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$A$ is $M_{n\times n}$ matrix and $x,y$ are column vector. I want to show that $||A(x-y)||_2\leq ||A||_2||x-y||_2$ with the Euclidean norm.

I know that for the Cauchy-Schwarz inequality, both vectors should be of the same dimension. Is it right to argue the above statement using the Cauchy-Schwarz inequality as it contains one matrix and another column vector?

Any help is appreciated.

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    $\begingroup$ It is Karl Hermann Amandus Schwarz, not Schwartz. $\endgroup$ – Martin R Apr 3 at 8:58
  • $\begingroup$ @MartinR Thanks a lot Sir AS I am doing this mistake from Long time .I will correct that ... Please Help me $\endgroup$ – MathLover Apr 3 at 9:01
  • $\begingroup$ You are not the only one – and I wonder why the name is so often written wrongly. $\endgroup$ – Martin R Apr 3 at 9:03
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    $\begingroup$ @MartinR: as far as I understand, Schwartz, like Leibnitz, come from erroneous transcriptions of French officers; they are "Frenchified" versions of the correct surnames. The fact that regions such as Alsace and Lorraine kept changing country means that there have been many such transcriptions. But the latter is just a conjecture of mine, I am no historian. $\endgroup$ – Giuseppe Negro Apr 3 at 9:11
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This is true for every linear operator $T: V\to W$ between two normed spaces $(V,\|\cdot \|_V)$ and $(W,\|\cdot \|_W)$ (we say that $T$ is linear if $T(x+y)=T(x)+T(y)$ and $T(\lambda x)=\lambda T(x)$ for all $x,y\in V$ and $\lambda\in \mathbb{K}$).

We say that a linear operator $T: V\to W$ is bounded if there exists $M>0$ such that $\|x\|_V\leq 1$ implies $\|T(x)\|_W\leq M$. In such cases, we define the operator norm by \begin{equation} \|T\|=\sup\{\|T(x)\|_W : x\in V, \;\|x\|_V\leq 1\} \end{equation} If $x\neq 0$ and $x\in V$, then $$\left\|T\left(\frac{x}{||x||_V}\right)\right\|_W\leq \sup\{\|T(x)\|_W : x\in V, \;\|x\|_V\leq 1\}=\|T\|,$$ since the point $x/\|x\|_V$ has norm $1$. Then, for $x\neq 0$ and $x\in V$ (the case $x=0$ is trivial), we have $$\|T(x)\|_W=\left\|T\left(x\right)\frac{\|x\|_V}{\|x\|_V}\right\|_W=\left\|\|x\|_VT\left(\frac{x}{\|x\|_V}\right)\right\|_W=\|x\|_V \left\|T\left(\frac{x}{\|x\|_V}\right)\right\|_W \leq \|T\|\|x\|_V$$

In particular, if $T:V\to W$ is a linear operator between normed spaces with finite dimension $n$ with the Euclidean norm whose associated matrix in some basis is $A$, then you have your result.

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  • $\begingroup$ Dear Sir , Is this true for Hilbert Schimdt norm on operator ? $\endgroup$ – MathLover Apr 3 at 9:34
  • $\begingroup$ Is it true, since Hilbert-Schimdt norm is indeed a norm $\endgroup$ – user326159 Apr 3 at 9:36
  • $\begingroup$ Nice Sir I got . I had only one doubt why Sup{$||Tx||| , ||x||\leq 1$}=||T||? last 7th line $\endgroup$ – MathLover Apr 3 at 9:41
  • $\begingroup$ Why Sir Norm T which is in Hilbert schimdt norm equals operator norm $\endgroup$ – MathLover Apr 3 at 9:49
  • $\begingroup$ This supremum is the definition of the norm operator $\endgroup$ – user326159 Apr 3 at 9:50

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