0

Let $A$ be any rectangular real matrix. I presume that the following is a standard result in linear algebra:

$$\operatorname{rowspace}(A) = \operatorname{rowspace}(A^T A)$$

Where can I find a proof of this?

This is not a homework exercise. I just ran into this question while doing some other computations and realized I wasn't able to find it in my linear algebra textbooks.

a06e
  • 6,665
  • What have you tried? Can you show few elementary steps? – Sourav Ghosh Mar 08 '23 at 13:36
  • @SouravGhosh I can show easily that $\mathbf{v} \in \operatorname{rowspace}(\mathbb{A}^{\top}\mathbb{A})$ implies $\mathbf{v} \in \operatorname{rowspace}(\mathbb{A})$. But for the other way I think I'm missing something. – a06e Mar 08 '23 at 14:20
  • Is this statement true for any rectangular matrix? Or we need $\mathbb{A}$ to be full rank or something? – a06e Mar 08 '23 at 14:22
  • Related: https://math.stackexchange.com/questions/215145/rank-of-product-of-a-matrix-and-its-transpose/682249, https://math.stackexchange.com/questions/349738/prove-operatornamerankata-operatornameranka-for-any-a-in-m-m-times-n – a06e Mar 08 '23 at 14:28
  • 1
    I think you essentially have the proof there: one is contained in the other, and by the linked posts the dimensions are the same (take transposes to translate them into ranks, if need be)? – Jakob Streipel Mar 08 '23 at 19:45
  • @prets You're saying that if $A,B$ are vector spaces of the same dimension, with $A \subset B$, then $A = B$ ? – a06e Mar 08 '23 at 19:47
  • Absolutely. An $n$-dimensional subspace of an $n$-dimensional space fills up the entire space. – Jakob Streipel Mar 08 '23 at 19:48
  • Ok that makes sense. Though I'd still like to see a "textbook" proof of this. Feels like an elementary fact. – a06e Mar 08 '23 at 19:49
  • 1
    See for example this question. – Jakob Streipel Mar 08 '23 at 19:54

0 Answers0