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An operator norm is defined as $\|A\|_S=\sup\{\|Av\|:v\in \Bbb R^n, \|v\|=1\}$. Where $\|\cdot\|$ is some norm on $\Bbb R^n$ and $A\in M_n(\Bbb F)$, space of square matrices of dimension $n$ over $\Bbb C$ or $\Bbb R$. I should prove that this operator norm is in fact a norm. I have a problem with understanding what is the role of $\sup$ there, how can I make a supremum of vectors and how do I prove the first axiom of norm, that $\|A\|=0$ iff $A=0$.

Thank you.

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It's not a supremum of vectors; each $||Av||$ is a number, so you have some subset of $\mathbb{R}$ that you're taking the supremum of, in the usual way. –  Matt Pressland Apr 25 '13 at 9:59
    
To show that $\| A \|=0$ implies $A=0$, try to prove that $Av=0$ for every $v\in \mathbb{R}^n$. –  flavio Apr 25 '13 at 10:01
    
@jiku1797 That is not true, there is a matrix $ \begin{pmatrix} 1 & 1 \\ 0 & 0 \\ \end{pmatrix}\begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}$ and product of this is zero –  user74200 Apr 25 '13 at 10:27
    
Remark: in finite dimension, every linear map is continuous and the closed unit ball is compact. So the sup is actually a max. It is not necessarily a max in infnite dimension. –  1015 Apr 25 '13 at 11:17
    
@user74200: But $\| \begin{pmatrix} 1 & 1\\ 0 &0 \end{pmatrix} \| \ne 0$. –  flavio Apr 29 '13 at 14:02

2 Answers 2

I have proven it. It is indeed a supremum of numbers, a subset of $\Bbb R$. We have to prove this: $\forall A\ne 0, A\in M_n(\Bbb F), \exists v\in \Bbb F^n: w=Av\ne 0$.

We can chose vector $v$ to be the first non-zero row of matrix $A$, then at least one component of $w$ is non-zero, if the matrix is real. If the matrix is complex, we chose vector $v$ to contain conjugate components of the first non-zero row of matrix $A$.

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if ||A||=0

then it means that

$||A(1,0,0,000,0)||=0 $ so first column of A is zero

and you also $||A(0,1,0,000,0)||=0$ so scond column of A s zero

,,,,,,,,,,,,,,,

||A(0,0,0,000,1)||=0 so end column of A s zero

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