Condition of matrix norm proof

Question

If

$Ax = b$

$A\hat x = \hat b$

Then show that:

$$\frac{1}{\operatorname{cond}(A)} \frac{\|b- \hat b\|}{\|b\|} \leq \frac{\|x- \hat x\|}{\|x\|}$$

where $\operatorname{cond}(A) = \|A\| \cdot \|A^{-1}\|$

I'm having some trouble proving this. I've gotten to:

$$\|b - \hat b\| \leq \|x-\hat x\| \cdot \|A\| $$

After this, what do I do?

Answer 1

You have the expression

$$ \frac{1}{cond(A)} \frac{\| b- \hat{b}\|}{\|b\|} \leq \frac{\|x - \hat{x} \|}{\|x\|} \tag{1}$$

$$ \frac{1}{\| A\| \| A^{-1}\|} \frac{\| b- \hat{b}\|}{\|b\|} \leq \frac{\|x - \hat{x} \|}{\|x\|} \tag{2}$$

$$ \frac{1}{\| A\| \| A^{-1}\|} \frac{\| Ax- A\hat{x}\|}{\|b\|} \leq \frac{\|x - \hat{x} \|}{\|x\|} \tag{3}$$ $$ \frac{1}{\| A\| \| A^{-1}\|} \frac{\| A(x-\hat{x})\|}{\|Ax\|} \leq \frac{\|x - \hat{x} \|}{\|x\|} \tag{4}$$

note that

$$ \| Ax \| \leq \| A \| \|x\|\tag{5} $$

what about now...

$$ cond(A) = \kappa(A) \geq 1 \tag{6} $$

if invertible.