If $P$ changes the basis from $e$ to $f$, then $P^{−1}$ changes the coordinates from $[v]_e$ to $[v]_f$

I was reading this article on the change of basis, and when I finally thought I understood the concept, it comes out another inconsistency, just because people write differently from place to place, and everything, even simple things like this, becomes confusing.

In that article, it's roughly written the following (with some my own additions of details and comments).

Suppose we have two arbitrary basis $f = \{f_1, f_2 \}$ and $e = \{ e_1, e_2 \}$ for $\mathbb{R}^2$, and an arbitrary vector $v$.

We know that $$v = c_1 e_1 + c_2 e_2$$ so $$[v]_e = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$$

That is $c_1$ and $c_2$ are the coordinates of $v$ with respect to the basis $e$. (I think therefore that $v$ is represented with respect to the standard basis). $v$ is a linear combinations of the two column vectors of $e$, and that previous expression can be written in matrix-vector product

$$v = E \cdot [v]_e = \begin{pmatrix} e_1 & e_2 \end{pmatrix} [v]_e$$

Similarly, $v$ can also be represented with respect to the basis $f$ as follows

$$v = F \cdot [v]_f = \begin{pmatrix} f_1 & f_2 \end{pmatrix} [v]_f$$

Conclusion:

$$v = E \cdot [v]_e = F \cdot [v]_f$$

(For now, nothing strange for me, assuming that $v$ is in practice the coordinates of a vector with respect to the standard basis).

Since $e$ is a basis, we can represent $f_1$ and $f_2$ as a linear combination of $e_1$ and $e_2$ as follows:

$$f_1 = a e_1 + b e_2 \\ f_2 = c e_1 + d e_2$$

(Well this makes sense for me).

The change of basis matrix from $E$ to $F$ is $$P = \begin{pmatrix} a & c \\ b & d\end{pmatrix}$$ (Note yet sure why, but I'm showing you now).

Note that $$E = \begin{pmatrix} e_1 & e_2 \end{pmatrix} \\ F = \begin{pmatrix} f_1 & f_2\end{pmatrix}$$

Assume $e_1 = \begin{pmatrix} e_{11} \\ e_{21} \end{pmatrix}$, $e_2 = \begin{pmatrix} e_{12} \\ e_{22} \end{pmatrix}$, $f_1 = \begin{pmatrix} f_{11} \\ f_{21} \end{pmatrix}$ and $f_2 = \begin{pmatrix} f_{12} \\ f_{22} \end{pmatrix}$. So, lets re-write $E$ and $F$

$$E = \begin{pmatrix} e_{11} & e_{12} \\ e_{21} & e_{22}\end{pmatrix} \\ F = \begin{pmatrix} f_{11} & f_{12} \\ f_{21} & f_{22}\end{pmatrix}$$

Now, take the following $$F = EP$$ or $$\begin{pmatrix} f_{11} & f_{12} \\ f_{21} & f_{22}\end{pmatrix} = \begin{pmatrix} e_{11} & e_{12} \\ e_{21} & e_{22}\end{pmatrix} * \begin{pmatrix} a & c \\ b & d\end{pmatrix}$$

If we multiply the two matrices on the right we obtain

$$\begin{pmatrix} f_{11} & f_{12} \\ f_{21} & f_{22}\end{pmatrix} = \begin{pmatrix} ae_{11} + be_{12} & ce_{11} + d e_{12} \\ ae_{21} + be_{22} & ce_{21} + d e_{22}\end{pmatrix}$$

Note that $$f_1 = \begin{pmatrix} f_{11} \\ f_{21}\end{pmatrix} = \begin{pmatrix} ae_{11} + be_{12} \\ ae_{21} + be_{22}\end{pmatrix} = \begin{pmatrix} e_{11} & e_{12} \\ e_{21} & e_{22} \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix}$$

which is the same thing as $$f_1 = a e_1 + b e_2$$

The same reasoning can be done for the second column of both matrices.

In other words, I've just shown that $$f_1 = a e_1 + b e_2 \\ f_2 = c e_1 + d e_2$$ is equivalent to $$F = E P$$

Now suppose $v$ has the known coordinates $[v]_e$ in the basis $e$, and $F = EP$, then $$v = E [v]_e = \left(F P^{-1} \right) \cdot [v]_e = F [v]_f$$

Take this part:

$$F P^{-1} \cdot [v]_e = F [v]_f$$

And remove from both sides $F$, we obtain $$P^{-1} \cdot [v]_e = [v]_f$$

In other words, if $P$ changes the basis from $E$ to $F$, then $P^{−1}$ changes the coordinates from $v_e$ to $v_f$, which was my doubt.

Apparently, change of basis is, as the name suggests, really a change of basis, i.e. it changes one basis into the other, i.e. it converts the vectors in one basis to the vectors in the other.

I've just realised that I solved my original problem related to this question, but I still have a few doubts.

In the literature, people usually say that $E$ and $F$ are actually the change of basis and not $P$, but it isn't true.

1. Why people keep saying that $E$ and $F$ are the change of basis? They should be called the change of coordinates of a vector represented in a basis $E$ or $F$ (respectively) to the standard basis's representation of that same vector, given of course the representation of the vector with respect to $E$ and respectively $F$ (of course if I understood correctly everything).
• I would like to add that more than $50 \%$ of the times math is difficult because of inconsistencies between notations of different resources or even in the same resource. – nbro Feb 28 '16 at 1:54

1 Answer

$e$ and $e$ are bases of $\mathbf R^2$ (or of any vector space, actually), and $P$ is called the change of basis matrix (from basis $e$ to basis $f$). One can oberve that $P$ is the matrix of the identity map, from $\mathbf R^2$ with basis $f$ to $\mathbf R^2$ with basis $e$ (note the bases swapping), and it gives an expression of the coordinates in basis $e$ in function of the coordinates in basis $f$: $$[v]_e=P[v]_f.$$ Furthermore, if a linear map has matrix $A_e$ in basis $e$, its matrix in basis $e$ is $$A_f=P^{-1}A_eP.$$ Note:

None of the bases has to be the standard basis, which is meaningless anyway for a general vector space.

• Now, wait, the bases are actually $e$ and $f$, i.e. two sets containing the lin. ind, etc. $E$ and $F$ are matrices (therefore not basis) whose columns are the vectors in the basis of $e$ and $f$ respectively. That why I'm suggesting to call them change of coordinates with respect to the standard basis. – nbro Feb 28 '16 at 1:39
• OK I didn't grasp well you (non standard) notations. I'll correct my answer – Bernard Feb 28 '16 at 1:41
• Let me just ask you a confirmation. $v$ in my explanations are the coordinates of a vector with respect to the standard basis (as I'm suggesting), right? – nbro Feb 28 '16 at 1:59
• You (or your teacher) decided that it's in the standard basis, but this is in no way necessary. What is the standard basis in a abstract vector space? – Bernard Feb 28 '16 at 2:58