me_n^t)=(-1)^{i+j}det(Delta_{ij}(A+h e_me_n^t))=$
$(-1)^{i+j}det(Delta_{ij}(A)+h Delta_{ij}(e_me_n^t))=
(-1)^{i+j}det(Delta_{ij}(A)+h uv^t)=$
By the matrix determinant lemma,
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h (v^tAdj(Delta_{ij}(A))u)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))uv^t)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$
$=Cof_{ij}(A)+h (-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$.
Thus, $d(Cof)_A(e_me_n^t)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t)))_{i,j}$.
Since $B=sum_{m,n=1}^db_{m,n}e_me_n^t$ then $d(Cof)_A(B)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(B)))_{i,j}$.
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
$endgroup$
$begingroup$
Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(Cof)_A(B)$ when $A$ is singular?
$endgroup$
– Asaf Shachar
Feb 13 '17 at 16:40
$begingroup$
$Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question.
$endgroup$
– Daniel
Feb 13 '17 at 16:42
$begingroup$
If $rank(A)leq d-2$ then $Cof(A)=0$ and your formula becomes $big(d(Cof)_A(B)big)^t circ A =0$. I cannot see a way to determine $big(d(Cof)_A(B)big)^t $ from the equality $XA=0$, there are infinitely many solutions.
$endgroup$
– Daniel
Feb 13 '17 at 16:54
$begingroup$
You are right. Thanks again!
$endgroup$
– Asaf Shachar
Feb 13 '17 at 17:00
$begingroup$
You are welcome.
$endgroup$
– Daniel
Feb 13 '17 at 17:03
add a comment |
$begingroup$
Let $Cof_{ij}(A)=(-1)^{i+j}det(Delta_{ij}(A))$, where $Delta_{ij}(A)$ is the submatrix obtained from $A$ by deleting row $i$ and column $j$. Notice that $Delta_{ij}(cdot)$ is a linear transformation. Let $Adj(D)$ be the adjugate matrix of $D$.
Let ${e_1,ldots,e_d}$ be the canonical basis of $mathbb{R}^d$.
Now, $d(Cof)_A(e_me_n^t)=lim_{hrightarrow 0}dfrac{Cof(A+h e_me_n^t)-Cof(A)}{h}$.
$Cof_{ij}(A+h e_me_n^t)=(-1)^{i+j}det(Delta_{ij}(A+h e_me_n^t))=$
$(-1)^{i+j}det(Delta_{ij}(A)+h Delta_{ij}(e_me_n^t))=
(-1)^{i+j}det(Delta_{ij}(A)+h uv^t)=$
By the matrix determinant lemma,
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h (v^tAdj(Delta_{ij}(A))u)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))uv^t)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$
$=Cof_{ij}(A)+h (-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$.
Thus, $d(Cof)_A(e_me_n^t)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t)))_{i,j}$.
Since $B=sum_{m,n=1}^db_{m,n}e_me_n^t$ then $d(Cof)_A(B)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(B)))_{i,j}$.
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
$endgroup$
$begingroup$
Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(Cof)_A(B)$ when $A$ is singular?
$endgroup$
– Asaf Shachar
Feb 13 '17 at 16:40
$begingroup$
$Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question.
$endgroup$
– Daniel
Feb 13 '17 at 16:42
$begingroup$
If $rank(A)leq d-2$ then $Cof(A)=0$ and your formula becomes $big(d(Cof)_A(B)big)^t circ A =0$. I cannot see a way to determine $big(d(Cof)_A(B)big)^t $ from the equality $XA=0$, there are infinitely many solutions.
$endgroup$
– Daniel
Feb 13 '17 at 16:54
$begingroup$
You are right. Thanks again!
$endgroup$
– Asaf Shachar
Feb 13 '17 at 17:00
$begingroup$
You are welcome.
$endgroup$
– Daniel
Feb 13 '17 at 17:03
add a comment |
$begingroup$
Let $Cof_{ij}(A)=(-1)^{i+j}det(Delta_{ij}(A))$, where $Delta_{ij}(A)$ is the submatrix obtained from $A$ by deleting row $i$ and column $j$. Notice that $Delta_{ij}(cdot)$ is a linear transformation. Let $Adj(D)$ be the adjugate matrix of $D$.
