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!
Best Answer
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.