Linear map preserving area? In Sean Dineen's book for multivariate calculus and geometry, there is an exercise which states:

If $T:\mathbb{R}^n \to \mathbb{R}^n$ is a linear mapping such that $$\|T(x)\|= \|x \| \quad \forall x \in \mathbb{R}^n,$$ then $T$ preserves the inner product, angle, area as well as length of curve. (Page $50,$ last problem of exercise $5$).

I have shown that $T$ preserves the inner product and angle, however I do not understand what is meant by area in this case and neither I am understanding how to show that the length isn't changing. Any help would be appreciated. I only know curves and haven't read what surfaces are yet, if that is relevant.. 
 A: Let $A$ be the matrix associated to $T$, that is, $T(x)=Ax$. 
From the standard result in multi-variable differentiation, $dT_x(v)=Av$. (In other words, the differentiation matrix $\frac{\partial T^i}{\partial x_j}$ is exactly $A$.)
What you have proved is $|Av|=|v|$ and $Ax\cdot Ay=x\cdot y$.
Regarding to the length part, let $\gamma(t)$ be a path in $\mathbb{R}^n$. What we want to show is that
\begin{align*}
\int_a^b \left|\frac{d\gamma}{dt}\right|dt
=
\int_a^b \left|\frac{d(T\circ\gamma)}{dt}\right|dt,
\end{align*}
which is clear since
\begin{align*}
\frac{d(T\circ\gamma)}{dt}=dT\left(\frac{d\gamma}{dt}\right)
=A\left(\frac{d\gamma}{dt}\right)
\end{align*}
and $|A(\gamma^{\prime})|=|\gamma^{\prime}|$.
For the area part (I guess it means volume here. If it means the area of the hypersurface, I think it would be more difficult.), we want to show that
\begin{align*}
\int_{\Omega} 1 dx = \int_{T(\Omega)} 1 dy,
\end{align*}
where $dx$ denotes $dx_1dx_2...dx_n$ and $dy$ denotes the same volume element.
By change of variable (with $y=T(x)$, injectivity of $T$ is left to you), we have
\begin{align*}
\int_{T(\Omega)}1dy&=
\int_{\Omega} \left|\det{
\left(\frac{\partial T^i}{\partial x_j}\right)}
\right| dx\\
&=\int_{\Omega} |\det{A}|dx.
\end{align*}
If we can show that $|\det{A}|=1$, then we are done. 
Observe that you have already shown that $Ax \cdot Ay=x\cdot y$, which means $(A^TAx) \cdot y=x\cdot y$ for every $x$ and $y$ in $\mathbb{R}^n$. Hope that you could derive the desired result from here. 
Intuitively (or geometrically), preserving length and preserving angle imply preserving the volume element. 
