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I have a homework, one of a few, that I should prove that the norm $||A||_S = \sup\{||Av||:||v||=1\}$ Where ||.|| is some norm on $R^n$ into which vector space belongs also $v$. I should mention that A is a square real or complex matrix with dimension $n$.

Wikipedia says it is a norm, however, first axiom in the definition of norm tells us that $||A||=0$ iff $A=0$. But we can find a matrix, for example: $$\begin{pmatrix}1 & 1\\ 0 & 0\end{pmatrix}$$ and vector $v=(1/\sqrt2, -1/\sqrt2)$ and then according to euclidean norm $||v||=1$ but $||Av||=0$ and clearly A is not zero. What am I missing?

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2 Answers 2

You have to take the sup over all $v$. Take a unit vector orthogonal to yours :)

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could you please explain it more descriptively? Thank you. –  user74200 Apr 24 '13 at 22:01
    
You took $v=(1,-1)/\sqrt2$. Take instead $v=(1,1)/\sqrt2$. Indeed, by Cauchy-Schwarz, this $v$ gives you the maximum possible $\|Av\|$. –  Ted Shifrin Apr 24 '13 at 23:22
    
Sorry, I don't seem to understand the role and function of the supremum there. I thought I know what the supremum means but as I thought about it, I found out that I am not sure what the supremum is. –  user74200 Apr 25 '13 at 8:40
    
The $\sup$ means to take the maximum (in this particular case) over all nonzero vectors $v$, not just plug in one vector and call it quits. –  hardmath Jan 15 at 13:47

As you have said $||A||$ is defined as the supremum over all possible values of $||Av||$ such that $||v||=1$. For the matrix you have given $||Av||=0$ for the particular vector $(1/\sqrt{2},-1/\sqrt{2})$, however take for instance $v'=(1/\sqrt{2},1/\sqrt{2})$. Then $||Av'||=\sqrt{2}$. Hence the supremum of $||Av||$ aver all vectors such that $||v||=1$ is at least as large as $\sqrt{2}$.

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