Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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 $SL' = SS^{-1}LS = LS$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.