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I'm aware that this question has been asked before by Zero. However, there is a step in the answers provided by others that I've yet to understand.

Let $T:\mathbb{R^n} \to \mathbb{R^n}$ be linear. If the range of $T$ is a subspace $Y$ of lower dimension, then $m(Y) = 0$.

I understand that the subspace $\mathbb{R}^k \subset \mathbb{R}^n$ of dimension $k < n$ spanned by standard bases ${e_1,...,e_k}$ has measure 0. But how does this extend to a subspace is spanned by other vectors ${b_1,...,b_k} = {U(e_1),...,U(e_k)}$ for some linear map $U$?

i.e. Prove measure of $T(X) = Y \subseteq U(\operatorname{span}({e_1,...,e_k}))$ is 0.

Below Zero's question, I believe somitra spoke of the Gram-Shmidt process to transform $b_i$ into an orthonormal basis, which seems relevant, but I have a rather limited background in linear algebra, so is there some elementary alternative?

amWhy
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Jia Cheng Sun
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    The Lebesgue measure is translation and rotation invariant. In general, if $T$ is a linear map, and $A$ a measurable set we have $m(T(A)) = |\det T| m(A)$. – copper.hat Dec 23 '19 at 03:27
  • Thanks! I was hoping to avoid rotations and determinants but I guess this is as elementary as it gets. – Jia Cheng Sun Dec 23 '19 at 03:35
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    There is no need for rotations and determinants, the dimension of a subspace is determined by the number of linearly independent vectors, not what they are. And it is trivial to construct thin covers of arbitrarily small volume for (bounded) parts of lower dimensional subspaces. – Conifold Dec 23 '19 at 04:06
  • @copper.hat Is it possible to prove the euclidean space spanned by any k independent vectors is a rotation of $R^k$? – Jia Cheng Sun Dec 23 '19 at 04:46
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    @JiaChengSun: If you mean $\mathbb{R}^k$ as a subset of $\mathbb{R}^n$ then yes, if not then I am not sure what you mean. In the former case, find an orthonormal basis $u_1,...,u_k$ for the span of the $k$ vectors and complete to form an orthonormal basis of $\mathbb{R}^n$. Let $U = [u_1 ... u_n]$ be the resulting orthogonal matrix and then $U \mathbb{R}^k$ is the subspace in question. – copper.hat Dec 23 '19 at 04:53

1 Answers1

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I am not sure if the below approach is correct. Rudin Real and Complex Analysis book states the result asked in the question within the proof of Theorem 2.20 regarding the Lebesgue measure on ${\boldsymbol{R^{k}}}$. Hence trying to use only results upto this portion of the book.

Part (a) of Theorem 2.20 states that $m\left(W\right)=\text{vol}\left(W\right)$ for every $k$-cell $W$. Using this, we can wrap or cover every compact (closed and bounded) set, $E$, in the lower subspace, $Y$, with dimension, $m<k$, with a suitable $k$-cell from the higher dimensional space. The additional parameters for this $k$-cell in the higher dimension can be $-\frac{\epsilon}{2},\frac{\epsilon}{2}$ with $\epsilon>0$. This gives $m\left(E\right)=\text{vol}\left(E\right)\leq\epsilon^{k-m}\text{volm}$. Here, volm is the volume of the wrapper of $E$, in the $m$-dimensional sub space, $Y$.

Since $\epsilon$ was arbitraty, $m\left(E\right)=\text{vol}\left(E\right)=0$. The entire space (and hence the subspace) is $\sigma-$compact so it can be written as the countable union of compact sets (which are closed and bounded and hence covered like the set $E$ above) all of which have measure zero, so the measure of the entire subspace is zero.

Related Questions: Sub-dimensional linear subspaces of $\mathbb{R}^{n}$ have measure zero.

Lebesgue measure of a subspace of lower dimension is 0

Any linear subspace has measure zero

Every subset of a subspace of $\mathbb{R}^n$ of dim $<n$ has measure 0

proof that the lebesgue measure of a subspace of lower dimension is 0.

texmex
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