How exactly is the magnitude of a vector of non-orthonormal basis calculated intuitively? I'll preface this question by stating that I am a beginner of sorts to the concept of vectors, and even more so for linear algebra in general.
From what I understand, for some given vector $(1,2)$, the magnitude for said vector for basis $\hat i $ and $\hat j$, in my textbook from my highschool I am studying from,  is given as the square root of the square of the components, 1 and 2. I can tell how it intuitively makes sense, taking into account how this vector in question can be decomposed into vectors that solely point along the basis , which with the vector itself, forms a right-angle triangle. Pythagoras's theorem leads the way for the rest
From what I figured, finding the components of my original vector along either of the basis vectors is as simple as projecting over them.
How would I find the magnitude for such a vector belonging to a non-orthonormal basis however (I'd prefer to understand it for a two-dimensional case for simplicity)? I'd explored a video about 'metric tensors', a particular matrix-like object from what I understand, that did the job of finding the magnitude given any general basis vector, from what little I could glean from it. However, I was left even more confused towards the end, since I couldn't quite understand how they worked. They'd also explained that the Pythagorean Law-based formula of finding the magnitude only works for basis vectors that are orthonormal
Here is what I assume the diagram for a vector of  some basis vectors, $\vec e1, \vec e2$ would look like:

The blue lines denote the decompositions along either basis
I can sort of tell why the Pythagoras-based formula to find the magnitude wouldn't work here.
The decomposition of the vector along $\vec e1$ does give me an accurate idea of HOW much of the vector is along $\vec e1$. But with the projection of the vector along $\vec e2$, something seems to be wrong
On second thought, something about my entire approach in of itself seems to be dreadfully wrong. I can't put my finger on what about it exactly that I am doing wrong
Even if I assume that I find the projections along either basis correctly, I am not sure how to move forward. I can't seem to think about how I could find the magnitude intuitively even if the projections along each basis vector were provided
I am sure I am getting something wrong. How exactly can one intuitively find the magnitude of some vector with respect to such orthonormal basis?
 A: $\newcommand{\brak}[1]{\left\langle#1\right\rangle}$Two snags that may explain why something seems wrong are:

*

*A vector space need not be a Cartesian space. For example, the set of polynomials of degree at most five on the interval $[-1, 1]$ is a real vector space of dimension six. Or, a plane tangent to a surface in Euclidean three-space may be viewed as a two-dimensional real vector space, but it has no natural choice of basis so we can't view its elements naturally as ordered pairs.

*The concept of length comes from additional structure that must be chosen, such as a "norm" or an "inner product."

A norm is a type of axiomization of length of vectors. Knowing the lengths of basis vectors in a normed vector space, however, does not determine the lengths of linear combinations. A "normed vector space" is therefore not sufficient to support the question as formulated.
By contrast, an inner product is an axiomization of the Euclidean dot product, and as a mathematical setting is (probably) implicitly what you had in mind when writing the question. The concept of an orthonormal basis only makes sense in an inner product space, a real vector space equipped with an inner product.
Precisely, an inner product on a real vector space $(V, +, \cdot)$ is a real-valued function $\brak{\ ,\ }$ on ordered paris of vectors satisfying the following axioms:

*

*(Symmetry) $\brak{v, u} = \brak{u, v}$ for all $u$, $v$ in $V$;

*(Bilinearity) $\brak{cu + v,w} = c\brak{u, w} + \brak{v, w}$ for all $u$, $v$, $w$ in $V$ and all real $c$;

*(Positive-definiteness) $\brak{v, v} \geq 0$ for all $v$ in $V$, with equality if and only if $v = 0$.

The length of a vector $v$ in an inner product space is defined to be $\|v\| = \sqrt{\brak{v, v}}$. If $(e_{j})_{j=1}^{n}$ is a basis of $V$ and $v = \sum_{j} v_{j} e_{j}$, then bilinearity guarantees
$$
\brak{v, v} = \sum_{j,k=1}^{n} v_{j} v_{k} \brak{e_{j}, e_{k}}.
$$
That formula (or really, its square root) is the answer to the question.
If the basis $(e_{j})$ is orthonormal, then $\brak{e_{j}, e_{k}} = \delta_{jk}$ is $1$ if $j = k$ and $0$ otherwise. In that situation, $\|v\| = \sqrt{\sum_{j} v_{j}^{2}}$, a generalized Pythagorean theorem.
