Understanding the notation in the definition of a tensor through an example. A $(p,q)$ tensor, $T$ is a MULTILINEAR MAP that takes $p$ copies of $V^*$ and $q$ copies of $V$ and maps multilinearly (linear in each entry) to $k:$
$$T: \underset{p}{\underbrace{V^*\times \cdots \times V^*}}\times \underset{q}{\underbrace{V\times\times \cdots  V\times V}} \overset{\sim}\rightarrow K\tag 1$$
The $(p,q)$ TENSOR SPACE is defined as a set:
$$\begin{align}T^p_q\,V &= \underset{p}{\underbrace{V\otimes\cdots\otimes V}} \otimes \underset{q}{\underbrace{V^*\otimes\cdots\otimes V^*}}:=\{T\, |\, T\, \text{ is a (p,q) tensor}\}\tag2\\[3ex]&=\{T: \underset{p}{\underbrace{V^*\times \cdots \times V^*}}\times \underset{q}{\underbrace{V\times \cdots \times V}} \overset{\sim}\rightarrow K\}\end{align}\tag3$$ 
is the set of all tensors where $T$ is (p,q), equipped this with pointwise addition and s-multiplication.

I can't find an example online to get an idea of what these expressions mean. I have followed, for example, all 25 lectures on tensors on this series, but these expressions are not even mentioned. I'd like to see an example that is not completely trivial, and that it could be have been dealt with using linear algebra - something with "arrow vectors" and matrices, perhaps, so that the linear functional(s) in $V^*$ and the vectors in $V$ are clearly spelled out, together with the operations entailed ($\otimes$).
If asking for an example is not a good question, a step-by-step translation in English of what these expressions are saying would be great.
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$Let $(\Basis_{j})_{j=1}^{n}$ denote the standard basis of $V = \Reals^{n}$ and let $(\Basis^{i})_{i=1}^{n}$ be the dual basis of $V^{*} = (\Reals^{n})^{*}$. (Where possible below, I've taken case to use the dummy indices $i$ and $j$ "globally".)


*

*The identity transformation $I_{n}:\Reals^{n} \to \Reals^{n}$ is
$$
\sum_{i,j=1}^{n} \delta_{i}^{j}\, \Basis_{j} \otimes \Basis^{i}
= \sum_{j=1}^{n} \Basis_{j} \otimes \Basis^{j}.
$$
Specifically, if $v = \sum\limits_{j=1}^{n} v^{j} \Basis_{j}$, then
$$
I_{n}(v) = \sum_{j=1}^{n} \Basis_{j} \otimes \Basis^{j}(v)
= \sum_{j=1}^{n} v^{j}\Basis_{j}
= v.
$$
Similarly, if $A = [a_{i}^{j}]$ is an $n \times n$ matrix, the tensor
$$
T = \sum_{i,j=1}^{n} a_{i}^{j}\, \Basis_{j} \otimes \Basis^{i} \in T_{1}^{1}\Reals^{n}
$$
is the linear operator whose standard matrix is $A$.
If $\Basis_{j}$ is written as an $n \times 1$ column matrix with a $1$ in the $j$th row and $0$'s elsewhere, then $\Basis^{i}$ is the $1 \times n$ row matrix with a $1$ in the $i$th column and $0$'s elsewhere, and the tensor product $\Basis_{j}^{i} := \Basis_{j} \otimes \Basis^{i}$ may be denoted with ordinary matrix multiplication, the outer product of a column and a row, the $n \times n$ matrix with a $1$ in the $(i, j)$-entry and $0$'s elsewhere.

*The Euclidean inner product is
$$
\Brak{\ ,\ } = \sum_{i=1}^{n} \Basis^{i} \otimes \Basis^{i}.
$$
If $u$ and $v$ are arbitrary vectors, then
$$
\Brak{u, v} = \sum_{i=1}^{n} \Basis^{i}(u)\, \Basis^{i}(v)
= \sum_{i=1}^{n} u^{i}\, v^{i}.
$$

*If $n = 2$, the determinant viewed as a bilinear function of two vectors in $\Reals^{2}$ is
\begin{align*}
\det &= \Basis^{1} \otimes \Basis^{2} - \Basis^{2} \otimes \Basis^{1} \in T_{2}^{0} \Reals^{2}; \\
\det(u, v) &= \Basis^{1}(u)\, \Basis^{2}(v) - \Basis^{2}(v)\, \Basis^{1}(u) \\
&= u^{1} v^{2} - u^{2} v^{1}.
\end{align*}

*Similarly, if $n = 3$, the ordinary cross product is
\begin{align*}
&(\Basis^{2} \otimes \Basis^{3} - \Basis^{3} \otimes \Basis^{2})\otimes \Basis_{1} \\
+ &(\Basis^{3} \otimes \Basis^{1} - \Basis^{1} \otimes \Basis^{3})\otimes \Basis_{2} \\
+ &(\Basis^{1} \otimes \Basis^{2} - \Basis^{2} \otimes \Basis^{1})\otimes \Basis_{3}
 \in T_{2}^{1} \Reals^{3}.
\end{align*}
