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?
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- 2$\begingroup$ The operator norm depend on a norm on the vector space. What are you calling euclidian norm for a matrix? $\endgroup$Davide Giraudo– Davide Giraudo2012-04-19 15:50:35 +00:00Commented Apr 19, 2012 at 15:50
- $\begingroup$ @ Davide Giraudo : By euclidean norm (2-norm) of a matrix, I mean the following $\left \| A \right \|_{2}=sup_{\left \| x \right \|=1}{\left \| Ax \right \|_{2}}$ $\endgroup$Boyan Klo– Boyan Klo2012-04-19 15:59:13 +00:00Commented Apr 19, 2012 at 15:59
- $\begingroup$ @BoyanKlo: What exactly is there to clarify? $\endgroup$Martin Wanvik– Martin Wanvik2012-04-19 16:31:26 +00:00Commented Apr 19, 2012 at 16:31
- $\begingroup$ @ Martin Wanvi : I just want to know if the claim: "operator norm is exactly the same as the 2-norm" correct! $\endgroup$Boyan Klo– Boyan Klo2012-04-19 16:38:04 +00:00Commented Apr 19, 2012 at 16:38
- 1$\begingroup$ @BoyanKlo: As you've defined those terms, no - there is no difference. Is your main difficulty with the equality $\sup_{\| x \| = 1} \| A x\| = \sup_{x \neq 0} \| Ax \|/\|x\|$? (On an unrelated note, you are not using comment replies correctly - for starters, you need to remove the space between the @-character and the username you're replying to. See meta.stackexchange.com/questions/43019/…) $\endgroup$Martin Wanvik– Martin Wanvik2012-04-19 16:55:56 +00:00Commented Apr 19, 2012 at 16:55
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$\newcommand{\norm}[1]{\left\Vert #1 \right\Vert}$ $\newcommand{\Norm}[1]{ ||| #1 ||| }$ 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.
