Mixing
Proving that $sum_{i=1}^3sum_{j=1}^3sum_{k=1}^3varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=det(A)varepsilon_{pqr}$.
$begingroup$
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
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
$endgroup$
add a comment |
$begingroup$
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
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
$endgroup$
$begingroup$
@daw fixed. Thanks
$endgroup$
– Gabriel Ribeiro
Jan 18 at 21:53
add a comment |
$begingroup$
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
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
$endgroup$
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
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
edited Jan 18 at 21:52
Gabriel Ribeiro
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
asked Jan 18 at 12:25
Gabriel RibeiroGabriel Ribeiro
1,456523
1,456523
$begingroup$
@daw fixed. Thanks
$endgroup$
– Gabriel Ribeiro
Jan 18 at 21:53
add a comment |
$begingroup$
@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
add a comment |
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 $epsilon_{pqr}=0$?
answered Jan 18 at 12:42
BotondBotond
5,9152832
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
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active
oldest
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votes
$begingroup$
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 $epsilon_{pqr}=0$?
answered Jan 18 at 12:42
BotondBotond
5,9152832
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $epsilon_{pqr}=0$?
answered Jan 18 at 12:42
BotondBotond
5,9152832
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $epsilon_{pqr}=0$?
answered Jan 18 at 12:42
BotondBotond
5,9152832
$endgroup$
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 $epsilon_{pqr}=0$?
answered Jan 18 at 12:42
BotondBotond
5,9152832
answered Jan 18 at 12:42
BotondBotond
5,9152832
answered Jan 18 at 12:42
BotondBotond
5,9152832
answered Jan 18 at 12:42
BotondBotond
5,9152832
5,9152832
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
@daw fixed. Thanks
$endgroup$
– Gabriel Ribeiro
Jan 18 at 21:53