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Problem: Let $A$ be a matrix with $m$ rows and $n$ columns. The matrix defines a linear operator, also denoted by $A$, \begin{align} A: \mathbb R^n & \longrightarrow \mathbb R^m\\ x & \longmapsto Ax \end{align} Assume that $n$ columns of matrix $A$ are independent (this is equivalent to say that operator $A$ is injective, see the 3rd property here), then operator $A$ is bounded below, i.e. there exists a positive constant $c>0$ such that $$||Ax||\geq c||x||, \hspace{0.5cm} \forall x\in \mathbb R^n.$$ where $||\cdot||$ is the Euclidean norm.

Motivation: Above problem is a special case (finite dimensional case) of Theorem 2.5 in this book, which stated that: "A continuous operator between Banach spaces is bounded below iff it is injective and has closed range." Regarding our problem, we know that:

  • The operator $A$ is continuous,
  • $\mathbb R^n$ and $\mathbb R^m$ are Banach spaces,
  • The operator $A$ trivially has closed range since we are working with finite-dimensional spaces.

Thus, the statement in our problem should hold true. In the proof of Theorem 2.5, they used the Open Mapping Theorem, which is a technical theorem! I am wondering are there any simpler proof with intuition and without Functional analysis for our special case. Thank you in advance!

Leonard Neon
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  • Another way to see this is to notice that the function $$x\mapsto |Ax|$$ is continuous and hence attains its infimum on the compact set ${x:|x|=1}$. – Ruy Dec 27 '20 at 15:58
  • @Ruy but the infimum can be $0$. While we expect that the infimum is strictly positive. – Leonard Neon Dec 27 '20 at 16:49
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    To say that the imfimum is attained is to say that there is some $x$ such that $|Ax|$ coincides with the infimum, hence it cannot be zero. – Ruy Dec 27 '20 at 16:53
  • Ah ok you right, I now understand your idea, thank you so much! – Leonard Neon Dec 27 '20 at 17:02

1 Answers1

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Yes. If there exists no $c$ such that $\|Ax\|\geq c\|x\|$ for all $x$, then there exists a sequence $x_1,x_2,\dotsc$ such that $\|x_i\|=1$ for all $i$, and $\|Ax_n\|\to 0$ as $n\to \infty$. Since $\mathbb{R}^n$ is finite-dimensional, its unit ball is compact, hence some subsequence $(x_{k_i})_i$ tends to a limit $x$ in the unit ball (by compactness). But $A$ is bounded and linear, hence continuous, so we get

$$\|Ax\|=\lim_i \|Ax_{k_i}\| =0$$

i.e. $Ax=0$ for an $x$ with $\|x\|=1$, which contradicts the fact that the columns of $A$ are independent.

mss
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