The real Matrix of change of basis. Not really. Only in $\mathbb{R}^n$.

Question

Suppose we have an $\mathbb{R}$-vector space $E$, $\text{dim}(E)=n$, and two bases $\alpha:=\{v_i\}$ and $\beta:=\{w_i\}$ of it.

We can consider the maps to $\mathbb{R}^n$ given by the coordinates $v\mapsto [v]_\alpha$ and $v\mapsto[v]_\beta$. Let $I:E\rightarrow E$ be the identity function and let $[I]_{\alpha}^{\beta}$ be the matrix of $I$ in the bases $\alpha$ and $\beta$. We have then, from the construction of matrix of a linear transformation, that $$[v]_{\beta}=[I]_{\alpha}^{\beta}\cdot[v]_{\alpha}.$$

This matrix $[I]_{\alpha}^{\beta}$ is often called the matrix of change of coordinates (from $\alpha$ to $\beta$), and sometimes the matrix of change of basis (from $\alpha$ to $\beta$). But let us call it here only the matrix of change of coordinates (from $\alpha$ to $\beta$) since that is what is doing.

Consider now $T:E\rightarrow E$ the linear transformation defined by $T(v_i):=w_i$, for all $i$. This linear transformation is sending the vectors of the basis $\alpha$ to the vectors of the basis $\beta$. Let us call it the transformation of change of basis (because that is what is doing).

Assume from this point on that $E=\mathbb{R}^n$. Then we have the standard basis $e:=\{e_i\}$, which has the property that coordinates in that basis look like the same vectors, i.e. $[v]_e=v$, for every $v\in E=\mathbb{R}^n$. Therefore the matrix $[T]_{e}^{e}$ of $T$ in the basis $e$ satisfies

$$w_i=[w_i]_e=[T(v_i)]_e=[T]_{e}^{e}\cdot[v_i]_e=[T]_{e}^{e}v_i$$

Therefore, this matrix $[T]_{e}^{e}$ is sending the vectors of the basis $\alpha$ to the vectors of the basis $\beta$. We could call it matrix of change of basis since that is what is doing.

Questions:

1. What is the relationship between $[T]_{e}^{e}$ and $[I]_{\alpha}^{\beta}$?
2. What is $[T]_{e}^{e}$ usually called?

Example:

Assume that $E=\mathbb{R}^2$, $\alpha=\{v_1=\begin{bmatrix}1\\1\end{bmatrix}, v_2=\begin{bmatrix}2\\1\end{bmatrix}\}$, and $\beta=\{w_1=\begin{bmatrix}2\\2\end{bmatrix}, w_2=\begin{bmatrix}3\\2\end{bmatrix}\}$.

Then we have

$$\begin{align}v_1&=\phantom{-}\frac{1}{2}w_1+0w_2\\v_2&=-\frac{1}{2}w_1+1w_2\end{align}$$

Therefore

$$[I]_{\alpha}^{\beta}=\begin{bmatrix}\frac{1}{2}&-\frac{1}{2}\\0&\phantom{-}1\end{bmatrix}$$

To compute $[T]_{e}^{e}$ we can evaluate $T(e_1)$ and $T(e_2)$, expand them in the basis $e$ and put the coefficients as columns in a matrix. Since $$\begin{align}e_1&=-1v_1+1v_2\\e_2&=2v_1-v_2\end{align}$$

we get that $$\begin{align}T(e_1)&=T(-v_1+v_2)&=-w_1+w_2\\T(e_2)&=T(2v_1-v_2)&=2w_1-w_2\end{align}$$

This tells us that

$$[T]_{e}^{e}=\begin{bmatrix}1&1\\0&2\end{bmatrix}.$$

Notice that when we compute the inverse of $[I]_{\alpha}^{\beta}$ in this case, we don't get $[T]_{e}^{e}$ as this answer seems to be implying. We get

$$([I]_{\alpha}^{\beta})^{-1}=\begin{bmatrix}2&1\\0&1\end{bmatrix}$$

Answer 1

(This is a much-revised answer).

