# How Orthogonal matrices preserve dot product and volume-proof

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

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