If $T:\mathbf R^n\to\mathbf R^n$ is a norm preserving linear transformation and $M$ is a $k$-manifold in $\mathbf R^n$, show that $M$ has the same volume as $T(M)$.
This is an exercise from Spivak. The volume is defined as follows:
Let $M$ be a $k$-dimensional manifold (or manifold-with-boundary) in $\mathbf{R}^n$, with an orientation $\mu$. If $x\in M$, then $\mu_x$ and the inner product $T_x$ we defined previously determine a volume element $\omega(x)\in\Lambda^k(M_x)$. We therefore obtain a nowhere-zero $k$-form $\omega$ on $M$, which is called the volume element on $M$ (determined by $\mu$) and denoted $dV$, even though it is not generally the differential of a $(k-1)$-form. The volume of $M$ is defined as $\int_M dV$, provided this integral exists, which is certainly the case if $M$ is compact. "Volume" is usually called length or surface area for one- and two-dimensional manifolds, and $dV$ is denoted $ds$ (the "element of length") or $dA$ [or $dS$] (the "element of [surface] area").
I don't know how to approach the problem. It looks like I need to find $\omega$, then take $T$ of the corresponding integral, etc. But I don't know how to do any of the steps from these strategy...
HINT: If $\omega$ is a volume element on $T(M)$, then show that because $T$ is an isometry ("norm-preserving") it will follow that $T^*\omega$ is a volume element on $M$. Recall that a $k$-form $\eta$ on the oriented $k$-dimensional submanifold $M$ is a volume element precisely when $\eta_x(v_1,\dots,v_k) = 1$ whenever $v_1,\dots,v_k$ is an oriented orthonormal basis for the tangent space $M_x$.