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Hi am studying orthogonal matrices and I am having difficulties in finding a proof and to show the following :

An n x n matrix is orthogonal if $A^T A = I $, Show that such matrices preserve volumes.

I found that it is related with the determinant. It says that the determinant of an orthogonal matrix is $\pm$1 and orthogonal transformations and isometries preserve volumes. However I do not know how to show it.

Also I would like to show that Orthogonal matrices preserve dot product and I found that:

$A\vec{x}$ $.$ $A\vec{y}$= $\vec{x}$.$\vec{y}$ then,

$A\vec{x}$ $.$ $A\vec{y}$= $A^T $$A\vec{x}$.$\vec{y}$ and because of orthogonality property, $A^T A = I $

this is $\vec{x}$.$\vec{y}$

In this case, I am not sure if this is correct or complete.

Can anyone help me on this?

Thanks

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  • $\begingroup$ You can just take "preserves volume" to literally mean $\det(A) = \pm 1$ which is easily provable from the definition ($A^TA=I$). $\endgroup$ – user137731 Feb 26 '17 at 2:51
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    $\begingroup$ As for $A$ preserving the dot product: $$Ax \cdot Ay := (Ax)^T(Ay) = (x^TA^T)Ay = x^T(A^TA)y = x^TIy = x^Ty =: x\cdot y$$ $\endgroup$ – user137731 Feb 26 '17 at 2:54
  • $\begingroup$ In the last part do you mean: $\vec{x}$.$\vec{y}$? I do not understand the literally mean of the determinant. I found something that says: that is related to the scalar triple product (a x b) . c- volume of a parallelepiped spanned by the vectors a,b and c. and by using the alternative notation for the derminant the volume of the parallelepiped spanned by a,b,c is |(axb).c | but I do not understand how it shows that preserves volume $\endgroup$ – user290335 Feb 26 '17 at 3:04
  • $\begingroup$ Not "literally mean of the determinant" as if "literally mean" is an operation. I'm just using English vernacular, but I'll rephrase: you should consider "preserving volumes" as the same thing as "$\det(A) = \pm 1$". $\endgroup$ – user137731 Feb 26 '17 at 3:07
  • $\begingroup$ Ah, so showing that preserve volume is the same as proving that determinant is 1 or -1? $\endgroup$ – user290335 Feb 26 '17 at 3:10
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Like said in the comments, for orthogonal matrices $A$ have $|\det(A)|=1$ (as a consequence of the multiplicativity of $\det$). The volume preserving property is a special case of a substitution theorem for integrals with multiple variables from analysis (https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables): When you transform an object $U$ with a linear map $\phi:\mathbb{R}^n\rightarrow\mathbb{R}^n,x\mapsto Ax$ you have $\phi'(x)=A$ and we can compute the volume of the transformed object $\phi(U)$ as follows: $$\operatorname{Vol}(\phi(U)) = \int_{\phi(U)}dx = \int_U \underbrace{|\det \phi'(x)|}_1 dx = \int_U dx = \operatorname{Vol}(U).$$ Thats why orthogonal matrices are volume preserving. (Btw orthogonal matrices only allow rotations (case $\det A=1$) or reflections (case $\det A=-1$), which makes the volume preserving property intuitively plausible)

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  • $\begingroup$ Thank you very much for your help $\endgroup$ – user290335 Mar 1 '17 at 14:57

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