How to prove the Cofactor Expansion Theorem for Determinant of a Matrix? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to prove the cofactor formula for determinants, using a different definition of the determinant?Matrix determinant using Laplace methodChange in determinant when multiplying row of a matrixMaximal determinant of a $1,−1$ matrix of size $n$ is $2^n−1$ times the maximal determinant of a $ 0,1$ matrix of size $n−1$.Super Simple Proof of Cofactor ExpansionFind the determinant after a certain row operation is applied to a matrix with known determinantOn the cofactor expansion of the determinantA conceptual definition of a Matrix DeterminantProve that the Laplace expansion for the determinant is the same for any choice of row or columnErroneous proof? Determinant = 0, when i =/= j of cofactor expansion
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How to prove the Cofactor Expansion Theorem for Determinant of a Matrix?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to prove the cofactor formula for determinants, using a different definition of the determinant?Matrix determinant using Laplace methodChange in determinant when multiplying row of a matrixMaximal determinant of a $1,−1$ matrix of size $n$ is $2^n−1$ times the maximal determinant of a $ 0,1$ matrix of size $n−1$.Super Simple Proof of Cofactor ExpansionFind the determinant after a certain row operation is applied to a matrix with known determinantOn the cofactor expansion of the determinantA conceptual definition of a Matrix DeterminantProve that the Laplace expansion for the determinant is the same for any choice of row or columnErroneous proof? Determinant = 0, when i =/= j of cofactor expansion
$begingroup$
I understand that the definition of a determinant of a matrix implies that you can expand over the first row over the matrix, but how does that itself imply that you can expand over any row the matrix. Is there a proper proof to this, perhaps using induction on the size of matrix or something? Thanks in advance.
linear-algebra matrices determinant
asked Apr 2 at 15:28
TimTim
795
$endgroup$
add a comment |
$begingroup$
I understand that the definition of a determinant of a matrix implies that you can expand over the first row over the matrix, but how does that itself imply that you can expand over any row the matrix. Is there a proper proof to this, perhaps using induction on the size of matrix or something? Thanks in advance.
linear-algebra matrices determinant
asked Apr 2 at 15:28
TimTim
795
$endgroup$
$begingroup$
What definition of determinant are you working with and what preliminaries do you have? In case you have proven that the space of determinant functions is one-dimensional, a rather slick proof can be given.
$endgroup$
– Thorgott
Apr 2 at 15:34
$begingroup$
The definition I've been given is as follows: $det(A) = a_11(-1)^1+1det(A_11) + ... + a_1k(-1)^1+kdet(A_1k)$ where $A$ is a $k$ by $k$ square matrix and $A_1i$ is the submatrix of A from deleting the first row and $i$th column.
$endgroup$
– Tim
Apr 2 at 15:40
add a comment |
$begingroup$
I understand that the definition of a determinant of a matrix implies that you can expand over the first row over the matrix, but how does that itself imply that you can expand over any row the matrix. Is there a proper proof to this, perhaps using induction on the size of matrix or something? Thanks in advance.
linear-algebra matrices determinant
asked Apr 2 at 15:28
TimTim
795
$endgroup$
I understand that the definition of a determinant of a matrix implies that you can expand over the first row over the matrix, but how does that itself imply that you can expand over any row the matrix. Is there a proper proof to this, perhaps using induction on the size of matrix or something? Thanks in advance.
linear-algebra matrices determinant
linear-algebra matrices determinant
asked Apr 2 at 15:28
TimTim
795
asked Apr 2 at 15:28
TimTim
795
asked Apr 2 at 15:28
TimTim
795
asked Apr 2 at 15:28
TimTim
795
asked Apr 2 at 15:28
TimTim
795
795
$begingroup$
What definition of determinant are you working with and what preliminaries do you have? In case you have proven that the space of determinant functions is one-dimensional, a rather slick proof can be given.
$endgroup$
– Thorgott
Apr 2 at 15:34
$begingroup$
The definition I've been given is as follows: $det(A) = a_11(-1)^1+1det(A_11) + ... + a_1k(-1)^1+kdet(A_1k)$ where $A$ is a $k$ by $k$ square matrix and $A_1i$ is the submatrix of A from deleting the first row and $i$th column.
$endgroup$
– Tim
Apr 2 at 15:40
add a comment |
$begingroup$
What definition of determinant are you working with and what preliminaries do you have? In case you have proven that the space of determinant functions is one-dimensional, a rather slick proof can be given.
$endgroup$
– Thorgott
Apr 2 at 15:34
$begingroup$
The definition I've been given is as follows: $det(A) = a_11(-1)^1+1det(A_11) + ... + a_1k(-1)^1+kdet(A_1k)$ where $A$ is a $k$ by $k$ square matrix and $A_1i$ is the submatrix of A from deleting the first row and $i$th column.
