Prove that if $\tau$ and $\sigma$ are two dyadic tensors then $\tau\cdot \sigma=\text{tr}(T\cdot S^T)$. So using orthonormal bases, I proved that if $\tau$ and $\sigma$ are dyadic tensors then its inner product is the trace of the product of the matrices $T$ and $S$ of its image in $\hom(V,V)$ under a certain isomorphism. So since the product of a matrices is referred to the composition of linear function and since the trace is an invariant I thought that the result is generally true so that it holds even if we does not use orthonormal bases but unfortunately I was not sure about the correctness of the this argument and so I proved to calculate the inner product of two dyadic tensors with not using an orthonormal basis and strangerly I did not able to obtain the same result so I ask it my above argumentations are correct and if not I ask to prove the result if it is true even with not using an orthonormal basis. To follow some explanations about the definition I use of inner product between tensors: naturally if you know it you can read immediately only the final calculation about the inner product of two tensors.
Theorem 1
If $V$ is a vector space equipped with an inner product then there exist a unique isomorphism
$$
\#:V\rightarrow V^*
$$
with is induced by the inner product in such a way that
$$
\big\langle\#(\vec v),\vec w\big\rangle=\vec v\cdot\vec w
$$
for any $v,w\in V$. Moreover under this isomorphism the image of the reciprocal basis $\text E:=\{\vec\epsilon_{_{1}},...,\vec\epsilon_{_{n}}\}$ of any basis $\mathscr E:=\{\vec e_{{_1}},...,\vec e_{_{n}}\}$ is the dual basis $\mathscr E^*:=\{\vec e^{\,_1}\,...,\vec e^{_{n}}\}$, that is
$$
\#(\vec\epsilon_{_i})=\vec e^{\,_{i}}
$$
for each $i=1,...,n$
If $\tau$ is a mixed simple tensor in $\mathscr T^p_q(V)$ say
$$
\tau:=\vec v_{_{1}}\otimes...\otimes\vec v_{_{p}}\otimes \vec v^{\,_{1}}\otimes...\otimes \vec v^{\,_{q}}
$$
for some $\vec v_{_{1}},...,\vec v{_{p}}\in V$ and for some $\vec v^{\,_{1}},...,\vec v^{\,_{q}}\in V^*$, then we define the pure covariant representation of $A$ to be the tensor
$$
\#^p_q(\tau):=\#(\vec v_{_{1}})\otimes...\otimes\#(\vec v_{_{p}})\otimes \vec 
 v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}}
$$
Theorem 2
The above condition define an isomorphism between $\mathscr T^p_q(V)$ and $\mathscr T_{p+q}$ whose inverse is given by the equality
$$
(\#^p_q)^{-1}(\vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{p}}\otimes \vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}})=\#^{-1}(\vec v^{\,_{1}})\otimes...\otimes\#^{-1}(\vec v^{\,_{p}})\otimes \vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}})
$$
for each simple tensor.
Theorem 3
If $\tau\in\mathscr T^p_q(V)$ is an arbitrary simple tensor, say
$$
\tau:=\vec v_{_{1}}\otimes...\otimes\vec v_{_{p}}\otimes \vec v^{\,_{1}}\otimes...\otimes\vec v^{\,_{q}}
$$
for some $\vec v_{_{1}},...,\vec v{_{p}}\in V$ and for some $\vec v^{\,_{1}},...,\vec v^{\,_{q}}\in V^*$, then for eanch pair of integers $(i,j)$, wherere $1\le i\le p$ and $1\le j\le q$, we seek a unique linear transformation called simple contraction
$$
C^i_j:\mathscr T^p_q(V)\rightarrow\mathscr T^{p-1}_{q-1}(V)
$$
such that
$$
C^i_j(\tau):=\langle v^{\,_j},\vec v_{_i}\rangle v_1\otimes...\otimes v_{_{i-1}}\otimes v_{_{i+1}}\otimes...\otimes v_{_p}\otimes\vec v^{\,_1}\otimes...\otimes\vec v^{\,_{j-1}}\otimes\vec v^{\,_{j+1}}\otimes...\otimes\vec v^{\,_{q}}
$$
for each all simple tensor $\tau$.
So using the contraction operation, we can define a double product for tensors. If $\tau\in\mathscr T^p_q(V)$ and $\sigma\in\mathscr T^q_p(V)$ then the tensor product $\tau\otimes\sigma$ is a tensor in $\mathscr T^{p+q}_{p+q}(V)$ so that if we apply the contraction
$$
C:=\underset{\text{q times}}{\underbrace{C^1_1\circ...\circ\ C^1_1}}\circ \underset{\text{p times}}{\underbrace{C^1_{q+1}\circ...\circ\ C^1_{q+1}}}
$$
to $\tau\otimes\sigma$ then the result is a scalar and moreover we observe that it has the same property of the dual operator $\langle\,\cdot\,\rangle$ so that we can define a isomorphism between $\mathscr T^p_q(V)$ and the dual of $\mathscr T^q_p(V)$ and so we can identify the two spaces. So we can define the dual product between two tensor through the equation