Doing my best to follow your notation, I'm going to say that $$ [s]_B $$ is the vector of coefficients $(c_1, \ldots, c_n)^t$ with the property that $c_1 b_1 + \ldots + c_n b_n = s$, where $B$ is the basis whose vectors are $b_1, b_2, \ldots$.

I'm going to write $e_i$ for the $i$th standard basis vector in $R^n$ -- the one with all entries 0 except the $i$th, which is 1. This unfortunately means that $$ [e_i]_E = e_i $$ where $E = e_1, e_2, \ldots$ is the standard basis.

Let $V$ be a matrix such that $$ V[e_i]_E = [v_i]_E $$

i.e., the $i$th column of $V$ contains the standard coordinates of $v_i$. In your example, $$ V = \begin{bmatrix} 1 & 2 \\ 1 & 1\end{bmatrix}. $$

Define $W$ analogously.

Then we have \begin{align} [e_i]_E &= V^{-1} [v_i]_E \text{ (1) }\\ [e_i]_E &= W^{-1} [w_i]_E \text{ (2) }\\ V[e_i]_E &= [v_i]_E \text{ (3) }\\ W[e_i]_E &= [w_i]_E \text{ (4) }\\ \end{align}

Your definition of $T_e^e$, is, in this notation, that it's the matrix such that \begin{align} [w_i]_E &= T_e^e [v_i]_E \\ W[e_i]_E &= T_e^e [v_i]_E \text{, by eq 4}\\ W[e_i]_E &= T_e^e V[e_1]_E \text{, by eq 3} \end{align} Since this holds for each $i$, we can laminate together all the $[e_i]_E$ into the identity matrix to get \begin{align} W I &= T_e^e VI \text{, so that}\\ W V^{-1} &= T_e&e. \end{align}

$$\newcommand{\Iab}{I_\alpha^\beta}$$ Now let's look at $\Iab$.

First, look at $v = v_i$, the $i$th entry of the first basis. In the first basis, its coordinates are just $(1, 0, 0, \ldots)^t$, i.e., $$ [v_i]_\alpha = [e_i]_E $$ That's a statement about equality of coordinate representations.

Second, according to equation (1) above, we have \begin{align} [e_i]_E &= V^{-1} [v_i]_E \end{align}

Combining, we get \begin{align} [v_i]_\alpha &= V^{-1} [v_i]_E \end{align} We can apply this to a linear combination of the $v_i$ to get \begin{align} [v]_\alpha &= V^{-1} [v]_E \text{ (5) } \end{align} and correspondingly, \begin{align} [v]_\beta &= W^{-1} [v]_E \text{ (6) } \end{align}

Now it all falls into place. with substitution, we get

\begin{align} [v]_\beta &= \Iab [v]_\alpha \text{, from the definition of $\Iab$}\\ W^{-1} [v]_E &= \Iab [v]_\alpha \text{, applying Eq. (6)}\\ W^{-1} [v]_E &= \Iab V^{-1} [v]_E \text{, applying Eq. (5)}\\ W^{-1} &= \Iab V^{-1} \text{, because prev. eqn is true for all $v$}\\ W^{-1}V &= \Iab \end{align}

Here's a matlab script verifying that these computations lead to the result you got when you worked it out by hand:

v1 = [1;1]
v2 = [2;1]
w1 = [2;2]
w2 = [3;2]
V = [v1, v2]
W = [w1, w2]
Tee = W * V^(-1)
Iab = (W^(-1)) * V

Summary: $\Iab = W^{-1} V$; $T_e^e = W V^{-1}$.

And just to make the comment-stream below continue to make sense, I'll suggest that putting the $\alpha$ up and the $\beta$ down leads to a nice situation in which "up" and "down" indices appear to cancel.

For your second question ("what is $T_e^e$ usually called?") I'd say "The matrix of the transformation taking the $v$s to the $w$s," at least in the case where the $v$s and $w$s are in $R^n$. That's the term I've heard used most often, anyhow. It may be that a poll of mathematicians would reveal something different.

One last thought: when one of the two bases actually is the standard basis, then either $W$ or $V$ is the identity, and the two matrices $\Iab$ and $T_e^e$ end up being inverses.