Area preserving transformation in a higher dimensional space is unitary.

Question

In $\mathbb{R}^3$, a linear operator $Q:\mathbb{R}^3 \to \mathbb{R}^3$ preserves the area of parallelograms: that is, given $x,y\in \mathbb{R}^3$, the area of a parallelogram formed by $x$ and $y$ is the same as the area of a parallelogram formed by $Qx$ and $Qy$. Show that $Q$ must be unitary (that is, $Q^T Q=I$, where $I$ is the identity.)

This is from Shilov's Linear Algebra, Chapter $8$, exercise $32$. I am self-studying and trying to do some exercises but I'm really stumped on this one. I don't even know how to show the hint given: to show that $Q$ must preserve right angles: if $(x,y)=0$, then $(Qx,Qy)=0$ also (where $(\dot{},\dot{})$ means dot product).

Answer 1

Hint : Area of parallelogram $\Delta = \|x\| \|y\| \sin \theta $, where $\theta $ is angle between the two vectors $\theta = \cos^{-1} \dfrac{x^T y}{\|x\| \|y\|}$

Answer 2

Hint: In $\Bbb R^3$, the area of the parallelogram is given by $\| x \times y \| = \|x\|\|y\| \sin \theta$, where $\theta$ is the angle between $x$ and $y$ and $\times$ denotes the cross product. Use Lagrange's identity: $$\|x\|^2\|y\|^2 - (x,y)^2 = \|x \times y \|^2 \\ \|x\|^2\|y\|^2 - (x,y)^2 = \|x\|^2\|y\|^2 \sin^2 \theta$$ where $\theta$ is the angle between $x$ and $y$. Also, by hypothesis, we have $$\|x\|\|y\|\sin \theta = \|Qx\|\|Qy\| \sin \varphi$$ where $\varphi$ is the angle between $Qx$ and $Qy$. Repeating Lagrange's identity, we get: $$\|Qx\|^2\|Qy\|^2 - (Qx,Qy)^2 = \|Qx\|^2\|Qy\|^2\sin^2 \varphi$$Suppose $(x,y) = 0$. Joining everything, we get: $$\|x\|^2\|y\|^2 = \|Qx\|^2\|Qy\|^2 - (Qx,Qy)^2$$ Try to do something to conclude that $(Qx,Qy) = 0$, for starters. I'm thinking about the problem, and if I come up with something else, I'll edit the answer here.

Answer 3

Rather than using the Lagrange identity for the area of a parallelogram and doing the algebraic work (often unsuccessfully), one could look at which transformations preserve the areas: rotations, reflections (unitary transformations) and those that don't: shear, scaling, projections.

Let's use the singular value decomposition $Q=UDV^*$, where U and V are unitary matrices and D is a diagonal matrix with nonnegative elements representing a non-uniform scaling. For Q to be area-preserving D has to be the identity matrix, and so Q -- unitary as a product of unitary matrices.

Answer 4

"one could look at which transformations preserve the areas: rotations, reflections (unitary transformations) and those that don't: shear, scaling, projections." (quoted from rych)

Shears preserve areas. The area of a rectangle/parallelogram is the base times the height, no matter how it is sheared parallel to its side.