Some confusion about tensor notation I'm learning about tensor notation but I've become a little confused by an example in my notes as it seemingly contradicts what I've read elsewhere. 
What I know:
Superscripts denote column vectors , so $P^i=\begin{pmatrix} p^1 \\p^2\\p^3\end{pmatrix}$
Subscripts denote column vectors , so $L_i=\begin{pmatrix} l_1 & l_2 &l_3\end{pmatrix}$
Matrices are denoted $M_i^j=\begin{pmatrix} m_1^1 & m^1_2 &m^1_3 \\m_1^2 & m^2_2 &m^2_3 \\ m_1^3 & m^3_2 &m^3_3 \end{pmatrix}$.
In the example in my notes it says :
$$ds^2=dx^2+dy^2+dz^2=\begin{pmatrix} dx & dy &dz\end{pmatrix}\begin{pmatrix} 1 & 0 &0 \\0 & 1 &0 \\ 0 & 0 &1 \end{pmatrix} \begin{pmatrix} dx \\dy\\dz\end{pmatrix}$$
Which makes perfect sense so far but then it expresses it in tensor notation and it leaves me with a few questions , it says 
$$ds^2=\delta_{ij}dx^idx^j$$
Where $\delta_{ij}=\begin{pmatrix} 1 & 0 &0 \\0 & 1 &0 \\ 0 & 0 &1 \end{pmatrix}$ and Einstein summation convention is assumed. 
My questions: 
1) why is it okay to write $dx^idx^j$ , I thought upper indices were just for row vectors but then we'd be multiplying a row vector by a row vector which obviously we cant do , so what am I misunderstanding. 
2) In a similar vain why do we write $\delta_{ij}$ , instead of $\delta_i^j$
 A: You should ask yourself: what does it mean to write $ds^2=dx^2+dy^2+dz^2$ in the first place? The symbol $dx^2$ denotes $dx\otimes dx$, namely the tensor product of the form $dx$ with itself. However, no one can stop you from taking the tensor product $dx\otimes dy$, for instance. Using the notations $dx,dy,dz$ is customarily reserved for the case of forms on a $3-$dimensional space with coordinates $x,y,z$. If we pass to an $n-$dimensional space with coordinates $x^1,\ldots, x^n$, then our  associated forms are $dx^1,\ldots, dx^n$. 
$(1)$ As for the first question, you should view $dx^idx^j(=dx^i\otimes dx^j)$ as being a bilinear operator $\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ operating by $(v,w)\mapsto dx^i(v)\cdot dx^j(w).$ The pair of upper indices should signal to you that this is a "$2-$form" in that it eats two vectors and spits out a scalar number. 
$(2)$ We write $\delta_{ij}$ because when we write $ds^2=\delta_{ij}dx^idx^j$ we secretly mean 
$$ ds^2=\delta_{ij}dx^idx^j=\sum_{i,j} \delta_{ij}dx^idx^j$$
and the convention is that in a sum like this the coefficients of the forms get lower indices and  the forms get upper indices. (This is called the Einstein summation convention.)