*

*$\langle\tau,\sigma\rangle:=C(\tau\otimes\sigma)$
for any $\tau\in\mathscr T^p_q(V)$ and for any $\sigma\in\mathscr T^q_p(V)$.
Theorem 4
The equality defined for any $\tau,\sigma\in\mathscr T^p_q(V)$

*

*$\tau\cdot\sigma:=\big\langle\#^p_q(\tau),(\#^{p+q})^{-1})\big(\#^p_q(\sigma)\big)\rangle$
define an inner product on $\mathscr T^p_q(V)$.
Definition 1
If $\tau\in\mathscr T^p_q(V)$ and $\sigma\in\mathscr T^r_s(V)$ an external contraction of order $l$ is a composition of simple contractions applied to $\tau\otimes\sigma$ each of whose contract an idices of $\tau$ with an indices of $\sigma$ respectively.
Theorem 5
If $\overline p\le p$ and $\overline q\le q$ then for any external contracion of order $(\overline p+\overline q)$ the function $J_C$ from $\mathscr T^p_q(V)$ in $\hom\big[\mathscr T^{\overline q}_{\overline p},\mathscr T^{p-\overline p}_{q-\overline q}\big]$ defined as
$$
[J_C(\tau)](\sigma):=C(\tau\otimes\sigma)
$$
for each $\tau\in\mathscr T^p_q$ and for each $\sigma\in\mathscr T^{\overline q}_{\overline p}$ is an isomorphism.
Corollary 5.1
Let be $\mathscr E:=\{\vec e_{_1},...,\vec e_{_n}\}$ a basis of $V$. So the spaces $\mathscr T^2(V)$ and $\hom(V,V)$ are isomorphic via the composition of an isomorphism $h$ from $\hom(V^*,V^{**})$ to $\hom(V,V)$ and an isomorphim $J_C$ from $\mathscr T^2(V)$ to $\hom(V^*,V^{**})$ such that
$$
\big[h(J_C(\tau)\big](\vec v):=T^{i,j}(\vec v\cdot\vec e_{_j})\vec e_{_i}
$$
for each $v\in V$ where the $n^2$ scalars $T^{i,j}$ are the components of $\tau$ with respect the basis of $\mathscr T^2(V)$ induced from $\mathscr E$.
Definition 2
The transpose operator is the endomorphism $T_{pq}$ of $\mathscr T^{p+q}$ defined throught the equation

*

*$T_{pq}(\vec v_{_1}\otimes...\otimes\vec v_{_p}\otimes\vec u_{_1}\otimes...\otimes\vec u_{_q}):=u_{_1}\otimes...\otimes\vec u_{_q}\otimes\vec v_{_1}\otimes...\otimes\vec v_{_p}$
for any simple tensor. Moreover the general transpose operator $T^{pq}_{pq}$ could be defined thorugh the equation

*

*$T^{pq}_{pq}(\tau):=\big((\#^p_q)^{-1}\circ T_{pq}\circ\#^p_q\big)(\tau)$
for each $\tau\in\mathscr T^p_q$
Theorem 6
The image of the transpose tensor of a tensor $\tau$ under the isomorphism $[h\circ J_C]$ above defined is the transpose function of the the image of $\tau$.
Now if $\mathscr E:=\{\vec e_{_1},...,\vec e_{_n}\}$ is a basis for $V$ then the componet of the image of a tensor $\tau$ and $\sigma$ under the isomorphism $h\circ J_C$ are the scalars $t_{i,k}:=T^{i,j}(\vec e_{_j}\cdot\vec e_{_k})$ and $s_{h,i}:=S^{hk}(\vec e_{_k}\cdot\vec e_{_i})$. So we observe that
$$
\tau\cdot\sigma:=\big\langle\#^p_q(\tau),(\#^{p+q})^{-1}\big(\#^p_q(\sigma)\big)\big\rangle=...=T^{i,j}S^{h,k}(\vec e_{_i}\cdot\vec e_{_h})(\vec e_{_j}\cdot\vec e_{_k})=\big(T^{i,j}(\vec e_{_j}\cdot\vec e_{_k})\big)\big(S^{h,k}(\vec e_{_i}\cdot\vec e_{_h})\big)
$$
so that if $\mathscr E:=$ is an orthonormal basis then
$$
\tau\cdot\sigma=...=\big(T^{i,j}(\vec e_{_j}\cdot\vec e_{_k})\big)\big(S^{h,k}(\vec e_{_i}\cdot\vec e_{_h})\big)=\big(T^{i,j}\delta_{j,k}\big)\big(S^{h,k}\delta_{i,h}\big)=T^{i,j}S^{i,j}=\text{tr}(T\cdot S^T)
$$
where $T$ and $S$ are the matrix of the image of $\tau$ and $\sigma$. So if $\mathscr E$ is not an orthonormal basis then  the result seems that does not hold. So what can you say about?
 A: First of all for practical reasons we define the matrices $L$ and $M$ through the equalities

