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Let $u=(u_1,u_2,\ldots,u_n)$ and $v=(v_1,v_2,\ldots,v_n)$ be two vectors in $\Bbb R^n$. Suppose $\left|v_i\right|>|u_i|$ for all $i$. Let $\| \cdot\|$ be any vector norm on $\Bbb R^n$.

Is it true that $||v||>||u||$?

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Let $A=\begin{bmatrix} 2 & -2 \\ 1 & 1 \end{bmatrix}$, and let $\|x\|_* = \|Ax\|_\infty = \max(2|x_1-x_2|, |x_1+x_2|)$.

Let $u=(0,1), v=(\frac{1}{2}, \frac{5}{4})$. Then $|v| > |u|$, but $\|u\|_* = 2$, $\|v\|_* = \frac{7}{4} <2$.

(Picture not quite to scale. And slightly incorrect, the lower green box edgeshould line up with $u_2$.)

enter image description here

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  • $\begingroup$ thanks so much! the problem i want to show is that $||D||=max{|D_{ii}|:1≤i≤n}$ where D is a diagonal matrix and $||.||$ is any induced norm. Now I know that this is incorrect. $\endgroup$ – Ryan Oct 18 '13 at 16:08
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The answer is no. In $\mathbb R^2$, define the transformation $P(x_1,x_2)=(2x_1-2x_2,x_1+x_2)$, and define the norm $$\|x\|_0=\|P(x)\|_2,$$ where $\|\cdot\|_2$ is the Euclidean norm. Then, this is indeed a norm: first, it is nonnegative, and $P$ is invertible, so $$\|x\|_0=0\Rightarrow \|P(x)\|_2=0\Rightarrow P(x)=0\Rightarrow x=0.$$ Also, $P$ is linear, so $$\|\lambda x\|_0=\|P(\lambda x)\|_2=\|\lambda P(x)\|_2=|\lambda|\|P(x)\|_2=|\lambda|\|x\|_0.$$ Finally, the triangle inequality also holds: $$\|x+y\|_0=\|P(x+y)\|_2=\|P(x)+P(y)\|_2\leq\|P(x)\|_2+\|P(y)\|_2=\|x\|_0+\|y\|_0.$$ Now, let $v=(1,1)$ and $u=\left(\frac{3}{4},0\right)$. Then $|v_i|>|u_i|$ for $i=1,2$, but $$\|v\|_0=\sqrt{4(v_1-v_2)^2+(v_1+v_2)^2}=2,$$ while $$\|u\|_0=\sqrt{4\frac{9}{16}+\frac{9}{16}}=\frac{3\sqrt{5}}{4}>2=\|v\|_0.$$

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  • $\begingroup$ You may want to tweak this example, it doesn't show what you want. $\endgroup$ – Empy2 Oct 18 '13 at 6:00
  • $\begingroup$ You were right, I edited it. $\endgroup$ – detnvvp Oct 18 '13 at 6:07

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