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A stretch transformation can be represented as:

$$ \begin{bmatrix} k & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

However, this changes the volume of any object which it operates on by a factor of $k$. I'm looking for a way to perform such a stretch while keeping the volume constant, but I'm not sure how exactly to go about doing this. I'm considering something like:

$$ \begin{bmatrix} k^{2/3} & 0 & 0 \\ 0 & k^{-1/3} & 0\\ 0 & 0 & k^{-1/3} \end{bmatrix} $$

Unfortunately, I can't find any good justification for why I should reduce each dimension by a factor of $k^{1/3}$. I also considered simply keeping the factor of $k$ in my volume and later renormalizing by dividing the volume of each object in the "new" coordinate system by $k$, however this doesn't address how I should treat a vector that transforms into the new coordinate system.

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1 Answer 1

up vote 5 down vote accepted

Your intuition indeed gives you the right result. The volume change factor after applying the matrix is precisely the determinant of the matrix. If you want a transformation which retains volume then you will want a determinant of $1$. The determinant of the original matrix is $k$ and what you've done is scale the matrix to give it determinant $1$.

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