# A Tensor Problem: A skew-symmetric tensor and another tensor

## Homework Statement

If $A_{ij}$ is a skew-symmetric tensor, and $B_{ij}$ is a second-order tensor, evaluate the expression

$$(B_{ij} B_{kl} + B_{il}B_{kj})A_{ik}$$

and express the final answer in its simplest form.

## Homework Equations

For a skew-symmetric tensor, $A_{ik}=-A_{ki}$

## The Attempt at a Solution

I'm stuck and unsure what's the first step. I notice that the expression in the bracket looks similar to what happens when two Levi-Civita symbols come together to form an expression of two pairs of the Kronecker delta. Other than that I'm quite lost. Can I get a tip please?

Related Calculus and Beyond Homework Help News on Phys.org
You are probably summing over all the repeated indices, right? Remember then that the dummy indices are arbitrary, and you can for example swap k and i if you feel like it. Using this, maybe you can write the expression into a form where you take BijBkl as a common factor, multiplying some expression containing the tensor A.

Here's what I have so far:

$$(B_{ij} B_{kl} + B_{il}B_{kj})A_{ik}\\ =B_{ij} B_{1l}A_{i1}+B_{ij}B_{2l}A_{i2}+B_{il} B_{1l}A_{i1}+B_{il}B_{2j}A_{i2}\\ =B_{1j} B_{1l}A_{11}+B_{2j} B_{1l}A_{21}+B_{1j}B_{2l}A_{12}+B_{2j}B_{2l}A_{22}+B_{1l} B_{1l}A_{11}+B_{2l} B_{1l}A_{21}+B_{1l}B_{2j}A_{12}+B_{2l}B_{2j}A_{22}\\ =2B_{1j} B_{1l}A_{11}+2B_{2l}B_{2j}A_{22}$$

So the final answer that I can give is... $$2\sum_i B_{ij} B_{il} A_{ii}$$ or in the Einstein summation, $$2B_{mj} B_{ml} A_{nn}$$ with nn no sum.

If this is correct, is there any other way to write this without the no sum?