operator normOperatorNorm2002-06-03 20:12:522008-09-17 20:00:13Definitionbounded linear mapunbounded linear mapbounded operatorunbounded operator% this is the default PlanetMath preamble. as your knowledge
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\def\op{{\rm op}}\subsection*{Definition}
Let $A\colon \V\to\W$ be a linear map between normed vector spaces $\V$ and
$\W$. To each such map (operator) $A$ we can assign a non-negative number
$\|A\|_{\op}$ defined by
\[
\|A\|_{\op} := \mathop{\sup_{\v\in\V}}_{\v\ne{\bf 0}} \frac{\|A\v\|}{\|\v\|},
\]
where the supremum $\|A\|_{\op}$ could be finite or infinite.
Equivalently, the above definition can be written as
\[
\|A\|_\op := \mathop{\sup_{\v\in\V}}_{\|\v\|=1} \|A\v\|
= \mathop{\sup_{\v\in\V}}_{0<\|\v\|\le1} \|A\v\|.
\]
By convention, if $\V$ is the zero vector space, any operator from $\V$
to $\W$ must be the zero operator and is assigned zero norm.
$\|A\|_{\op}$ is called the the \emph{operator norm} (or the \emph{induced norm})
of $A$, for reasons that will be clear in the next \PMlinkescapetext{section}.
\subsection*{Operator norm is in fact a norm}
{\bf Definition -} If $\|A\|_{\op}$ is finite, we say that $A$ is a
\emph{\PMlinkescapetext{bounded}}. Otherwise, we say that $A$ is \emph{\PMlinkescapetext{unbounded}}.
It turns out that, for bounded operators, $\|\cdot\|_{\op}$ satisfies all the properties of a norm
(hence the name \emph{operator norm}). The proof follows immediately from the definition:
\begin{description}
\item[Positivity:]
Since $\|A\v\|\ge 0$, by definition
$\|A\|_\op \ge 0$. Also, $\|A\v\| = 0$ identically only if $A=0$.
Hence $\|A\|_\op = 0$ only if $A = 0$.
\item[Absolute homogeneity:]
Since $\|\lambda A\v\|=|\lambda| \|A\v\|$, by definition
$\|\lambda A\|_\op = |\lambda| \|A\|_\op$.
\item[Triangle inequality:]
Since $\|(A+B)\v\|=\|A\v+B\v\|\le \|A\v\| + \|B\v\|$, by definition
$\|A+B\|_\op \le \|A\|_\op + \|B\|_\op$.
\end{description}
The set $L(\V,\W)$ of bounded linear maps from ${\mathsf V}$ to ${\mathsf W}$ forms a vector space and $\|\cdot\|_{\op}$ defines a norm in it.
\subsection*{Example}
Suppose that $\V=(\R^n,\|\cdot\|_p)$ and $\W=(\R^n,\|\cdot\|_p)$, where
$\|\cdot\|_p$ is the vector p-norm. Then the operator norm
$\|\cdot\|_\op = \|\cdot\|_p$ is the matrix p-norm.