# Proof of Matrix Norm (Inverse Matrix)

Show for any induced matrix norm and nonsingular matrix A that $$\left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1}$$ where $$\left\|A^{-1}\right\| = \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}\\ \left\|A\right\| = \max_{\left\|x\right\|=1}\{\left\|Ax\right\|\}.$$

I am not sure how to show that: $$\begin{split} \left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1}\\ \text{or}\\ \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} ≥ (\max_{\left\|x\right\|=1}\{\left\|Ax\right\|\})^{-1} \end{split}$$

• i am still learning how to write equations in here, is there a faster way? Ill put what Ive tried in the actual question portion, it doesn't seem to be working here – Megan Sep 24 '14 at 1:41
• Please don't vote this question down, I am still learning how to use the special text. Thanks. – Megan Sep 24 '14 at 1:53
• @ Megan: welcome to MSE! – Kim Jong Un Sep 24 '14 at 2:11
• Thanks! I used to use stack exchange all the time for coding, so excited to discover the math stack exchange. :) – Megan Sep 24 '14 at 2:18

Use $\lVert AB\rVert \leq \lVert A\rVert \lVert B\rVert$, as the induced norm is in particular submultiplicative. So that $\lVert I_n\rVert \leq \lVert A\rVert \lVert A^{-1}\rVert$.
Suppose $|y|=1$ is such that $|A|=|Ay|$. Then $x=Ay/|A|$ also has norm $1$ so it follows that $$|A^{-1}|\geq |A^{-1}x|=|A^{-1}Ay/|A||=\frac{|y|}{|A|}=\frac{1}{|A|}=|A|^{-1}.$$
Proof: $$\left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1}$$ where $$\left\|A^{-1}\right\| = \max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\}\\ \left\|A\right\| = \max_{\left\|x\right\|=1}\{\left\|Ax\right\|\}\\ \left\|A\right\|^{-1} = (\max_{\left\|x\right\|=1}\{\left\|Ax\right\|\})^{-1} = (\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\})$$ and $$\max_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} ≥(\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\})$$ therefore: $$\left\|A^{-1}\right\| ≥ (\left\|A\right\|)^{-1}$$
• Are you supposing here that $\min_{\left\|x\right\|=1}\{\left\|A^{-1}x\right\|\} = \|A\|^{-1}$? That's a very interesting supposition, and you should probably prove that as a part of your proof – Omnomnomnom Sep 24 '14 at 3:27