$endgroup$
– Tim
Apr 2 at 15:40
$begingroup$
What definition of determinant are you working with and what preliminaries do you have? In case you have proven that the space of determinant functions is one-dimensional, a rather slick proof can be given.
$endgroup$
– Thorgott
Apr 2 at 15:34
$begingroup$
What definition of determinant are you working with and what preliminaries do you have? In case you have proven that the space of determinant functions is one-dimensional, a rather slick proof can be given.
$endgroup$
– Thorgott
Apr 2 at 15:34
$begingroup$
The definition I've been given is as follows: $det(A) = a_11(-1)^1+1det(A_11) + ... + a_1k(-1)^1+kdet(A_1k)$ where $A$ is a $k$ by $k$ square matrix and $A_1i$ is the submatrix of A from deleting the first row and $i$th column.
$endgroup$
– Tim
Apr 2 at 15:40
$begingroup$
The definition I've been given is as follows: $det(A) = a_11(-1)^1+1det(A_11) + ... + a_1k(-1)^1+kdet(A_1k)$ where $A$ is a $k$ by $k$ square matrix and $A_1i$ is the submatrix of A from deleting the first row and $i$th column.
$endgroup$
– Tim
Apr 2 at 15:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Below is a proof I found here. The idea is to do induction: since the minors are smaller matrices, one can calculate them via the desired row.
One first checks by hand that the determinant can be calculated along any row when $n=1$ and $n=2$.
$newcommanddetname,det$
$newcommandmatrixentry[2]#1_#2$
$newcommandsubmatrix[3]#3)$
For the induction, we use the notation $submatrixAi_1,i_2j_1,j_2$ to denote the $(n-2)times (n-2)$ matrix obtained from $A$ by removing the rows $i_1$ and $i_2$, and the columns $j_1$ and $j_2$.
For the calculation of the minors there will be a column missing, so one needs to be careful with the signs. For that we use
$$
epsilon_ell j=begincases 0,& ell <j \ 1,& ell>jendcases
$$
As mentioned above, the idea is that one calculates the minors along the $i^rm th$ row, which is ok by inductive hypothesis.
beginalign*
detnameA
&=
sum_j=1^n(-1)^1+jmatrixentryA1jdetnamesubmatrixA1j
\
&=
sum_j=1^n(-1)^1+jmatrixentryA1j
sum_substack1leqellleq n\ellneq j
(-1)^i-1+ell-epsilon_ell jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_j=1^nsum_substack1leqellleq n\ellneq j
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^nsum_substack1leq jleq n\jneqell
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^j-epsilon_ell j
matrixentryA1jdetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^epsilon_ell j+j
matrixentryA1jdetnamesubmatrixAi,1ell,j
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAielldetnamesubmatrixAiell
endalign*
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Below is a proof I found here. The idea is to do induction: since the minors are smaller matrices, one can calculate them via the desired row.
One first checks by hand that the determinant can be calculated along any row when $n=1$ and $n=2$.
$newcommanddetname,det$
$newcommandmatrixentry[2]#1_#2$
$newcommandsubmatrix[3]#3)$
For the induction, we use the notation $submatrixAi_1,i_2j_1,j_2$ to denote the $(n-2)times (n-2)$ matrix obtained from $A$ by removing the rows $i_1$ and $i_2$, and the columns $j_1$ and $j_2$.
For the calculation of the minors there will be a column missing, so one needs to be careful with the signs. For that we use
$$
epsilon_ell j=begincases 0,& ell <j \ 1,& ell>jendcases
$$
As mentioned above, the idea is that one calculates the minors along the $i^rm th$ row, which is ok by inductive hypothesis.
beginalign*
detnameA
&=
sum_j=1^n(-1)^1+jmatrixentryA1jdetnamesubmatrixA1j
\
&=
sum_j=1^n(-1)^1+jmatrixentryA1j
sum_substack1leqellleq n\ellneq j
(-1)^i-1+ell-epsilon_ell jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_j=1^nsum_substack1leqellleq n\ellneq j
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^nsum_substack1leq jleq n\jneqell
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^j-epsilon_ell j
matrixentryA1jdetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^epsilon_ell j+j
matrixentryA1jdetnamesubmatrixAi,1ell,j
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAielldetnamesubmatrixAiell
endalign*
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
$begingroup$
Below is a proof I found here. The idea is to do induction: since the minors are smaller matrices, one can calculate them via the desired row.
One first checks by hand that the determinant can be calculated along any row when $n=1$ and $n=2$.
$newcommanddetname,det$
$newcommandmatrixentry[2]#1_#2$
$newcommandsubmatrix[3]#3)$
For the induction, we use the notation $submatrixAi_1,i_2j_1,j_2$ to denote the $(n-2)times (n-2)$ matrix obtained from $A$ by removing the rows $i_1$ and $i_2$, and the columns $j_1$ and $j_2$.
