Why do we have $(\vec{u}\cdot\vec{n})\vec{n}=\vec{u}^{T}(\vec{n}\otimes\vec{n})=(\vec{n}\otimes\vec{n})\vec{u}$? In this question the answer gives a nice illustration about the tangential projection $P = I - \vec{n}\otimes\vec{n}$. 
I understand everything but the identity it mentioned: 

I've applied the identity
  $(\vec{u}\cdot\vec{n})\vec{n}=\vec{u}^{T}(\vec{n}\otimes\vec{n})=(\vec{n}\otimes\vec{n})\vec{u}$

I have no background in tensor analysis, thus I hope someone can help me to understand in details how the transpose $\vec{u}^{T}$ and the tensor product $\vec{n}\otimes\vec{n}$ are derived here. Thanks! 
Let me verify the identity as follows: 
For the sake of simplicity, let us consider 2-d case, we assume row vector $\vec{u} = \begin{bmatrix}{u_1 \; u_2}\end{bmatrix}$ and column vector $\vec{n} = \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}$, then $\vec{u}\cdot\vec{n} = (u_1 n_1 + u_2 n_2)$ which is a scalar, so the column vector $(\vec{u}\cdot\vec{n})\vec{n} = (u_1 n_1 + u_2 n_2)\begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}$ is well-defined on the LHS. 
However, $\vec{u}^T = \begin{bmatrix}{u_1 \; u_2}\end{bmatrix}^T = \begin{bmatrix}{u_1 \\ u_2}\end{bmatrix}$, and $\vec{n}\otimes\vec{n} = \vec{n}\vec{n}^T = \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix} \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}^{T} = \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix} \begin{bmatrix}{n_1 \; n_2}\end{bmatrix} = \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix}$, thus $\vec{u}^T(\vec{n}\otimes\vec{n}) = \begin{bmatrix}{u_1 \\ u_2}\end{bmatrix} \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix}$ is ill-defined. 
In addition, $(\vec{n}\otimes\vec{n})\vec{u} = \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix} \begin{bmatrix}{u_1 \; u_2}\end{bmatrix}$ is also ill-defined. 
I realized that $\vec{u}$ should be a column vector in order to make the dot product $\vec{u} \cdot \vec{n}$ well-defined. 
However, even if we assume $\vec{u}$ a column vector, the identity $\vec{u}^{T}(\vec{n}\otimes\vec{n}) = (\vec{n}\otimes\vec{n})\vec{u}$ does not hold, because LHS will be a row vector while RHS a column vector.
 A: There is a clash of conventions here. Matrices are usually intended
to be a way to conveniently express a linear transformations relative
to bases of the domain and codomain. More precisely, suppose that
$\,f:X\to Y\,$ is a linear transformation from vector space $\,X\,$ to
vector space $\,Y,\,$ where $\,(x_1,x_2,\dots,x_n)\,$ is an orthonormal
basis for $\,X\,$ and $\,(y_1,y_2,\dots,y_m)\,$ is an orthonormal basis
for $\,Y.\,$ Now given $\,v\in X\,$ and $\,w\in Y,\,$ consider the linear transformation $\,w\otimes v:X\to Y\,$ for all $\,u\in X\,$ defined by
$\,(w\otimes v)(u) := (u\cdot v)w.\,$ Since, by linearity, and because
$\,u=\sum_{i=1}^n (u\cdot x_i)\,$ for all $\,u\in X$,
$$ f(u)=\sum_{i=1}^n (u\cdot x_i)f(x_i) =
\sum_{i=1}^n (u\cdot x_i)\sum_{j=1}^m (f(x_i)\cdot y_j)y_j\\
 =\sum_{i=1}^n\sum_{j=1}^m a_{i,j}(u\cdot x_i)y_j.$$
Thus, the $\,a_{i,j} := f(x_i)\cdot y_j\,$ are the coefficients of the
expansion of $\,f = \sum_{i=1}^n\sum_{j=1}^m a_{i,j}(y_j\otimes x_i).$
Conventionally, the coefficients $\,a_{i,j}\,$ are arranged in a
rectangular array with rows and columns. Again, by convention, the result
$\,f(u)=v\,$ is written as $\,Au=v\,$ where $\,u\,$ and $\,v\,$ are
regarded as column vectors using the rules of matrix multiplication.
This is just a convention and for convenience. Now, taking transposes, $\,v^T=u^TA^T\,$ expresses the same result.
Similarly we have $\,(w\otimes v)^T=(v\otimes w).$
Now, restrict to $\,X\,$ where $u\in X$ and $n\in X\;$ By definition,
$\,(u\cdot n)n = (n\otimes n)u\,$ which is one part of the equality.
Now, taking transposes, $\,(u\cdot n)n^T=u^T(n\otimes n)\,$ which is
just the same result.
A: I don't think the middle term is equal to the other two. It should be more obvious using bra-ket notation:
$$\begin{array}{ccccc}
(\vec{u}\cdot\vec{n})\vec{n} &=&(\vec{n}
\otimes\vec{n})\vec{u} &≠&\vec{u}^{T}(\vec{n}\otimes\vec{n})
\\ ↕   &&  ↕ && ↕
\\ \big(⟨u∣n⟩\big)|n⟩  &=& \big(|n⟩⟨n|\big)|u⟩ &≠& ⟨u| \big(|n⟩⟨n|\big)
\end{array}$$
