# Matrix norms, sup,max and headache

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• September 16th 2010, 02:48 AM
Mollier
Matrix norms, sup,max and headache
Hi,

I am reading (again) about matrix norms and have a few questions.

The definition I have says that given any norm $||*||$ on the space $\mathbb{R}^n$ of $n$-dimensional vectors with real entries, the subordinate matrix norm on the space $\mathbb{R}^{n x n}$ of $n x n$ matrices with real entries is defined by

$||A|| = \max_{x\in\mathbb{R}^n\backslash\{0\}}\frac{||Ax|| }{||x||}.$

I've also read that it can be defined as,

$||A|| = \sup_{||x||=1}||Ax||.$

The way I "understand" the second definition is that we take all the vectors $x$ in $\mathbb{R}^n$ whose norm is one, and multiply them with the matrix A to create a set of vectors, say,

$\{v_1,v_2,\cdots\}$

I believe that there is an infinite number of vectors with norm one..
We then take the norm of all these new vectors to get a set of real numbers. We now find the smallest real number that is larger than all the real numbers in this set, and use it as the norm of the matrix $A$. By the definition of $\sup$, I believe that this number that is larger than all numbers in our set is not part of the set.

As for the first definition, I am tempted to say that we take all vectors $x$ in $\mathbb{R}^n$ and divide then by $||x||$ to make a unit vector.We then multiply this unit vector by $A$ to get the same set of vectors ( $v_i$) as before. We take the norm of these vectors and get a set of numbers. Since $\max$ is involved I guess that this implies that the upper bound is in this set, and not outside of it as it is with $\sup$... I do not understand this..

I read somewhere that since $||x||_p$ is a scalar, we have that

$||A||_p = \sup_{x\neq 0}\frac{||Ax||_p}{||x||} = \sup_{x\neq 0}||\frac{Ax}{||x||_p}||_p$,

not sure how that works either..

By the way, I've also seen the matrix norm defined as,
$<br /> ||A|| = \sup_{x\in\mathbb{R}^n\backslash\{0\}}\frac{||Ax|| }{||x||}.$

I am confused. Hope someone can take the time and explain this to me, thanks.

• September 16th 2010, 04:04 AM
Mollier
After a bit more research I've found that if

$||A||=\sup_{||x||=1}||Ax||$ the domain of the continuous function $||Ax||$ of $x$ is closed and bounded and therefore the function will achieve its maximum and minimum values. We may then replace $\sup$ with $\max$..

Sorry for all the ranting..
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