Let ${e_1,ldots,e_d}$ be the canonical basis of $mathbb{R}^d$.
Now, $d(Cof)_A(e_me_n^t)=lim_{hrightarrow 0}dfrac{Cof(A+h e_me_n^t)-Cof(A)}{h}$.
$Cof_{ij}(A+h e_me_n^t)=(-1)^{i+j}det(Delta_{ij}(A+h e_me_n^t))=$
$(-1)^{i+j}det(Delta_{ij}(A)+h Delta_{ij}(e_me_n^t))=
(-1)^{i+j}det(Delta_{ij}(A)+h uv^t)=$
By the matrix determinant lemma,
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h (v^tAdj(Delta_{ij}(A))u)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))uv^t)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$
$=Cof_{ij}(A)+h (-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$.
Thus, $d(Cof)_A(e_me_n^t)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t)))_{i,j}$.
Since $B=sum_{m,n=1}^db_{m,n}e_me_n^t$ then $d(Cof)_A(B)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(B)))_{i,j}$.
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
$endgroup$
Let $Cof_{ij}(A)=(-1)^{i+j}det(Delta_{ij}(A))$, where $Delta_{ij}(A)$ is the submatrix obtained from $A$ by deleting row $i$ and column $j$. Notice that $Delta_{ij}(cdot)$ is a linear transformation. Let $Adj(D)$ be the adjugate matrix of $D$.
Let ${e_1,ldots,e_d}$ be the canonical basis of $mathbb{R}^d$.
Now, $d(Cof)_A(e_me_n^t)=lim_{hrightarrow 0}dfrac{Cof(A+h e_me_n^t)-Cof(A)}{h}$.
$Cof_{ij}(A+h e_me_n^t)=(-1)^{i+j}det(Delta_{ij}(A+h e_me_n^t))=$
$(-1)^{i+j}det(Delta_{ij}(A)+h Delta_{ij}(e_me_n^t))=
(-1)^{i+j}det(Delta_{ij}(A)+h uv^t)=$
By the matrix determinant lemma,
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h (v^tAdj(Delta_{ij}(A))u)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))uv^t)$
$=(-1)^{i+j}det(Delta_{ij}(A))+(-1)^{i+j}h tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$
$=Cof_{ij}(A)+h (-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t))$.
Thus, $d(Cof)_A(e_me_n^t)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(e_me_n^t)))_{i,j}$.
Since $B=sum_{m,n=1}^db_{m,n}e_me_n^t$ then $d(Cof)_A(B)=((-1)^{i+j} tr(Adj(Delta_{ij}(A))Delta_{ij}(B)))_{i,j}$.
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
answered Feb 13 '17 at 16:33
DanielDaniel
4,19911022
4,19911022
$begingroup$
Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(Cof)_A(B)$ when $A$ is singular?
$endgroup$
– Asaf Shachar
Feb 13 '17 at 16:40
$begingroup$
$Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question.
$endgroup$
– Daniel
Feb 13 '17 at 16:42
$begingroup$
If $rank(A)leq d-2$ then $Cof(A)=0$ and your formula becomes $big(d(Cof)_A(B)big)^t circ A =0$. I cannot see a way to determine $big(d(Cof)_A(B)big)^t $ from the equality $XA=0$, there are infinitely many solutions.
$endgroup$
– Daniel
Feb 13 '17 at 16:54
$begingroup$
You are right. Thanks again!
$endgroup$
– Asaf Shachar
Feb 13 '17 at 17:00
$begingroup$
You are welcome.
$endgroup$
– Daniel
Feb 13 '17 at 17:03
add a comment |
$begingroup$
Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(Cof)_A(B)$ when $A$ is singular?
$endgroup$
– Asaf Shachar
Feb 13 '17 at 16:40
$begingroup$
$Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question.
$endgroup$
– Daniel
Feb 13 '17 at 16:42
$begingroup$
If $rank(A)leq d-2$ then $Cof(A)=0$ and your formula becomes $big(d(Cof)_A(B)big)^t circ A =0$. I cannot see a way to determine $big(d(Cof)_A(B)big)^t $ from the equality $XA=0$, there are infinitely many solutions.
