Proving that $sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=det(A)varepsilon_{pqr}$.












0












$begingroup$


I want to prove the following identity: if $A$ is a $3times 3$ matrix, then
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=det(A)varepsilon_{pqr},$$
where $varepsilon$ is the Levi-Civita symbol.



I know that (Leibniz formula)
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{1i}a_{2j}a_{3k}=det(A).$$
However I cant see why
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{pqr}varepsilon_{ijk}a_{1i}a_{2j}a_{3k}.$$






linear-algebra differential-geometry







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  • $begingroup$
    @daw fixed. Thanks
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 21:53
















0












$begingroup$


I want to prove the following identity: if $A$ is a $3times 3$ matrix, then
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=det(A)varepsilon_{pqr},$$
where $varepsilon$ is the Levi-Civita symbol.



I know that (Leibniz formula)
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{1i}a_{2j}a_{3k}=det(A).$$
However I cant see why
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{pqr}varepsilon_{ijk}a_{1i}a_{2j}a_{3k}.$$






linear-algebra differential-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    @daw fixed. Thanks
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 21:53














0












0








0


0



$begingroup$


I want to prove the following identity: if $A$ is a $3times 3$ matrix, then
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=det(A)varepsilon_{pqr},$$
where $varepsilon$ is the Levi-Civita symbol.



I know that (Leibniz formula)
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{1i}a_{2j}a_{3k}=det(A).$$
However I cant see why
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{pqr}varepsilon_{ijk}a_{1i}a_{2j}a_{3k}.$$






linear-algebra differential-geometry







share|cite|improve this question











$endgroup$




I want to prove the following identity: if $A$ is a $3times 3$ matrix, then
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=det(A)varepsilon_{pqr},$$
where $varepsilon$ is the Levi-Civita symbol.



I know that (Leibniz formula)
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{1i}a_{2j}a_{3k}=det(A).$$
However I cant see why
$$sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{pqr}varepsilon_{ijk}a_{1i}a_{2j}a_{3k}.$$






linear-algebra differential-geometry




linear-algebra differential-geometry






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edited Jan 18 at 21:52







Gabriel Ribeiro

















asked Jan 18 at 12:25









Gabriel RibeiroGabriel Ribeiro

1,451522




1,451522












  • $begingroup$
    @daw fixed. Thanks
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 21:53


















  • $begingroup$
    @daw fixed. Thanks
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 21:53
















$begingroup$
@daw fixed. Thanks
$endgroup$
– Gabriel Ribeiro
Jan 18 at 21:53




$begingroup$
@daw fixed. Thanks
$endgroup$
– Gabriel Ribeiro
Jan 18 at 21:53










1 Answer
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$begingroup$

It's easy to see that it's true for $p=1, q=2, r=3$. From this, we can see that it's true when $epsilon_{pqr}=1$. Can you see why? If $epsilon_{pqr}=-1$, we will get $-det(A)$. To see why, let $p=2, q=1, r=3$. The lhs is now $epsilon_{ijk}a_{2i}a_{1j}a_{3k}=epsilon{ijk}a_{1j}a_{2i}a_{3k}=-epsilon_{jik}a_{1j}a_{2i}a_{3k}$. What happens when $epsilon_{pqr}=0$?






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$endgroup$













  • $begingroup$
    If $varepsilon_{pqr}=0$, we have two equal indices. We can suppose $p=q$. Then, $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=-varepsilon_{jik}a_{pi}a_{pj}a_{rk}=-varepsilon_{ijk}a_{pi}a_{pj}a_{rk}$. We conclude that $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=0$. Nice!
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 13:04










  • $begingroup$
    @GabrielRibeiro exactly :)
    $endgroup$
    – Botond
    Jan 18 at 13:23











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1 Answer
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1 Answer
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active

oldest

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active

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1












$begingroup$

It's easy to see that it's true for $p=1, q=2, r=3$. From this, we can see that it's true when $epsilon_{pqr}=1$. Can you see why? If $epsilon_{pqr}=-1$, we will get $-det(A)$. To see why, let $p=2, q=1, r=3$. The lhs is now $epsilon_{ijk}a_{2i}a_{1j}a_{3k}=epsilon{ijk}a_{1j}a_{2i}a_{3k}=-epsilon_{jik}a_{1j}a_{2i}a_{3k}$. What happens when $epsilon_{pqr}=0$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $varepsilon_{pqr}=0$, we have two equal indices. We can suppose $p=q$. Then, $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=-varepsilon_{jik}a_{pi}a_{pj}a_{rk}=-varepsilon_{ijk}a_{pi}a_{pj}a_{rk}$. We conclude that $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=0$. Nice!
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 13:04










