Decomposing B_{ij} into Symmetric and Antisymmetric Tensors

[JohanL]

show that $$B_{ij}$$ can be written as the sum of a symmetric tensor
$$B^S_{ij}$$ and an antisymmetric tensor $$B^A_{ij}$$

i don't know how to do this one.
for a symmetric tensor we have
$$B^S_{ij} = B^S_{ji}$$

and for an antisymmetric tensor we have
$$B^A_{ij} = -B^A_{ji}$$

the only thing my book says is that the sum should be a tensor of the same type.

 

[robphy]

Hint:
define the "symmetric part of B" to be
$$(B_{ij})^S = \frac{1}{2}\left(B_{ij}+ B_{ji}\right)$$
... quite analogous to defining the "real part of a complex number z" as (z+z*)/2. [Check for yourself that this "symmetric part" is truly symmetric.]

I'm sure you can finish the rest.

 

[thepatientmental]

To decompose a tensor B_{ij} into its symmetric and antisymmetric parts, we can use the following formula:

B_{ij} = \frac{1}{2}(B_{ij} + B_{ji}) + \frac{1}{2}(B_{ij} - B_{ji})

Let's call the first term on the right-hand side B^S_{ij} and the second term B^A_{ij}. It is clear that B^S_{ij} is symmetric, since it is the sum of two symmetric tensors. Similarly, B^A_{ij} is antisymmetric, since it is the difference of two antisymmetric tensors.

To show that B_{ij} can be written as the sum of B^S_{ij} and B^A_{ij}, we can simply substitute the expressions for B^S_{ij} and B^A_{ij} into the original formula:

B_{ij} = \frac{1}{2}(B_{ij} + B_{ji}) + \frac{1}{2}(B_{ij} - B_{ji})

= \frac{1}{2}B_{ij} + \frac{1}{2}B_{ji} + \frac{1}{2}B_{ij} - \frac{1}{2}B_{ji}

= B_{ij}

Thus, we have shown that B_{ij} can be decomposed into its symmetric and antisymmetric parts, B^S_{ij} and B^A_{ij}. This decomposition is useful in many applications, as it allows us to break down a complex tensor into simpler, more manageable components.