Show that non zero vectors exist $\textbf{b}$ and $\delta \textbf{b}$ satisfying these functions.

Question

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.

Answer 1

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.