How to prove the formula for two dots contraction of two tensors

Question

there is a formula $${\bf{T}} \cdot \cdot \,{\bf{E}} = {T_{ij}}{E_{ji}} = tr({\bf{TE}})$$ I`m trying to prove it$$\eqalign{ & {T_{ij}}{E_{ji}} = \left( {{{{\bf{\hat e}}}_i} \cdot {\bf{T}}{{{\bf{\hat e}}}_j}} \right)\left( {{{{\bf{\hat e}}}_j} \cdot {\bf{E}}{{{\bf{\hat e}}}_i}} \right) = \left( {{{{\bf{\hat e}}}_i} \cdot {\bf{T}}{{{\bf{\hat e}}}_j}} \right){{{\bf{\hat e}}}_j} \cdot {\bf{E}}{{{\bf{\hat e}}}_i} = {{{\bf{\hat e}}}_i} \cdot \underbrace {{{{\bf{\hat e}}}_j} \otimes {\bf{T}}{{{\bf{\hat e}}}_j} \cdot {\bf{E}}}_A{{{\bf{\hat e}}}_i} = {A_{ii}} = tr({\bf{A}}) \cr & {\bf{A}} = {{{\bf{\hat e}}}_j} \otimes {\bf{T}}{{{\bf{\hat e}}}_j} \cdot {\bf{E}} = {{\bf{T}}^T}{\bf{E}} \cr} $$ Where is my mistake?Please help

Answer 1

EDIT: The question was about the coordinate formula for the trace of a product of two matrices.

The product of two matrices is given by $$(AB)_{ik} = \sum_{j} A_{ij}B_{jk}$$ So their trace is the sum of entries like the above when $i=k$, or: $$tr(AB) = \sum_{i,j} A_{ij}B_{ji}$$

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Do I understand your question correctly? I think what you are saying is that you got $tr(T^TE)$, but you were expecting to get $tr(TE)$. If this is the case, make sure you inspect the definition of the double dot product from whatever reference you are using carefully, and also inspect what type of tensor $T,E$ are in your problem.

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Beware: There are two definitions/conventions for the double dot product. According to wikipedia , this seems to be a common problem

"There are two ways to define the double dot product, one must be careful when deciding which convention to use." Wikipedia, https://en.wikipedia.org/wiki/Dyadics#Product_of_dyadic_and_dyadic

The two formulas are

1. $ab:cd = (a\cdot d)(b\cdot c)$
2. $ab:cd = (a\cdot c)(b\cdot d)$

and can be found in the chart at the link above. The difference is in transposing $c$ and $d$, so you can see how this is the same issue as you are having.

Basically,

• if you start with two tensors $T_1,T_2$ both of type $(m,n)$ for $m\neq n$ then $T_1T_2$ doesn't make sense (you can't multiply matrices if the columns and rows don't match in the middle). $T_1^TT_2$ does make sense though.
• On the other hand, if $T_1$ is of type $(m,n)$ and $T_2$ is of type $(n,m)$ then $T_1T_2$ is the right expression.

In the case that $n=m$ everything is completely ambiguous, and the definitions do not agree - so you must be careful to be clear on the definition/convention chosen in your context!

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Here are the two definitions.

Definition 1. Let $T_1 \in V^*\otimes W$ and $T_2\in V\otimes W^*$ be tensors of opposite type. Then there is a tensor $T_1: T_2 \in V^*$ given by contraction and equal to $T_1:T_2 = tr(V\overset{T_2}\to W \overset{T_1}\to V) = tr(T_1T_2)$.

or,

Definition 2. Let $T_1, T_2 \in V^*\otimes W$ be two tensors of the same type, then there is a tensor $T_1:T_2$ given by contraction and equal to $T_1:T_2 = tr(V\overset{T_2}\to W \overset{T_1^T}\to V) = tr(T_1^TT_2)$.

You may note that in the special case that $V = W$, again everything is ambiguous and you need to carefully check which definition has been made.

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Checking the formula. I suppose if your asking specifically about tensor contraction, then definition 1 is probably the right definition. In this case there is an identity/definition (see Wikipedia)

$$ab:cd = (a\cdot d)(b\cdot c)$$

which you can apply:

\begin{align} T:E &= (\sum T_{ij}e_i^*\otimes e_j):(\sum E_{kl}e_k^*\otimes e_\ell)\\ &= \sum T_{ij}E_{kl} (e_i\cdot e_\ell)(e_j\cdot e_k)\\ &= \sum T_{ij}E_{ji}\\ &= tr(TE). \end{align}

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Mistakes in your work. It looks like you have sought to do something similar to the above. I can't really understand what you mean in your definition of $A$. You are taking the dot product of two tensors, $e_j\otimes Te_j$ and $E$? I cannot follow this step.