What is the difference between the operator norm and the Euclidean norm? I am quite confused whether the operator norm is the same as the Euclidean norm (2-norm). I know that :$\left \| A \right \|=\sup_{x\neq 0}\frac{\left \| Ax \right \|}{\left \| x \right \|}$. In wikipedia, I found that the $p$-norm is given by $\left \| A \right \|_p=\sup_{x\neq 0}\frac{\left \| Ax \right \|_{p}}{\left \| x \right \|_{p}}$, so for $p=2$, we get that the $2$-norm is exactly the same as the operator norm. Is the claim: "the operator norm is the same as the 2-norm" correct?
 A: $\newcommand{\norm}[1]{\left\Vert #1 \right\Vert}$
$\newcommand{\Norm}[1]{ ||| #1 ||| }$
To make the distinction clear, I will write $\norm{\cdot}$ to denote the norm of a vector, and $\Norm{\cdot}$ to denote the norm of the operator.
You start with the vector norm $\norm{\cdot}$, it is given to you. Then, given the vector norm, there is an operator norm $\Norm{\cdot}$ induced by the vector norm $\norm{\cdot}$ given by
$$
\Norm{A} = \sup_{x \neq 0} \frac{\norm{Ax}}{\norm{x}}.
$$
There is not just one operator norm. For every vector norm, we may use the preceding definition to define a norm on the operators. If you start with the Euclidean norm $\norm{\cdot}_2$ on vectors, you may define the corresponding Euclidean norm on operators $\Norm{\cdot}_2$. If you start with the $p$- norm $\norm{\cdot}_p$ on vectors, you will define a different operator norm (assuming $p \neq 2$, otherwise of course it will be the same one).
As a side note, I write "different," but for finite dimensional vector spaces, it has been shown that all norms are, in essence, equivalent. See https://planetmath.org/allnormsonfinitedimensionalvectorspacesareequivalent if interested.
