# Prove a matrix inequality involving norms.

I have the following problem. If $$\|Ax\| \geq \theta \|x\|$$ for a square matrix A, $\theta$ a positive real number, $x$ a vector and a natural norm $\| \cdot \|$. Prove that $A$ is invertible and $$\|A^{-1}\| \leq \frac{1}{\theta}$$. My attempt, let $x=A^{-1}y$ so $$\|y\| \geq \theta \|A^{-1} y \|$$ and $$\|A^{-1} \| = \sup_{\|y\|=1} \| A^{-1} y \| 1/ \theta \leq \sup_{\|y \|=1} 1/ \theta \|y\|=1/ \theta$$. Is this proof OK? Also I can't seem to figure out how to prove that $A$ is invertible in the first place, how should I approach this?

• You need to assume that $\theta>0$. Is the matrix square. Can you show that $A$ is injective? – copper.hat Dec 2 '17 at 22:03
• The matrix is square – Dimtsol Dec 2 '17 at 22:05
• Your attempt is wrong since you write $x=A^{-1}y$ before proving that $A$ is invertible. – José Carlos Santos Dec 2 '17 at 22:05
• So am I, but am I injective? – copper.hat Dec 2 '17 at 22:05

Your result is true if $A \in M(n\times n)$. This way, $|A(x)| \ge \theta |x|$ implies that $A$ in injective. By the rank-nullity theorem the dimension of image is $n$ so the $A$ is also surjective. Also, $|A(x)| \ge \theta |x|$ implies $\|A\| \ge \theta$. Finally, as $A\cdot A^- = I$ we have $$\|A\|\|A^-\| =1.$$ Then, $\|A^{-1}\| \le 1/\theta.$
• Thanks for the answer,can you show why $|Ax| \geq \theta |x|$ implies $A$ being injective? – Dimtsol Dec 2 '17 at 22:38
• Suppose $Ax=Ay$. Then, $0=|Ax-Ay|=|A(x-y)|\ge \theta |x-y|\ge 0$. So we must have $|x-y|=0$. This means $x=y$. – shall.i.am Dec 3 '17 at 7:45