Show that non zero vectors exist $\textbf{b}$ and $\delta \textbf{b}$ satisfying these functions. Show that for non-singular $A \in R^{m\times m}$ there exists non-zero vectors $\textbf{b}$ and $\delta \textbf{b}$ in $R^m$ such that the following equations hold: $A \textbf{x} = \textbf{b}$,  $A (\textbf{x} \delta \textbf{x}) = \textbf{b} + \delta \textbf{b}$, and $\frac{\Vert \delta \textbf{x} \Vert_p}{\Vert \textbf{x} \Vert_p} = \kappa_p(A) \frac{\Vert \delta \textbf{b} \Vert_p}{\Vert \textbf{b} \Vert_p}$, where $p = 1, 2, \infty$ and $\kappa_p = \Vert A \Vert_p \Vert A^{-1} \Vert_p$
We are encouraged to use the following information:
$$\Vert B \Vert_p = \max \limits_{0 \neq \textbf{z} \in R^m}\frac{\Vert B\textbf{z} \Vert_p}{\Vert \textbf{z} \Vert_p}$$
and that there always exists a particular nonzero $\textbf{z} \in R^m$ such that 
$$\Vert B \Vert_p = \frac{\Vert B\textbf{z} \Vert_p}{\Vert \textbf{z} \Vert_p}$$
We also know that $B = A$ and $B = A^{-1}$. 
My Thoughts
So first I wanted to establish what I am trying to prove. From what I understand I am supposed to use
$$\Vert B \Vert_p = \frac{\Vert B\textbf{z} \Vert_p}{\Vert \textbf{z} \Vert_p}$$
and somehow manipulate it to have the form $Ax = b$. If I can do that then I should have no problem with $A (x+dx) = b + db$. The closest I have gotten is
$$
\Vert \textbf{z}  \Vert_p = \frac{\Vert B\textbf{z} \Vert_p}{\Vert B\Vert_p} = \frac{\Vert A^{-1}\textbf{z} \Vert_p}{\Vert A^{-1}\Vert_p} 
$$
Is it ok for me to just say then that $A^{-1} \textbf{z} = \textbf{x}$ for some $\textbf{x} \in R^m$ and that $\textbf{z} = \textbf{b}$?
For the second part I would do the same as the above but have  $\textbf{z = z + dz}$ and instead say that $A^{-1} \textbf{(z + dz)} = \textbf{x + dx}$. This simplifies to $A^{-1} \textbf{dz} = \textbf{dx}$
For the third part,
$$
\Vert \delta \textbf{z}  \Vert_p = \frac{\Vert B \delta\textbf{z} \Vert_p}{\Vert B\Vert_p}
=\frac{\Vert A^{-1} \delta\textbf{z} \Vert_p}{\Vert A^{-1} \Vert_p}
$$
If I divide both sides by the x p-norm I get
$$
\frac{\Vert \delta \textbf{z}  \Vert_p \Vert A^{-1} \Vert_p }{\Vert \textbf{x}  \Vert_p}
=\frac{\Vert A^{-1} \delta\textbf{z} \Vert_p}{\Vert \textbf{x}  \Vert_p}
$$
I know that 
$$
\Vert \textbf{b} \Vert_p  = \Vert A \textbf{x} \Vert_p \leq \Vert A \Vert_p \Vert\textbf{x} \Vert_p \Rightarrow \frac{\Vert \textbf{b} \Vert_p} { \Vert A \Vert_p} \leq \Vert\textbf{x} \Vert_p 
$$
This can be rewritten as (I substituted z = b)
$$
\frac{\Vert \delta \textbf{z} \Vert_p \Vert A^{-1} \Vert_p }{\frac{\Vert \textbf{b} \Vert_p} { \Vert A \Vert_p}  }
\leq \frac{\Vert A^{-1} \delta\textbf{z} \Vert_p}{\Vert \textbf{x}  \Vert_p}
\Rightarrow 
\frac{\Vert \delta \textbf{z} \Vert_p  }{\Vert \textbf{b} \Vert_p  } \kappa_p(A)
\leq \frac{\Vert A^{-1} \delta\textbf{z} \Vert_p}{\Vert \textbf{x}  \Vert_p}
$$
We showed elsewhere that 
$$
\frac{\Vert \delta \textbf{z} \Vert_p  }{\Vert \textbf{b} \Vert_p  } \kappa_p(A)
\geq \frac{\Vert A^{-1} \delta\textbf{z} \Vert_p}{\Vert \textbf{x}  \Vert_p}
$$
so this should prove the equality. 
Any feedback on this proof would be very valuable. I just want to make sure I have all the pieces. Also, I'm not even sure if I did it right. 
Note: Proofs are my greatest weakness.
 A: Something goes wrong in your step following "This can be rewritten as": the inequality is wrong and should be
$$\frac{\Vert \delta \textbf{z} \Vert_p \Vert A^{-1} \Vert_p }{\frac{\Vert \textbf{b} \Vert_p} { \Vert A \Vert_p}  }
\geq \frac{\Vert A^{-1} \delta\textbf{z} \Vert_p}{\Vert \textbf{x}  \Vert_p}.$$
In any case, there is no need to use inequalities to solve this problem. The key is to slowly transform what you know into a form where you can infer the right $\bf{x}$ and $\delta \bf{x}$.
First, you know that $\mathbf{b} = A \bf{x}$ and $\delta \mathbf{b} = A \delta \bf x$, and you also have a formula for $k_p$. Plugging it in, you are trying to find a $\bf x$ and $\delta  \bf x$ with
$$\frac{\|\delta \mathbf{x}\|_p}{\|\mathbf{x}\|_p} = \|A\|_p\|A^{-1}\|_p \frac{\|A\delta \mathbf{x}\|_p}{\|A\mathbf{x}\|_p}.$$
You can now plug in your second property of matrix norms: there exists a $\mathbf z$ with
$$\|A\|_p = \frac{\|A\mathbf{z}\|_p}{\|\mathbf{z}\|_p},$$
and likewise for $A^{-1}$ and a vector $\mathbf{w}$.
Thus you are trying to find a $\bf x$ and $\delta  \bf x$ with
$$\frac{\|\delta \mathbf{x}\|_p}{\|\mathbf{x}\|_p} = \frac{\|A\mathbf{z}\|_p}{\|\mathbf{z}\|_p}\frac{\|A^{-1}\mathbf{w}\|_p}{\|\mathbf{w}\|_p} \frac{\|A\delta \mathbf{x}\|_p}{\|A\mathbf{x}\|_p},$$
or, separating the knowns and unknowns,
$$\frac{\|\delta \mathbf{x}\|_p}{\|\mathbf{x}\|_p} \frac{\|A\mathbf{x}\|_p}{\|A\delta \mathbf{x}\|_p}= \frac{\|A\mathbf{z}\|_p}{\|\mathbf{z}\|_p}\frac{\|A^{-1}\mathbf{w}\|_p}{\|\mathbf{w}\|_p} .$$
Comparing the LHS and RHS, you should get a strong urge to set $\mathbf{x} = \mathbf{z}$. We still don't know what we need for $\delta\mathbf{x}$, but after plugging in $\mathbf{x}=\mathbf{z}$ and simplifying,
$$\frac{\|\delta \mathbf{x}\|_p}{\|A\delta \mathbf{x}\|_p}= \frac{\|A^{-1}\mathbf{w}\|_p}{\|\mathbf{w}\|_p},$$
and we see that $\delta \mathbf{x} = A^{-1}\mathbf{w}$ will work like a charm.
