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I'm a bit confused about the use of $M^{-1}RM$ where $R$ is a transformation matrix.

Actually I was looking at the script here which reads and renders bvh files. But, I could not understand the meaning of line 483 which does similar job (where $M$ is transformation matrix of rest bone and $R$ is rotation matrix of pose bone).

Or, can anyone tell me what does $A^{-1}BA$ represent in general?

Thanks in advance, Mihir Gokani

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up vote 9 down vote accepted

If $A$ and $B$ are two matrices, then $ABA^{-1}$ is a matrix that acts similar to $B$, except on a different basis. $B$ and $ABA^{-1}$ are called similar.

If $v$ is a vector, we can compare the action of $B$ on $v$ to the action of $ABA^{-1}$ on $Av$. Namely, $ABA^{-1}(Av)=A(Bv)$. Thus, if adopt the notation $v'=Av$, we have $ABA^{-1}v'=(Bv)'$

We can understand things better if we think about what exactly a matrix is. Suppose that we have a basis of our space $v_1, \ldots, v_n$, and that $B$ representing a linear transformation $T_B$ in terms of the basis. This means that $T_B(v_j)=\sum_i b_{ij} v_i$. Phrased differently, the $j$th column of $B$ contains the coefficients required to express $T_B(v_j)$ in terms of the $v_i$.

So what do the entries in $ABA^{-1}$ express? Because $ABA^{-1}(Av)=A(Bv)$, we have $ABA^{-1}(Av_j)=A(Bv_j)=A(\sum_i b_{ij} v_i)=\sum_i b_{ij} Av_i$. Thus, $ABA^{-1}$ acts on the basis $Av_1,Av_2,\ldots Av_n$ just like $B$ acts on $v_1,\ldots v_n$. Thus, we can view them as both representing the same linear transformation $T_B$, just with respect to different bases.

If $v$ had been an eigenvector of $B$ with eigenvalue $\lambda$, for example, then $Av$ would be be an eigenvalue of $ABA^{-1}$ with eigenvalue $\lambda$. Similarly, $ABA^{-1}$ will have many of the same invariants that $B$ does, including characteristic polynomial (which implies they will have the same determinant, trace, and rank). In linear algebra, the properties of a linear transformation that we care about are exactly the ones that depend on the transformation but NOT the choice of basis, which means they are the ones that are the same for $B$ and $ABA^{-1}$ for every possible invertible $A$.

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Thanks for the explanation – mg007 Aug 8 '11 at 17:10

It represents the same transformation with respect to a different basis. See

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Thanks for the link! – mg007 Aug 8 '11 at 16:43

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