The full title question I could not ask due to character limitations: Why can't we define the scalar magnitude $S$ of the symmetric tensor $[S_{ij}]$ as the projection of $q_j$ onto a line parallel to $p_i$ divided by the vector magnitude $p_i$?
The question asked another way: Why can't we define the scalar magnitude $S$ of the symmetric tensor $[S_{ij}]$ in a particular direction as Eqn(1)? $$\tag{1} S=\frac{|\mathbf{p}|}{|\mathbf{q}|\cos \theta}=\frac{|\mathbf{p}|}{q_\parallel}$$
where $\cos \theta$ is the angle between the two vectors $p_i$ and $q_j$, which are mapped to each other through Eqn(2): $$\tag{2} p_i=S_{ij}q_j$$
In Eqn(1) $q_\parallel$ is the component of $\mathbf{q}$ that is parallel with $\mathbf{p}$. In Eqn(2) $S_{ij}=S_{ji}$ since the tensor is symmetric.
And perhaps my question is really: What is the consequence of trying to define the scalar magnitude $S$ of $[S_{ij}]$ as Eqn(1) and the explanation for why we cannot do so?
The motivation for my question and my attempt to answer the question are as follows.
Motivation for question: In the literature, the definition for the scalar magnitude $S$ of $[S_{ij}]$ as given in Eqn(2), is given by taking the projection of $\mathbf{p}$ onto a line parallel to $\mathbf{q}$ divided by the vector magnitude $|\mathbf{q}|$: $$\tag{3} S=\frac{p_\parallel}{|\mathbf{q}|}=\frac{|\mathbf{p}|\cos \theta}{|\mathbf{q}|}=\frac{\mathbf{p}\cdot \mathbf{q}/|\mathbf{q}|}{|\mathbf{q}|}=\frac{\mathbf{p}\cdot \mathbf{q}}{|\mathbf{q}|^2}$$
We can now write Eqn(3) in terms of $S_{ij}$: $$\tag{4} S=\frac{\mathbf{p}\cdot \mathbf{q}}{|\mathbf{q}|^2}=\frac{p_i q_i}{|\mathbf{q}|^2}$$ Invoking (2) into (4) $$\tag{5} S=\frac{S_{ij}q_j q_i}{|\mathbf{q}|^2}$$ Since $q_i$ is $|\mathbf{q}|l_i$, where $l_1, l_2, l_3$ are the direction cosines of the orientation in which $\mathbf{q}$ is oriented, then, $$\tag{6} S=\frac{S_{ij}l_i l_j |\mathbf{q}||\mathbf{q}|}{|\mathbf{q}|^2}=S_{ij}l_i l_j$$ which is the result I see published in literature.
My attempt: If I try to follow the same procedures as above but with the definition given by Eqn(1) I get the following equations [mirroring Eqns(3) through (6)]: $$\tag{3*} S=\frac{|\mathbf{p}|}{q_\parallel}=\frac{|\mathbf{p}|}{|\mathbf{q}|\cos \theta}=\frac{|\mathbf{p}|}{\mathbf{q}\cdot\mathbf{p}/|\mathbf{p}|}=\frac{|\mathbf{p}|^2}{\mathbf{q}\cdot\mathbf{p}}$$
$$\tag{5*} S=\frac{|\mathbf{p}|^2}{\mathbf{q}\cdot\mathbf{p}}=\frac{|\mathbf{p}|^2}{q_i p_i}=\frac{|\mathbf{p}|^2}{q_i S_{ij}q_j}=\frac{|\mathbf{p}|^2}{S_{ij}q_i q_j}$$
$$\tag{6*} S=\frac{|\mathbf{p}|^2}{S_{ij}q_i q_j}=\frac{|\mathbf{p}|^2}{S_{ij}l_i l_j|\mathbf{q}||\mathbf{q}|}=\frac{|\mathbf{p}|^2}{S_{ij}l_i l_j|\mathbf{q}|^2}$$
At this point I'm not sure if the steps I took to obtain Eqn(6*) are permissible, and if they are, how to interpret the result and the explanation for why it does not work to act as the scalar magnitude of $S$.
Perhaps an explanation along the lines of why Eqn(6) works with Eqn(2), and Eqn(6*) does not, is required for me to grasp this? Or perhaps the explanation lies within the transformation law for $[S_{ij}]$? I'm not sure and just throwing some thoughts out there for all of you.
If the matrix $S$ is $$ \pmatrix{0 & 0 & 1 \\ 1 & 0 & 0\\ 0 & 1 & 0} $$ and the $p_i$ are the standard unit vectors, then the $q_i$ will be those same vectors with index increased by $1$, i.e., $p_1 = e_1$, but $q_1 = e_2$.
For each $i$, the number $\theta$ (which you should have called $\theta_{ii}$, I suppose) will be 90 degrees or $\pi/2$ radians. In any case, its cosine will be zero, so the thing computed in equation 1 will be infinite. That seems like a Bad Thing.
So maybe that's one reason that this isn't the definition folks use.