Lebesgue Measure of lower dimension is zero. I am currently studing Rudin's Real and Complex Analysis book. In page 52 it’s written if T is a linear transformation from $R^k$ to $R^k$ and range of T is a subspace Y of lower dimension,  then m(Y)=0.
I have seen several proves regarding this using the fact that m(T(E))=$\Delta$(T)m(E)  from Theorem 2.20(e).  But I have to prove m(Y) =0 for proving the theorem  2.20(e) itself. Is there any another way of doing it without using linear transformation,determinant or rotation?
 A: This is a proof using 'linear trasformations', not in the sense you cited : Is sufficient prove it for the hyperplane $P_{0} := \{(y_{1},\cdots,y_{d}) : y_1 = 0\} \cong \mathbb{R}^{d-1} \subset \mathbb{R}^{d}$, using the invariancy of isometry of Lebesgue measure, the completeness of measure. Given $\epsilon > 0$ let $Q_{n} = [-\frac{\epsilon}{(2n)^{d-1}2^{n+1}},\frac{\epsilon}{(2n)^{d-1}2^{n+1}}]\times [-n,n]\times \cdots \times [-n,n]$. We have that $m(Q_{n}) = \frac{2\epsilon}{(2n)^{d-1}2^{n+1}} (2n)^{d-1} = \frac{\epsilon}{2^{n}}$.
Since $P_{0} \subset \bigcup\limits_{n=1}^{\infty}Q_{n}$, by sub-additivity of measure $m(P_{0}) \leq \sum\limits_{n =1}^{\infty}\frac{\epsilon}{2^{n}} = \epsilon$, which leads to our thesis given the arbitrary of $\epsilon$.
$\textbf{Edit :}$ Let $S$ be an isometry and let $\tau_v$ be a translation, (already known Lebesgue measure invariant, denoted by $\lambda$). Define the operator $S_*$ which associate a measure $\mu \to \mu \circ S^{-1}$, same for $\tau_{v_*}$. Observe
$\tau_{v_*} \circ (S_* \lambda) = (S_* \lambda) \circ \tau_{-v_*} = \lambda \circ S^{-1} \circ \tau_{-v_*} = \lambda \circ (\tau_v \circ S)^{-1} = (\tau_v \circ S)_* \lambda = (S \circ \tau_{S^{-1}v})_* \lambda =  (S)_*(\tau_{S^{-1}v})_* \lambda = S_* \lambda$
i.e $S_*\lambda$ is invariant for translations so $S_*\lambda = c \lambda$ for some constant $c$. But denoted with $B$ the euclidean ball we get $S_*\lambda(B) = \lambda(B)$ so $c = 1$, in other words $S_*\lambda =  \lambda$ .
Here you are using this fact since every proper subspace lies in some hyperplane.
