I'm a bit confused about what an operator norm is.

Can someone explain it to me using a simple matrix?

What value would you use for the Euclidean norm (x) when it is multiplied by A? (before taking the supremum).

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- Apr 16th 2010, 07:13 AMnotgoodatmathOperator norm
I'm a bit confused about what an operator norm is.

Can someone explain it to me using a simple matrix?

What value would you use for the Euclidean norm (x) when it is multiplied by A? (before taking the supremum). - Apr 16th 2010, 07:32 PMGiraffro
I'm not sure which definition you're using for the operator norm, but the definition I use is if $\displaystyle (X, \|\cdot\|_X), (Y, \|\cdot\|_Y)$ are normed vector spaces and $\displaystyle T : X \to Y$ is linear and bounded, define $\displaystyle \|T\|_{B(X,Y)} := \inf(\{M \in [0,\infty) : \forall x \in X, \|T(x)\|_Y \leq M \|x\|_X\}$. I assume you're using the formula $\displaystyle \|T\|_{B(X,Y)} = \sup(\{ \|T(x)\|_Y : x \in X, \|x\| \leq 1\})$.

Define $\displaystyle A := \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ and let $\displaystyle (x,y)^t \in \mathbb{R}^2$ be given such that $\displaystyle \|(x,y)^t\| \leq 1$. Then $\displaystyle \|A (x,y)^t\| = \|(y, -x)^t\| = \|(x,y)^t\|$. Therefore $\displaystyle \|A\|_{B(X,Y)} = \sup(\{\|(x,y)^t\| : (x,y)^t \in \mathbb{R}^2, \|(x,y)^t\| = 1\}) = 1$.