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.
Best Answer
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: $$ \begin{bmatrix} 3&-1\\ -2&1\\ \end{bmatrix} \begin{bmatrix} v^1\\ v^2\\ \end{bmatrix} = \begin{bmatrix} 3v^1-v^2\\ -2v^1+v^2\\ \end{bmatrix}. $$