1
$\begingroup$

This came up in an exercise about differential equations and to be honest, I do not quite see the connection to them! The last exercise was about fundamental systems and now this was given:

Let $B,R:\mathbb{R}\rightarrow\mathbb{R}^{n\times n}$ be continuous, matrix valued functions with $\lim\limits_{h \rightarrow 0} \frac{||R(h)||}{|h|} = 0$.

Show with a carefully chosen way of representing the determinant that there exists a function $w:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\det(B(t)+R(t))=\det(B(t))+w(t)$$ and $$\frac{||w(h)||}{|h|} = 0$$

Ok, first of, $\frac{||R(h)||}{|h|} = 0$ looks like you're supposed to use the differential somehow somewhere. But this exercise doesn't really point you in any direction. You could use the Laplace representation, but I'm not sure what I am looking for, so this makes it quite hard to know what to use..

What can I look at to construct such function?

$\endgroup$
  • $\begingroup$ Which norm are you using? $\endgroup$ – Hamed Jun 22 '17 at 18:33
  • 1
    $\begingroup$ @Hamed It doesn't matter which norm you use: all norms on a finite-dimensional vector space are equivalent. $\endgroup$ – Robert Israel Jun 22 '17 at 18:44
2
$\begingroup$

Of course the definition of $w(t)$ is

$$ w(t) = \det(B(t) + R(t)) - \det(B(t)) $$

The question is, to show that $$0 = \lim_{h \to 0} \frac{|w(h)|}{|h|} = \lim_{h \to 0} \frac{|\det(B(h) + R(h)) - \det(B(h))|}{|h|}$$

Take $M = \sup_{|h| \le 1} (\|B(h)\| + \|R(h)\|)$, which is finite because $B$ and $R$ are continuous. Since $\det$ is a smooth function, in fact a polynomial in the entries of the matrix, there is some $K$ such that $$|\det(X + Y) - \det(X)| \le K \|Y\|$$ for all matrices $X$ and $Y$ with $\|X\| \le M$ and $\|Y\| \le M$. For any $\epsilon > 0$, there is $\delta$ such that $0 < \delta < 1$ and if $|h| < \delta$, $$\frac{\|R(h)\|}{|h|} < \frac{\epsilon}{K}$$ Then for such $h$, $$\frac{|\det(B(h)) + R(h)) - \det(B(h))|}{|h|} \le K \frac{\|R(h)\|}{|h|} < \epsilon$$ and you're done.

$\endgroup$
  • $\begingroup$ That was a lot more doable than I thought, I guess, I gave up too quick. Your help is much appreciated! $\endgroup$ – Jonathan Jun 23 '17 at 6:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.