Math, asked by Anonymous, 7 months ago

Consider the norms ||.||  _{1} and ||.||  _{ \infty } on R  ^{n} prove that \large ||x|| = \frac {1}{3} ||x|| _{1} + \frac {2}{3} ||x|| _{ \infty} defines as a norm on R [tex] ^{n} .​

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Answered by Anonymous
5

An $M\times N$ matrix ${\bf A}$ can be considered as a particular kind of vector ${\bf x}={\bf A}\in R^{m,n}$, and its norm is any function that maps ${\bf A}$ to a real number $\vert\vert{\bf A}\vert\vert$ that satisfies the following required properties:

Positivity:

\begin{displaymath}

\vert\vert{\bf A}\vert\vert\ge 0,\;\;\;\;\;\;\vert\vert{\bf A}\vert\vert=0\;\;\;\mbox{iff}\;\;\;{\bf A}={\bf0}

\end{displaymath}

Homogeneity:

\begin{displaymath}

\vert\vert a{\bf A}\vert\vert=\vert a\vert\;\vert\vert{\bf A}\vert\vert

\end{displaymath}

Triangle inequality:

\begin{displaymath}

\vert\vert{\bf A}+{\bf B}\vert\vert\le \vert\vert{\bf A}\ve...

...\vert{\bf B}\vert\vert\le \vert\vert{\bf A}-{\bf B}\vert\vert

\end{displaymath}

In addition to the three required properties for matrix norm, some of them also satisfy these additional properties not required of all matrix norms:

Subordinance:

\begin{displaymath}

\vert\vert{\bf A}{\bf x}\vert\vert\le \vert\vert{\bf A}\vert\vert \cdot \vert\vert{\bf x}\vert\vert

\end{displaymath}

Submultiplicativity:

\begin{displaymath}

\vert\vert{\bf A}{\bf B}\vert\vert\le \vert\vert{\bf A}\vert\vert \cdot \vert\vert{\bf B}\vert\vert

\end{displaymath}

We now consider some commonly used matrix norms.

Element-wise norms

If we treat the $M\times N$ elements of ${\bf A}$ are the elements of an $MN$-dimensional vector, then the p-norm of this vector can be used as the p-norm of ${\bf A}$:

\begin{displaymath}\vert\vert{\bf A}\vert\vert _p=\left\{ \sum_{i=1}^M\sum_{j=1}^N \vert a_{ij}\vert^p\right\}^{1/p}\end{displaymath}

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