  • $begingroup$
    @GabrielRibeiro exactly :)
    $endgroup$
    – Botond
    Jan 18 at 13:23
















1












$begingroup$

It's easy to see that it's true for $p=1, q=2, r=3$. From this, we can see that it's true when $epsilon_{pqr}=1$. Can you see why? If $epsilon_{pqr}=-1$, we will get $-det(A)$. To see why, let $p=2, q=1, r=3$. The lhs is now $epsilon_{ijk}a_{2i}a_{1j}a_{3k}=epsilon{ijk}a_{1j}a_{2i}a_{3k}=-epsilon_{jik}a_{1j}a_{2i}a_{3k}$. What happens when $epsilon_{pqr}=0$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If $varepsilon_{pqr}=0$, we have two equal indices. We can suppose $p=q$. Then, $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=-varepsilon_{jik}a_{pi}a_{pj}a_{rk}=-varepsilon_{ijk}a_{pi}a_{pj}a_{rk}$. We conclude that $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=0$. Nice!
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 13:04










  • $begingroup$
    @GabrielRibeiro exactly :)
    $endgroup$
    – Botond
    Jan 18 at 13:23














1












1








1





$begingroup$

It's easy to see that it's true for $p=1, q=2, r=3$. From this, we can see that it's true when $epsilon_{pqr}=1$. Can you see why? If $epsilon_{pqr}=-1$, we will get $-det(A)$. To see why, let $p=2, q=1, r=3$. The lhs is now $epsilon_{ijk}a_{2i}a_{1j}a_{3k}=epsilon{ijk}a_{1j}a_{2i}a_{3k}=-epsilon_{jik}a_{1j}a_{2i}a_{3k}$. What happens when $epsilon_{pqr}=0$?






share|cite|improve this answer









$endgroup$



It's easy to see that it's true for $p=1, q=2, r=3$. From this, we can see that it's true when $epsilon_{pqr}=1$. Can you see why? If $epsilon_{pqr}=-1$, we will get $-det(A)$. To see why, let $p=2, q=1, r=3$. The lhs is now $epsilon_{ijk}a_{2i}a_{1j}a_{3k}=epsilon{ijk}a_{1j}a_{2i}a_{3k}=-epsilon_{jik}a_{1j}a_{2i}a_{3k}$. What happens when $epsilon_{pqr}=0$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 18 at 12:42









BotondBotond

5,8182832




5,8182832












  • $begingroup$
    If $varepsilon_{pqr}=0$, we have two equal indices. We can suppose $p=q$. Then, $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=-varepsilon_{jik}a_{pi}a_{pj}a_{rk}=-varepsilon_{ijk}a_{pi}a_{pj}a_{rk}$. We conclude that $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=0$. Nice!
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 13:04










  • $begingroup$
    @GabrielRibeiro exactly :)
    $endgroup$
    – Botond
    Jan 18 at 13:23


















  • $begingroup$
    If $varepsilon_{pqr}=0$, we have two equal indices. We can suppose $p=q$. Then, $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=-varepsilon_{jik}a_{pi}a_{pj}a_{rk}=-varepsilon_{ijk}a_{pi}a_{pj}a_{rk}$. We conclude that $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=0$. Nice!
    $endgroup$
    – Gabriel Ribeiro
    Jan 18 at 13:04










  • $begingroup$
    @GabrielRibeiro exactly :)
    $endgroup$
    – Botond
    Jan 18 at 13:23
















$begingroup$
If $varepsilon_{pqr}=0$, we have two equal indices. We can suppose $p=q$. Then, $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=-varepsilon_{jik}a_{pi}a_{pj}a_{rk}=-varepsilon_{ijk}a_{pi}a_{pj}a_{rk}$. We conclude that $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=0$. Nice!
$endgroup$
– Gabriel Ribeiro
Jan 18 at 13:04




$begingroup$
If $varepsilon_{pqr}=0$, we have two equal indices. We can suppose $p=q$. Then, $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=-varepsilon_{jik}a_{pi}a_{pj}a_{rk}=-varepsilon_{ijk}a_{pi}a_{pj}a_{rk}$. We conclude that $varepsilon_{ijk}a_{pi}a_{pj}a_{rk}=0$. Nice!
$endgroup$
– Gabriel Ribeiro
Jan 18 at 13:04












$begingroup$
@GabrielRibeiro exactly :)
$endgroup$
– Botond
Jan 18 at 13:23




$begingroup$
@GabrielRibeiro exactly :)
$endgroup$
– Botond
Jan 18 at 13:23


















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