Is the area of $T(\Omega)=|\det A|\,(\text{area of }\Omega)$?

Question

We are given that $\Omega$ is a parallelogram in $\mathbb{R}^3$ and $\left\{ T(\vec{x}) = A\vec{x} \mid \mathbb{R}^3 \mapsto \mathbb{R}^3\right\}$ is a linear transformation.

From the definition of the expansion factor of $T$, which is $\frac{\text{area of } T(\Omega)}{\text{area of }\Omega}=|\det A\,|\,,$ I'm tempted to say yes, but how would I conclusively prove that no counterexamples exist?

For example, there should be a linear transformation $\left\{T(\vec{x})=A\vec{x}\mid\mathbb{R}^3\mapsto\mathbb{R}^3\right\}$ such that $\frac{\text{area of }T(\Omega)}{\text{area of }\Omega}\neq\frac{|\det A|\det B|}{|\det B|}\,,$ but I'm told that "the linear transformation $T(\vec{x})=A\vec{x}$ expands the area of all parallelograms by the same factor, namely, $|\det A|\,.$ Surely there must be an exception?

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EDIT: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$ Best (and simplest) contradiction I have found thus far:

Let $T(\vec{x})=A\vec{x}=2I_3(\vec{x})$, i.e. $A$, the matrix corresponding to $T$, is $\,\begin{pmatrix}2&0&0\\0&2&0\\0&0&2\end{pmatrix}\Rightarrow|\det(A)|=8$

However, if $\Omega$ is the square defined by $\vec{\Basis_{1}}, \vec{\Basis_{2}}$, then $T(\Omega)$ is the square defined by $2\vec{\Basis_{1}}, 2\vec{\Basis_{2}}$, such that the area of $T(\Omega)$ must be 4.

Does anyone have any thoughts on what the general form of this proof might look like?

Answer 1

If $T$ has distinct real eigenvalues, then the "scale factor for area" isn't the same for all unit squares, and so cannot generally be equal to $\det T$.

For example, assume $T$ is diagonal in the standard basis, say $A = \operatorname{diag}[a, b, c]$. The unit square with edges $e_{1} = (1, 0, 0)$ and $v = (0, \cos\theta, \sin\theta)$ maps to the rectangle with edges $$ T(e_{1}) = ae_{1},\quad T(v) = (0, b\cos\theta, c\sin\theta), $$ whose area is $|a|\sqrt{b^{2} \cos^{2}\theta + c^{2} \sin^{2}\theta}$. If $b \neq c$, the area depends on $\theta$, i.e., on the spatial orientation of the square. Particularly, if $c = 0$, the preceding area expression reduces to $|ab\cos\theta|$, while $\det A = abc = 0$.

If $A = aI_{3}$, the scalar factor for area is $a^{2}$ for an arbitrary parallelogram, but again the scale factor is not $|\det A|$.

If you ask about parallelipipeds instead of parallelograms, then the "scale factor for volume" is indeed equal to $|\det A|$ independently of the parallelipiped.

Tangential remark: This failure of area to scale independently of the spatial orientation of a parallelogram (in space) explains why there's no elementary formula for the surface area of a general ellipsoid, despite the fact there's a simple formula for the area of a sphere, or even for a spheroid. A similar remark holds for lengths of segments in the plane, and how they stretch under linear transformations. The circumference of a circle is a polynomial function of the radius; the circumference of a non-circular ellipse is not even an elementary function of the semi-axes.

Answer 2

The area $\Vert a_1\wedge a_2\Vert$ can be found as $\sqrt{\det(A^tA)}$ (see this answer), where $A=\big(a_1 \enspace a_2\big)$. In your example $A=\big(Te_1 \enspace Te_2\big)=2\big(e_1 \enspace e_2\big)$, $A^tA=4I_2$, and $\sqrt{\det(A^tA)}=4.$