*

*$$L_{i,j}:=(\vec e^{\,_i}\cdot\vec e^{\,_j})$$

*$$M_{i,j}:=(\vec e_{_i}\cdot\vec e_{_j})$$
for each $i,j=1,...,n$ and then let be $S$ and $T$ the matrices whose elements are the component of two dyadic tensor $\sigma$ and $\tau$. Now using the theorem $5$ it is possible to prove that the dyadic pure covariant, mixed and pure contravariant tensors are isomorphic to the set of endomorphisms through the applications $f$, $g$ and $h$ and in particular if $\tau$ is a pure covariant tensor then
$$
\big[f(\tau)\big]\big(\vec v)=\big(\,e^{_{i,h}}T_{h,j}v^{_j}\,\big)\vec e_{_h}
$$
for each $\vec v\in V$ and if $\tau$ is a mixed tensor then
$$
\big[g(\tau)\big](\vec v)=\big(\,T^i_jv^{_j}\,\big)\vec e_{_i}
$$
for each $\vec v\in V$ and finally if $\tau$ is a pure contravariant tensor then
$$
\big[(h(\tau)\big](\vec v)=\big(\,T^{i,h}e_{_{h,j}}v^{_j}\,\big)\vec e_{_i}
$$
for each $\vec v\in V$ so that the matrices associated to the image of a dyadic tensor $\tau$ are respectively $LT$ for a pure covariant tensor, $T$ for a mixed tensor and $TM$ for a pure contravariant tensor so that generally the component of the imge of a dyadic tensor are different from the same tensor whether the basis used is not orthonormal because in this case the matrix $L$ and $M$ are the identity matrix. So if $\tau$ and $\sigma$ are pure covariant tensors then
$$
\tau\cdot\sigma\equiv C\Big(\big(T_{i,j}\vec e^{\,i}\otimes\vec e^{_j}\big)\otimes\big(S_{h,k}\vec e^{\,_h}\otimes\vec e^{\,_k}\big)\Big)=T_{i,j}S_{h,k}C\big(\vec e^{\,_i}\otimes\vec e^{\,_j}\otimes\vec e^{\,_h}\otimes\vec e^{\,_k}\big)=T_{i,j}S_{h,k}e^{_{i,h}}e^{_{j,k}}=T_{i,j}e^{_{j,k}}S^t_{k,h}e^{_{h,i}}=\text{tr}\big(TLS^tL)
$$
while if $\tau$ and $\sigma$ are mixed tensors then
$$
\tau\cdot\sigma:=\Big\langle G^p_q(\tau),(G^p_q)^{-1}\big(G^p_q(\sigma)\big)\Big\rangle=T^i_jS^h_k\Big\langle G^p_q(\vec e_{_i}\otimes\vec e^{\,_j}),(G^p_q)^{-1}\big(G^p_q(\vec e_{_h}\otimes\vec e^{\,_k})\big)\Big\rangle=T^i_jS^h_k\Big\langle G\vec e_{_i}\otimes\vec e^{\,_j},\vec e_{_h}\otimes G^{-1}\vec e^{\,_k}\Big\rangle=T^i_jS^h_hC\big(\vec e_{_h}\otimes G^{-1}\vec e^{\,_k}\otimes G\vec e_{_i}\otimes\vec e^{\,_j}\big)=T^i_jS^h_ke_{_{i,h}}e^{\,_{j,k}}=T^i_je^{\,_{j,k}}\tilde{S}^k_he_{_{h,i}}=\text{tr}\big(TLS^tM\big)
$$
where $\tilde{S}^k_h$ are the scalar of the transpose matrix of $S$ and finally if $\tau$ and $\sigma$ are pure contravariant tensor then
$$
\tau\cdot\sigma:=\Big\langle G^p_q(\tau),(G^p_q)^{-1}\big(G^p_q(\sigma)\big)\Big\rangle=...=T^{i,j}S^{h,k}e_{_{i,h}}e_{_{j,k}}=T^{i,j}e_{_{j,k}}S^{k,h}_te_{_{h,i}}=\text{tr}\big(TMS^tM\big)
$$
where $S^{k,h}_t$ are the scalar of the transpose matrix of $S$. So since if the basis adoped is not orthonormal then the matrix $L$ and $M$ are different from the identity so we conclude that the result is generally false using a not orthonormal basis.
