Relation between the determinant of a linear mapping and norm of a multiplied vector Let $S,T:\mathbb{R}^n \to \mathbb{R}^n$ be linear mappings with $\|Sv\|\le \|Tv\|$ for all $v\in\mathbb{R}^n$.
Is it generally true that $|\det(S)|\le |\det(T)|$?
 A: The fact that $\|Sv\|_2\leq\|Tv\|_2$ for all $v$ means that $v^*S^*Sv\leq v^*T^*Tv$ for all $v$, that is, $T^*T-S^*S$ is positive semidefinite. 
It is known that [see, e.g., Corollary 4.3.12 in Matrix Analysis by Horn and Johnson]:

If $A$ and $B$ are Hermitian and $B$ is positive semidefinite, then
  $$
\lambda_i(A)\leq\lambda_i(A+B), \quad i=1,\ldots,n,
$$
  where $\lambda_i(\cdot)$ denotes the $i$th eigenvalue (ordered in the ascending or descending order).

Using 
$$A:=S^*S\text{ and }B:=T^*T-S^*S$$ (hence $A+B=T^*T$) in the fact above gives $\lambda_i(S^*S)\leq\lambda_i(T^*T)$ and hence (since the eigenvalues of $S^*S$ and $T^*T$ are nonnegative) $$0\leq\det(S^*S)\leq\det(T^*T).$$ Now since for a square $X$, $$\det(X^*X)=\det(X^*)\det(X)=\overline{\det(X)}\det(X)=|\det(X)|^2,$$ we get $$|\det(S)|\leq|\det(T)|.$$
A: If $S$ is singular, then $|\det S| = 0 \leq |\det T|$ is trivial. While, if $S$ is invertible, then also $T$ is invertible, since $Tv=0 \Rightarrow Sv=0 \Rightarrow v=0$.
So we can call $A=ST^{-1}$. Note that $A$ is invertible, $\forall ||v|| \leq 1, ||Av|| \leq 1$. So all we need to prove is that $|\det A| \leq 1$ (because $|\det A| = \frac{|\det S|}{|\det T|}$).
Denote by $\mu$ the Lebesgue measure on $\mathbb{R}^n$.
Our hypothesis is $$\forall v, ||Av|| \leq ||v||$$
I call $B = \{ v \in \mathbb{R}^n : ||v|| \leq 1\} $ the ball of radius $1$. So our hypothesis implies that $A(B) \subseteq B$, where $A(B) = \{ Av : v \in B\}$.
Now, we have $\mu(A(B)) \leq \mu(B) $, so all we need to show is that $|\det A|\mu(B) = \mu(A(B))$.
But this is clear because $A$ is an invertible transformation, so
$$\mu(A(B)) = \int_{A(B)} d \mu (x) = \int_{A^{-1}A(B)} |\det A| d \mu (Ax) =$$
$$= \int_B |\det A| d \mu(Ax) = |\det A| \mu(B) $$
