Levi-Civita symbol and Kronecker delta

[typhoonss821]

Hello everyone, I am stuck when I study Levi-Civita symbol.
The question is how to prove

$$\varepsilon_{ijk}\varepsilon_{lmn} = \det \begin{bmatrix} \delta_{il} & \delta_{im}& \delta_{in}\\ \delta_{jl} & \delta_{jm}& \delta_{jn}\\ \delta_{kl} & \delta_{km}& \delta_{kn}\\ \end{bmatrix}$$

where $$\varepsilon_{ijk}$$ represents Levi-Civita symbol and $$\delta_{il}$$ represents kronecker symbol.

Thank you very much^^

 

[slider142]

Have you already established the identity $\epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$?

 

[typhoonss821]

Yes I have, but I don't know how to relate it to determinant...

 

[Landau]

Well, you could just write out that determinant and see what happens.

 

[pmacias]

The Levi-Civita symbol and Kronecker delta are both important mathematical concepts in the field of linear algebra and tensor calculus. The Levi-Civita symbol, denoted by \varepsilon_{ijk}, is a permutation symbol used to represent the orientation of a coordinate system. On the other hand, the Kronecker delta, denoted by \delta_{il}, is a special kind of tensor that takes the value of 1 when the two indices are equal and 0 otherwise.

To prove the given equation, we can start by expanding the determinant on the right-hand side. We have:

\det \begin{bmatrix}
\delta_{il} & \delta_{im}& \delta_{in}\\
\delta_{jl} & \delta_{jm}& \delta_{jn}\\
\delta_{kl} & \delta_{km}& \delta_{kn}\\
\end{bmatrix} = \delta_{il}\delta_{jm}\delta_{kn} + \delta_{im}\delta_{jn}\delta_{kl} + \delta_{in}\delta_{jl}\delta_{km} - \delta_{in}\delta_{jm}\delta_{kl} - \delta_{il}\delta_{jn}\delta_{km} - \delta_{im}\delta_{jl}\delta_{kn}

Next, we can use the properties of the Kronecker delta to simplify this expression. Since the Kronecker delta takes the value of 1 when the two indices are equal, all terms with repeated indices will be equal to 1. Therefore, we can rewrite the above expression as:

\det \begin{bmatrix}
1 & 0& 0\\
0 & 1& 0\\
0 & 0& 1\\
\end{bmatrix} = 1 + 1 + 1 - 1 - 1 - 1 = 0

This shows that the determinant on the right-hand side is equal to 0. Now, let's consider the left-hand side of the equation. We can use the properties of the Levi-Civita symbol to rewrite it as:

\varepsilon_{ijk}\varepsilon_{lmn} = \begin{cases}
1, & \text{if } (i,j,k) \text{ and } (l,m,n) \text{ are even permutations of each other} \\
-1, & \text

 

[CellCoree]

I can provide some insight into the proof of this equation. First, let's define the Levi-Civita symbol and Kronecker delta.

The Levi-Civita symbol, denoted by \varepsilon_{ijk}, is a mathematical symbol used in vector calculus and tensor analysis to represent the sign of a permutation of three objects. It takes on the value of +1, -1, or 0 depending on the order of the indices i, j, k. For example, \varepsilon_{123} = +1, \varepsilon_{231} = -1, and \varepsilon_{112} = 0.

The Kronecker delta, denoted by \delta_{ij}, is a mathematical symbol that represents the identity matrix. It takes on the value of 1 when the two indices i and j are equal, and 0 when they are not equal. For example, \delta_{11} = 1, \delta_{12} = \delta_{21} = 0.

Now, to prove the given equation, we can use the properties of the Levi-Civita symbol and Kronecker delta. First, we can rewrite the left side of the equation as:

\varepsilon_{ijk}\varepsilon_{lmn} = \varepsilon_{i1}\varepsilon_{j2}\varepsilon_{k3}\varepsilon_{l1}\varepsilon_{m2}\varepsilon_{n3}

Next, we can use the antisymmetry property of the Levi-Civita symbol to rearrange the indices:

\varepsilon_{ijk}\varepsilon_{lmn} = \varepsilon_{ij1}\varepsilon_{k3}\varepsilon_{l1}\varepsilon_{m2}\varepsilon_{n3}

Then, we can use the property that the Levi-Civita symbol is equal to 0 when two indices are equal:

\varepsilon_{ijk}\varepsilon_{lmn} = \varepsilon_{ij1}\varepsilon_{k3}\varepsilon_{l1}\varepsilon_{m2}\varepsilon_{n3} = \varepsilon_{ij1}\varepsilon_{k3}\varepsilon_{l1}\varepsilon_{m2}\varepsilon_{n3}\delta_{l