Definition of the operator norm

Question

I know that this definition is correct, is the bottom one also fine?

$A \in L(\mathbb {R}^m,\mathbb {R}^n)$

$\Vert A \Vert _{op}:=\sup \{{\vert Ax\vert } \big \vert \,x \in \mathbb {R}^m,\vert \,\vert x\vert \leq 1\} $

$\vert Ax\vert$ is the Euclidiean norm in $\mathbb {R}^n$ and $\vert x\vert$ is the Eucldidien norm in $\mathbb {R}^m$. $\Vert A \Vert _{op}$ is the operator norm.

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$A \in L(\mathbb {R}^n)$, $\Vert A \Vert _{op}:=\sup \{\frac {\vert Ax\vert }{\vert x\vert }\; \big \vert \,\vert x\vert \geq 1\}$

Answer 1

The correct equivalent of the definition is $\Vert A \Vert_{op}:= \sup \left\{\frac {\vert Ax\vert }{\vert x\vert } \,:\, x\neq 0\right\}$. See for instance A course in functional analysis, Conway, page 27.

Answer 2

Let $\|x\|<1$. Linearity implies that $\|A(x/\|x\|)\|=\frac{1}{\|x\|}\|Ax\|>\|Ax\|$. Thus your first definition reduces to $\|A\|_{op}=\sup\{\|Ax\|:\|x\|=1\}$.

Now for any $\|x\|>0$ note that $\|\frac{x}{\|x\|}\|=1$, which again by linearity means that $\sup\{\frac{\|Ax\|}{\|x\|}:x\neq 0\}=\sup\{\|Ax\|:\|x\|=1\}$, and these are the two standard definitions used for $\|A\|_{op}$.