Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – Matthew 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

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 $A$.

share|cite|improve this answer

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 second column of $A$ s zero


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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