An Injective Nondegenerate Square Matrix is Invertible

We work in the setting of Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background, whose notation is fixed for every dimension and is used here with a natural number $n$ satisfying $1\le n$: the real numbers and sequences, and the Euclidean norm $\lVert\,\cdot\,\rVert$, dot product, distance $d_{E}$, topology, the notions of open, closed and bounded subsets, and the closed balls $\bar{B}(x,r)$, are as fixed there.

Let $A$ be a real matrix with $n$ rows and $n$ columns, let $Ah$ denote the matrix-vector product, and put

$$A(\mathbb{R}^{n})=\{\,Ah\;:\;h\in\mathbb{R}^{n}\,\}.$$

Then the following hold.

1. (Lower bound) (i) Suppose the map $h\mapsto Ah$ is injective. Then there is $c\in\mathbb{R}$ with $0<c$ such that $\lVert Ah\rVert\ge c\,\lVert h\rVert$ for every $h\in\mathbb{R}^{n}$.

2. (The image is convex, and closed when the map is injective) (ii) The set $A(\mathbb{R}^{n})$ is convex and contains the zero vector. If moreover $h\mapsto Ah$ is injective, then $A(\mathbb{R}^{n})$ is a closed subset of $\mathbb{R}^{n}$.

3. (Surjectivity) (iii) Suppose the map $h\mapsto Ah$ is injective and that there is no $\nu\in\mathbb{R}^{n}$ with $\lVert\nu\rVert=1$ such that $\nu\cdot(Ah)=0$ for every $h\in\mathbb{R}^{n}$. Then $A(\mathbb{R}^{n})=\mathbb{R}^{n}$.

4. (Invertibility and a bound for the inverse) (iv) Under the hypotheses of claim 3 the matrix $A$ is invertible, its inverse $A^{-1}$ being unique by Uniqueness of the Matrix Inverse, and

$$\lVert A^{-1}k\rVert\le c^{-1}\lVert k\rVert\qquad\text{for every }k\in\mathbb{R}^{n},$$

where $c$ is any constant as in claim 1.