$endgroup$
– Daniel
Feb 13 '17 at 16:54
$begingroup$
You are right. Thanks again!
$endgroup$
– Asaf Shachar
Feb 13 '17 at 17:00
$begingroup$
You are welcome.
$endgroup$
– Daniel
Feb 13 '17 at 17:03
$begingroup$
Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(Cof)_A(B)$ when $A$ is singular?
$endgroup$
– Asaf Shachar
Feb 13 '17 at 16:40
$begingroup$
Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(Cof)_A(B)$ when $A$ is singular?
$endgroup$
– Asaf Shachar
Feb 13 '17 at 16:40
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$Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question.
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– Daniel
Feb 13 '17 at 16:42
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$Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question.
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– Daniel
Feb 13 '17 at 16:42
$begingroup$
If $rank(A)leq d-2$ then $Cof(A)=0$ and your formula becomes $big(d(Cof)_A(B)big)^t circ A =0$. I cannot see a way to determine $big(d(Cof)_A(B)big)^t $ from the equality $XA=0$, there are infinitely many solutions.
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– Daniel
Feb 13 '17 at 16:54
$begingroup$
If $rank(A)leq d-2$ then $Cof(A)=0$ and your formula becomes $big(d(Cof)_A(B)big)^t circ A =0$. I cannot see a way to determine $big(d(Cof)_A(B)big)^t $ from the equality $XA=0$, there are infinitely many solutions.
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– Daniel
Feb 13 '17 at 16:54
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You are right. Thanks again!
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– Asaf Shachar
Feb 13 '17 at 17:00
$begingroup$
You are right. Thanks again!
$endgroup$
– Asaf Shachar
Feb 13 '17 at 17:00
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You are welcome.
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– Daniel
Feb 13 '17 at 17:03
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You are welcome.
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– Daniel
Feb 13 '17 at 17:03
add a comment |
$begingroup$
As mentioned by Daniel, when $operatorname{rank}(A) le d-2$, $operatorname{Cof}(A)=0$, hence equation $(*)$ becomes
$$ big(d(operatorname{Cof})_A(B)big)^T A=0$$
However, the equation $XA=0$ has infinitely many solutions, as can be seen as follows:
Look at the linear map $T:M_d to M_d$, defined by $T(X)=X^TA$. $T$ is not surjective, since every matrix in its image has rank smaller than $d-1$. Thus, $T$ is not injective.
$endgroup$
add a comment |
$begingroup$
As mentioned by Daniel, when $operatorname{rank}(A) le d-2$, $operatorname{Cof}(A)=0$, hence equation $(*)$ becomes
$$ big(d(operatorname{Cof})_A(B)big)^T A=0$$
However, the equation $XA=0$ has infinitely many solutions, as can be seen as follows:
Look at the linear map $T:M_d to M_d$, defined by $T(X)=X^TA$. $T$ is not surjective, since every matrix in its image has rank smaller than $d-1$. Thus, $T$ is not injective.
$endgroup$
add a comment |
$begingroup$
As mentioned by Daniel, when $operatorname{rank}(A) le d-2$, $operatorname{Cof}(A)=0$, hence equation $(*)$ becomes
$$ big(d(operatorname{Cof})_A(B)big)^T A=0$$
However, the equation $XA=0$ has infinitely many solutions, as can be seen as follows:
Look at the linear map $T:M_d to M_d$, defined by $T(X)=X^TA$. $T$ is not surjective, since every matrix in its image has rank smaller than $d-1$. Thus, $T$ is not injective.
$endgroup$
As mentioned by Daniel, when $operatorname{rank}(A) le d-2$, $operatorname{Cof}(A)=0$, hence equation $(*)$ becomes
$$ big(d(operatorname{Cof})_A(B)big)^T A=0$$
However, the equation $XA=0$ has infinitely many solutions, as can be seen as follows:
Look at the linear map $T:M_d to M_d$, defined by $T(X)=X^TA$. $T$ is not surjective, since every matrix in its image has rank smaller than $d-1$. Thus, $T$ is not injective.
edited Jan 29 at 17:00
community wiki
2 revs, 2 users 92%
Asaf Shachar
add a comment |
add a comment |
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