Definition of Norm : Suppose $A$ be an $m\times n$ matrix. Then $\|A\|$ is defined by : \begin{align} \|A\|= &\sup\{\|Ax\| : \|x\|=1\}\\ = &\sup\{\|Ax\| : \|x\|\leq 1\} \end{align} These two definitions are equivalent. But why? Please anyone explain..
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since the norm of a vector is homogenous. – Ittay Weiss Nov 13 '13 at 13:50
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4Possible duplicate of Operator norm. Alternative definition and this. – Dec 19 '16 at 11:00
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Let $\alpha = \sup\{\|Ax\| : \|x\| = 1\}$ and $\beta = \sup\{\|Ax\| : \|x\| \leq 1\}$. Clearly, $\alpha \leq \beta$.
Now, for any $0\neq x \in X$ such that $\|x\| \leq 1$, consider $y = x/\|x\|$. Then $\|y\| = 1$. Then $$ \|Ay\| \leq \alpha $$ But $$ \|Ay\| = \frac{\|Ax\|}{\|x\|} \geq \|Ax\| $$ Hence, $\|Ax\| \leq \alpha$, which implies that $\beta \leq \alpha$
Prahlad Vaidyanathan
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