For the calculation of the minors there will be a column missing, so one needs to be careful with the signs. For that we use
$$
epsilon_ell j=begincases 0,& ell <j \ 1,& ell>jendcases
$$
As mentioned above, the idea is that one calculates the minors along the $i^rm th$ row, which is ok by inductive hypothesis.
beginalign*
detnameA
&=
sum_j=1^n(-1)^1+jmatrixentryA1jdetnamesubmatrixA1j
\
&=
sum_j=1^n(-1)^1+jmatrixentryA1j
sum_substack1leqellleq n\ellneq j
(-1)^i-1+ell-epsilon_ell jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_j=1^nsum_substack1leqellleq n\ellneq j
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^nsum_substack1leq jleq n\jneqell
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^j-epsilon_ell j
matrixentryA1jdetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^epsilon_ell j+j
matrixentryA1jdetnamesubmatrixAi,1ell,j
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAielldetnamesubmatrixAiell
endalign*
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
add a comment |
$begingroup$
Below is a proof I found here. The idea is to do induction: since the minors are smaller matrices, one can calculate them via the desired row.
One first checks by hand that the determinant can be calculated along any row when $n=1$ and $n=2$.
$newcommanddetname,det$
$newcommandmatrixentry[2]#1_#2$
$newcommandsubmatrix[3]#3)$
For the induction, we use the notation $submatrixAi_1,i_2j_1,j_2$ to denote the $(n-2)times (n-2)$ matrix obtained from $A$ by removing the rows $i_1$ and $i_2$, and the columns $j_1$ and $j_2$.
For the calculation of the minors there will be a column missing, so one needs to be careful with the signs. For that we use
$$
epsilon_ell j=begincases 0,& ell <j \ 1,& ell>jendcases
$$
As mentioned above, the idea is that one calculates the minors along the $i^rm th$ row, which is ok by inductive hypothesis.
beginalign*
detnameA
&=
sum_j=1^n(-1)^1+jmatrixentryA1jdetnamesubmatrixA1j
\
&=
sum_j=1^n(-1)^1+jmatrixentryA1j
sum_substack1leqellleq n\ellneq j
(-1)^i-1+ell-epsilon_ell jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_j=1^nsum_substack1leqellleq n\ellneq j
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^nsum_substack1leq jleq n\jneqell
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^j-epsilon_ell j
matrixentryA1jdetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^epsilon_ell j+j
matrixentryA1jdetnamesubmatrixAi,1ell,j
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAielldetnamesubmatrixAiell
endalign*
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
$endgroup$
Below is a proof I found here. The idea is to do induction: since the minors are smaller matrices, one can calculate them via the desired row.
One first checks by hand that the determinant can be calculated along any row when $n=1$ and $n=2$.
$newcommanddetname,det$
$newcommandmatrixentry[2]#1_#2$
$newcommandsubmatrix[3]#3)$
For the induction, we use the notation $submatrixAi_1,i_2j_1,j_2$ to denote the $(n-2)times (n-2)$ matrix obtained from $A$ by removing the rows $i_1$ and $i_2$, and the columns $j_1$ and $j_2$.
For the calculation of the minors there will be a column missing, so one needs to be careful with the signs. For that we use
$$
epsilon_ell j=begincases 0,& ell <j \ 1,& ell>jendcases
$$
As mentioned above, the idea is that one calculates the minors along the $i^rm th$ row, which is ok by inductive hypothesis.
beginalign*
detnameA
&=
sum_j=1^n(-1)^1+jmatrixentryA1jdetnamesubmatrixA1j
\
&=
sum_j=1^n(-1)^1+jmatrixentryA1j
sum_substack1leqellleq n\ellneq j
(-1)^i-1+ell-epsilon_ell jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_j=1^nsum_substack1leqellleq n\ellneq j
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^nsum_substack1leq jleq n\jneqell
(-1)^j+i+ell-epsilon_ell j
matrixentryA1jmatrixentryAielldetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^j-epsilon_ell j
matrixentryA1jdetnamesubmatrixA1,ij,ell
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAiell
sum_substack1leq jleq n\jneqell
(-1)^epsilon_ell j+j
matrixentryA1jdetnamesubmatrixAi,1ell,j
\
&=
sum_ell=1^n(-1)^i+ellmatrixentryAielldetnamesubmatrixAiell
endalign*
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
answered Apr 2 at 21:03
Martin ArgeramiMartin Argerami
130k1184185
130k1184185
add a comment |
add a comment |
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$begingroup$
What definition of determinant are you working with and what preliminaries do you have? In case you have proven that the space of determinant functions is one-dimensional, a rather slick proof can be given.
$endgroup$
– Thorgott
Apr 2 at 15:34
$begingroup$
The definition I've been given is as follows: $det(A) = a_11(-1)^1+1det(A_11) + ... + a_1k(-1)^1+kdet(A_1k)$ where $A$ is a $k$ by $k$ square matrix and $A_1i$ is the submatrix of A from deleting the first row and $i$th column.
$endgroup$
– Tim
Apr 2 at 15:40