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I am interested in finding the ratio of area formed by transformed and original sides of a parallelogram, given by: $$\frac{\|Ma\times Mb\| }{\| a\times b \|}$$ $M$ is a $3 \times 3$ matrix and $ a, b$ are vectors with 3 components each ($a,b$ are sides of the original parallelogram and $Ma, Mb$ are sides after transformation). I was thinking of using the identity: $$(M a)\times (M b)=\det(M) (M^{\rm T})^{-1}(a\times b)$$

which is proved here but its not working out. Is there any way to find this ratio?

EDIT

Motivation: we can write the ratio of volumes of parallelepipeds in terms of only $\det (M)$ i.e. $$\frac{\|Ma.(Mb\times Mc)\|}{\|a.(b\times c)\|}=\det(M)$$ so I was thinking whether a similar expression in terms of only $M$ was possible.

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zed111
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  • What exactly is not working out with the identity you mention? – Travis Willse Feb 02 '15 at 19:33
  • From that identity, I can not find the ratio in terms of only M for a general a and b – zed111 Feb 03 '15 at 05:25
  • As the linked posting indicates, when $M$ is invertible, $(Ma)\times(Mb) = \operatorname{cof}(M)(a\times b)$. Therefore you shouldn't expect that $|(Ma)\times(Mb)|/|a\times b|$ is a constant, unless $\operatorname{cof}(M)$ (and hence $M$) is a scalar multiple of the identity matrix. – user1551 Feb 03 '15 at 08:16
  • @user1551 My $M$ is always invertible. Also see the EDIT. – zed111 Feb 03 '15 at 08:27

2 Answers2

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The ratio in general doesn't just depend on $M$, it depends on $a$ and $b$ too.

For example, for $$M = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ Then, for $a = {\bf i}$, $b = {\bf j}$, the ratio is $$\frac{||M{\bf i} \times M{\bf j}||}{{\bf i} \times {\bf j}} = \frac{||{\bf i} \times {\bf j}||}{||{\bf i} \times {\bf j}||} = 1 ,$$ but for $a = {\bf i}$, $b = {\bf k}$, the ratio is $$\frac{||M{\bf i} \times M{\bf k}||}{{\bf i} \times {\bf j}} = \frac{||{\bf i} \times {\bf 0}||}{||{\bf i} \times {\bf j}||} = \frac{||\bf 0||}{||{\bf k}||} = \frac{0}{1} = 0 .$$

Of course, in the special case $M = \lambda I$, the ratio is $\lambda^2$, which is independent of $a, b$.

Travis Willse
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Following this formula the ratio can be written as $$\sqrt\frac{\det((MA)^TMA)}{\det(A^TA)}=\sqrt\frac{\det (A^T(M^TM)A)}{\det(A^TA)}$$ where $A=\pmatrix{a & b}$.

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rych
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