Prove that det $A$ = det $A^T$
$begingroup$
I know some ways to prove this, but now I'm looking for another one using this:
$det A=det A^T$
$detleft(frac{ A+A^T}{2} +frac{ A-A^T}{2}right)= det left(frac{ A^T+A}{2} +frac{ A^T-A}{2}right)$
So I have something like:
$det(B+C) = det(B-C)$ where $B$ is symmetric and $C$ is skew-symmetric.
Can it be finished somehow using this?
linear-algebra
edited Dec 20 '18 at 13:56
amWhy
1
asked Dec 20 '18 at 11:37
nonamenoname
223
$endgroup$
add a comment |
$begingroup$
I know some ways to prove this, but now I'm looking for another one using this:
$det A=det A^T$
$detleft(frac{ A+A^T}{2} +frac{ A-A^T}{2}right)= det left(frac{ A^T+A}{2} +frac{ A^T-A}{2}right)$
So I have something like:
$det(B+C) = det(B-C)$ where $B$ is symmetric and $C$ is skew-symmetric.
Can it be finished somehow using this?
linear-algebra
edited Dec 20 '18 at 13:56
amWhy
1
asked Dec 20 '18 at 11:37
nonamenoname
223
$endgroup$
add a comment |
$begingroup$
I know some ways to prove this, but now I'm looking for another one using this:
$det A=det A^T$
$detleft(frac{ A+A^T}{2} +frac{ A-A^T}{2}right)= det left(frac{ A^T+A}{2} +frac{ A^T-A}{2}right)$
So I have something like:
$det(B+C) = det(B-C)$ where $B$ is symmetric and $C$ is skew-symmetric.
Can it be finished somehow using this?
linear-algebra
edited Dec 20 '18 at 13:56
amWhy
1
asked Dec 20 '18 at 11:37
nonamenoname
223
$endgroup$
I know some ways to prove this, but now I'm looking for another one using this:
$det A=det A^T$
$detleft(frac{ A+A^T}{2} +frac{ A-A^T}{2}right)= det left(frac{ A^T+A}{2} +frac{ A^T-A}{2}right)$
So I have something like:
$det(B+C) = det(B-C)$ where $B$ is symmetric and $C$ is skew-symmetric.
Can it be finished somehow using this?
linear-algebra
linear-algebra
edited Dec 20 '18 at 13:56
amWhy
1
asked Dec 20 '18 at 11:37
nonamenoname
223
edited Dec 20 '18 at 13:56
amWhy
1
asked Dec 20 '18 at 11:37
nonamenoname
223
edited Dec 20 '18 at 13:56
amWhy
1
edited Dec 20 '18 at 13:56
amWhy
1
edited Dec 20 '18 at 13:56
amWhy
1
1
asked Dec 20 '18 at 11:37
nonamenoname
223
asked Dec 20 '18 at 11:37
nonamenoname
223
asked Dec 20 '18 at 11:37
nonamenoname
223
223
add a comment |
add a comment |
1 Answer
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$begingroup$
That's not the approach you want. If $A$ is an $ntimes n$ matrix, $det A=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{isigma(i)}$ is a sum over all permutations $sigma$ of $1,,2,,cdots,,n$, where the Levi-Civita symbol $epsilon_sigma=pm 1$ is the permutation parity of $sigma$. Similarly, $det A^T=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{sigma(i)i}$. Since each permutation has an inverse of the same parity, $det A=det A^T$.
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
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$begingroup$
That's not the approach you want. If $A$ is an $ntimes n$ matrix, $det A=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{isigma(i)}$ is a sum over all permutations $sigma$ of $1,,2,,cdots,,n$, where the Levi-Civita symbol $epsilon_sigma=pm 1$ is the permutation parity of $sigma$. Similarly, $det A^T=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{sigma(i)i}$. Since each permutation has an inverse of the same parity, $det A=det A^T$.
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
$begingroup$
That's not the approach you want. If $A$ is an $ntimes n$ matrix, $det A=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{isigma(i)}$ is a sum over all permutations $sigma$ of $1,,2,,cdots,,n$, where the Levi-Civita symbol $epsilon_sigma=pm 1$ is the permutation parity of $sigma$. Similarly, $det A^T=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{sigma(i)i}$. Since each permutation has an inverse of the same parity, $det A=det A^T$.
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
$endgroup$
add a comment |
$begingroup$
That's not the approach you want. If $A$ is an $ntimes n$ matrix, $det A=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{isigma(i)}$ is a sum over all permutations $sigma$ of $1,,2,,cdots,,n$, where the Levi-Civita symbol $epsilon_sigma=pm 1$ is the permutation parity of $sigma$. Similarly, $det A^T=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{sigma(i)i}$. Since each permutation has an inverse of the same parity, $det A=det A^T$.
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
$endgroup$
That's not the approach you want. If $A$ is an $ntimes n$ matrix, $det A=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{isigma(i)}$ is a sum over all permutations $sigma$ of $1,,2,,cdots,,n$, where the Levi-Civita symbol $epsilon_sigma=pm 1$ is the permutation parity of $sigma$. Similarly, $det A^T=sum_sigmaepsilon_sigmaprod_{i=1}^nA_{sigma(i)i}$. Since each permutation has an inverse of the same parity, $det A=det A^T$.
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
answered Dec 20 '18 at 14:15
J.G.J.G.
27.1k22843
27.1k22843
add a comment |
add a comment |
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