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If $L' = S^{-1}LS$ where $L$ is a linear map wrt basis $B_1$, and $S$ has columns of coefficients of another basis $B_2$ wrt $B_1$. Could someone please explain what this means? Is $L'$ acting on vectors expressed in terms of $B_2$ and gives them in terms of $B_2$? Thx.

Specifically could someone explain what is going on with each $S$ and $S^{-1}$? like changing to some basis or changing back or something...?

Added: Suppose one wants to feed in vectors wrt one set of basis and want the map to spit out transformed vector in another basis, what does one do?

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Simeone's not in today, so I'll take a shot instead :). $L'$ represents the same transformation as $L$, only with respect to a different basis. Then, as you said, $L'$ is defined on the basis $B_2$. Is that your question? – gary Sep 12 '11 at 20:43
Thx, gary. :) Could you possibly explain what each S or $S^{-1}$ are doing? – kiddo Sep 12 '11 at 20:46
I'm still looking for a nice answer to your original. for the addition, given bases $B_1,B_2$ , there is an invertible matrix $M$ taking basis elements $b_i$ in $B_i$ to $b_j$ in $B_j$. Then, if , say, $L(v_i)= v_j$, if you want an expression in terms of a basis $B_k$ , then you can write $v_j$ (as a linear combination)in terms of the basis $B_j$ , as $v_j=a_1b_{j1}+a_2b_{j2}+...+a_nb_{jn}$, then use the map taking $b_jn$ to $b_{kn}$ – gary Sep 12 '11 at 21:19
Just a comment: once you know what $S$ is doing, you can tell what $S^{-1}$ is doing, as just inverting $S$.What is going on is that $SL'=LS$. But I don't know if I can give you a better answer at this point. – gary Sep 12 '11 at 21:23
up vote 0 down vote accepted

Say the basis vectors in $B_1$ and $B_2$ are $e_{B_1}^1, e_{B_1}^2, \dots e_{B_1}^n$ and $e_{B_2}^1, e_{B_2}^2, \dots, e_{B_2}^n$ respectively.

From your definition of the matrix $S = (s_{ij})$, the first column is $e_{B_2}^1$ expressed in $B_1$, or in general, the $i$:th column is $e_{B_2}^i$ expressed in $B_1$, thus:

$$Se_{B_2}^j = \sum_{i = 1}^n s_{ij} e_{B_1}^i$$

Say you have have the $B_2$-coefficients of a vector $v = \sum_{k=1}^n \alpha_k e_{B_2}^k$. Applying $S$ to this vector yields:

$$ \begin{align*} Sv &= S \left( \sum_{k=1}^n \alpha_k e_{B_2}^k \right) = \sum_{k=1}^n S (\alpha_k e_{B_2}^k ) = \sum_{k=1}^n \sum_{i=1}^n \alpha_k s_{ik} e_{B_1}^i = \\ &= \sum_{i=1}^n \sum_{k=1}^n \alpha_k s_{ik} e_{B_1}^i = \sum_{i=1}^n \left( \sum_{k=1}^n \alpha_k s_{ik} \right) e_{B_1}^i \end{align*} $$ and you get the coefficients $\sum_{k=1}^n \alpha_k s_{ik}$ for the vector in the basis $B_1$.

So, yes, $S$ maps the coefficient vector for a vector in basis $B_2$ to the coefficient vector in the basis $B_1$. Per definition, $S^{-1}$ does the opposite. In $S^{-1}TS$, your first map from basis $B_2$ to basis $B_1$, then applies the linear transformation, then change back from $B_1$ to $B_2$.

Say, you have your transformation $L' = S^{-1}LS$, which maps vectors expressed in $B_2$ to vectors expressed in $B_2$. If you want to feed in vectors expressed in $B_2$, but get vectors expressed in $B_1$, you can just skip the changing back part and use $LS$. If you want to feed it vectors in $B_1$ and get vectors in $B_2$, you don't need to change basis before applying $L$, so $S^{-1}L$ will do.

You can see this as changing basis once more before or after. You know that your change of basis matrix is $S$, and $L'$ takes vectors expressed in $B_2$ and gives you vectors expressed in $B_2$. If you want the output to be in $B_1$, apply $S$ after $L'$, i.e. use $SL' = SS^{-1}LS = LS$.

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You basically say the answer in your question. Your map $L'$ acts on vectors given wrt basis $B_2$. The map $S$ rewrites the vector in terms of basis $B_1$, so that you can apply $L$. Then the map $S^{-1}$ rewrites the result in terms of basis $B_2$ again.

I find this easiest to visualize when $S$ represents a rotation of the given space. Perhaps $L$ represents a stretching in the $x$ direction, but I want to stretch my space in the $y$ direction. Then the easiest thing to do is to secretly rotate my space so that the $y$ vector points in the $x$ direction, apply my $L$ stretch, and the rotate back. The result is that I just stretched in the $y$ direction.

For your second question, you can just leave off the matrix $S^{-1}$. The matrix $S$ rewrites your vector in the new basis, and then $L$ acts on it in that basis, giving you an answer in terms of the new basis.

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