Tranpose of a mixed tensor

Question

If $(-)'$ denotes the transpose operation then for a type (1,1) mixed tensor ${T^a}_b$, $\left({T^a}_b\right)'={T^b}_a$. Although this seems right I would like to be able to show this starting with a bilinear form $T:V^{*} \times V \rightarrow \mathbb{R}$ where $V$ is a $n$-dimensional vector space and $V^{*}$ is its dual. The set of bilinear maps with this property is a vector space which we can denote as $\mathcal{T}^1_1$. Any bilinear map $T \in \mathcal{T}^1_1$ can be written as, $$ T(u^{*},v)={T^a}_b e_a \otimes e^b (u^{*},v) $$ where $e_a$ (resp. $e^b$) are the basis vectors (resp. covectors) of $V$ (resp. $V^{*}$) and the tensor product is defined as, $$ e_a \otimes e^b (u^{*},v) := u^{*}(e_a)e^b(v) $$ If there is a formal definition of the transpose of a bilinear form in $\mathcal{T}^1_1$ then it should be possible to show that $\left({T^a}_b\right)'={T^b}_a$.