# Change of Basis

1. Apr 12, 2015

### devd

Suppose a change of basis from basis $B$ to basis $C$ is represented by the matrix $S$.
That is, $S$ is the transformation matrix from $B$ to $C$.

Now if $t$ is a given linear transformation, $t:~V\rightarrow V$, with eigenvectors $\epsilon_i$, say, and $T$ is the representation of $t$ in $B$, then, the representation of t in $C$ is $STS^{-1}$.

Now, if the representation of $\epsilon_i$ in basis $B$ be $v^B_i$, then the representation of $\epsilon_i$ in basis $C$, $v^C_i=Sv^B_i$.

But shouldn't the vectors themselves transform in an opposite sense to the transformation of the basis, that is, shouldn't it be that $v^C_i=S^{-1}v^B_i$ ? I'm getting really confused. Please, can someone clarify?

2. Apr 13, 2015

### RUber

I agree with what you have here.
Remember that S:B-->C. So what would it mean to take the inverse on something in B? As in:
$v^C_i=S^{-1}v^B_i$.

3. Apr 14, 2015

### Fredrik

Staff Emeritus
Let's write $B=(e_1,\dots,e_n)$, $C=(f_1,\dots,f_n)$. Let $T$ be the linear operator such that $Te_i=f_i$ for all $i$. I won't be writing any summation sigmas. There's always a sum over the index or indices that appear exactly twice. We have
\begin{align}
([x]_B)_j e_j =x=([x]_C)_i f_i =([x]_C)_i Te_i = ([x]_C)_i (Te_i)_j e_j =([x]_C)_i ([T]_B)_{ji} e_j=([T]_B[x]_C)_j e_j.
\end{align} This implies that $[x]_B=[T]_B[x]_C$. If we define $S=([T]_B)^{-1}$, we can write this as $[x]_C=S[x]_B$, or equivalently as
$$([x]_C)_i =S_{ij} ([x]_B)_j.$$ We also have
$$f_i=Te_i =(Te_i)_j e_j =([T]_B)_{ji} e_j =(S^{-1})_{ji} e_j.$$

4. Apr 14, 2015