covariant and contravariant components and change of basis

I encountered the following in reading about covariant and contravariant:

In those discussions, you may see words to the effect that covariant components transform in the same way as basis vectors (“co” ≈ “with”), and contravariant components transform in the opposite way to basis vectors (“contra” ≈ “against”). As you’ll see later in this chapter, there’s plenty of truth in that description, but there’s also a major pitfall. That’s because the “transformation” of basis vectors usually refers to the conversion of the basis vectors in the original (non-rotated) coordinate system to the different basis vectors which point along the coordinate axes in the new (rotated) system, whereas the “transformation” of vector components refers to the change in the components of the same vector referred to two different sets of coordinate axes.

Later on it shows the following:

$$\begin{pmatrix} \text{Components of} \\ \text{same vector} \\ \text{in new system} \end{pmatrix} = \begin{pmatrix} \text{Inverse} \\ \text{transformation} \\ \text{matrix} \end{pmatrix} \begin{pmatrix} \text{Components of} \\ \text{vector in} \\ \text{original system} \end{pmatrix}$$

$$\begin{pmatrix} \text{New basis} \\ \text{vectors} \end{pmatrix} = \begin{pmatrix} \text{Direct} \\ \text{transformation} \\ \text{matrix} \end{pmatrix} \begin{pmatrix} \text{Original basis} \\ \text{vectors}\end{pmatrix}$$

These confuse me because from change of basis we have $B'= BP$ and $[v]_{B'}=P[v]_B$. The $[v]_{B'}=P[v]_B$ is the first of the aforementioned equations. But the second of them I don't understand since the book as shown above has the direct transformation matrix on the left side and not the right side. We could write $B'= BP = (BPB^{-1})B$ but if $BPB^{-1}$ is the direct transformation matrix then the $P$ in $[v]_{B'}=P[v]_B$ does not make sense as the inverse transformation matrix as the inverse of $BPB^{-1}$ is not $P$.

So then later on when I read the following:

you can combine superscripted (contravariant) components with subscripted (covariant) basis vectors

I don't know how to reconcile it with what I already know about change of basis.

To understand, WLOG let's take a basis change given by $$b_1={B^1}_1e_1+{B^2}_1e_2$$ $$b_2={B^1}_1e_1+{B^2}_2e_2$$ where $\{e_1,e_2\}$ is an old basis and $\{b_1,b_2\}$ is the new. Relation which can be succintily expressed as $b_i={B^s}_ie_s$ (here we see how bases covariate ).

Agree that the matrix of such data is $$[B]=\begin{bmatrix} {B^1}_1, {B^1}_2\\ {B^2}_1, {B^2}_2\\ \end{bmatrix}$$ Then, to get the new components of a vector $v=v^1e_1+v^2e_2$, you will see that $$v_b=[B]^{-1}v_e,$$ (here we see how components contravariate ), $v_e$ is a column arrange from the old components; $v_b$ is the data on the new components of the very same vector $v$.

Unfolded is $$ \begin{bmatrix} w^1\\ w^2\\ \end{bmatrix} \ =\ \begin{bmatrix} {B^1}_1, {B^1}_2\\ {B^2}_1, {B^2}_2\\ \end{bmatrix}^{-1} \begin{bmatrix} v^1\\ v^2\\ \end{bmatrix}$$ such that $v=w^1b_1+w^2b_2$ in the new basis.

Take an explicit example to illuminate even more: Let $$b_1=e_1+2e_2,$$ $$b_2=e_1+3e_2,$$ be a basis change. Its change-of-basis matrix is $[B]= \begin{bmatrix} 1& 1\\ 2&3\\ \end{bmatrix} $.

Now solving for $e_i$ we get $$e_1=3b_1-b_2,$$ $$e_2=-2b_1+b_2.$$ Which substitution on $v$ gives: $$v=v^1(3b_1-b_2)+v^2(-2b_1+b_2).$$ This simplifies into $$v=(3v^1-2v^2)b_1+(-v^1+v^2)b_2.$$

Now follow with your eyes the $[B]^{-1}v_e$ product: $$ \begin{bmatrix} 3&-1\\ -2&1\\ \end{bmatrix} \begin{bmatrix} v^1\\ v^2\\ \end{bmatrix} = \begin{bmatrix} 3v^1-v^2\\ -2v^1+v^2\\ \end{bmatrix}. $$

Let me try to explain using a "foreign" vector space other than $\mathbb{R}^3$, such as the space of polynomials of degree $\leq 2$. Of course, all finite-dimensional real vector spaces of the same dimension are isomorphic, but hopefully this will eliminate some of the confusion that arises when everything in sight is $\mathbb{R}^3$.

So let's say I have a basis $B$ for these polynomials, say $\{1, x, x^2\}$. I can write any random element of my vector space, like $3x^2+2x+1$, as a linear combination of those basis elements: $1\cdot 1 + 2\cdot x + 3\cdot x^2$, so I might represent that polynomial as the vector $(1,2,3)$ in that basis.

Suppose I have a second basis $B'$ for these polynomials; let's say $\{x+1, x-1, x^2-1\}$. There are now two kinds of transformations relating $B$ to $B'$ I might be interested in.

1. How do I express the new basis vectors as linear combinations of the old? In other words, what is the matrix $P$ with $$\left[\begin{array}{ccc}x+1&x-1&x^2-1\end{array}\right] = \left[\begin{array}{ccc}1&x&x^2\end{array}\right]P?$$ In this case it is $$P=\left[\begin{array}{ccc}1 & -1 & -1\\1 & 1 & 0\\0 & 0 & 1\end{array}\right].$$ Notice that the components of $B$ and $B'$ are different polynomials. There is no relationship between the components of $B$ and $B'$ other than the fact that both bases span the whole space.

2. Given a vector in basis $B$, how do I write the same vector in basis $B'$? For instance, for the example of $3x^2 + 2x+1$ we saw that I could write it as $(1,2,3)$ in $B$. The same polynomial is $(3,-1,3)$ in the basis $B'$. And the two are related by $$\left[\begin{array}{c}3\\-1\\3\end{array}\right] = P^{-1}\left[\begin{array}{c}1\\2\\3\end{array}\right]$$ This can be seen to work in general by first converting the component vector $v$ into the full polynomial $Bv$, the using the relationship $B' = BP$ to get that that polynomial can also be written as $B'P^{-1}v$ and so the coefficients of the polynomial in the basis $B'$ are $P^{-1}v$.

The two key points are that a) in transformation 1, the objects being transformed are basis polynomials, while in transformation 2, the objects are vectors of real coefficients, and b) that in transformation 2, the old coefficients and new coefficients represent the same polynomial, written in a new way. Compare to transformation 1, where the first basis polynomial of $B$ is not the same polynomial as the first basis polynomial of $B'$.

By the way: the language of covariant/contravariant vectors is too entrenched in physics to go anywhere unfortunately, but if you pick up a good book on Riemannian geometry and read at least the first few chapters up through the discussion of tangent space, cotangent space, and the musical isomorphisms, it may give you a new perspective on what is going on with all those upper/lower indices.