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Which two of the vectors $u=(-2,2,1)^T$, $v=(1,4,1)^T$, and $w=(0,0,-1)^T$ are closets to each other in distance for (a) the Euclidean norm? (b) the infinity norm? (c) the 1 norm?

I believe I know how to solve this, but I was hoping someone could confirm or deny this for me. For (a) I took $||u-v||_1$, $||u-w||_1$, and $||v-w||_1$, but what do I take for the infinity and 1 norm? How do the 1-norm and the Euclidean norm differ?

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2 Answers 2

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Let $v=(x,y,z)$. Then:

$$||v||_1=|x|+|y|+|z|$$

$$||v||_2=\sqrt{x^2+y^2+z^2}$$

$$||v||_{\infty}=\max{(|x|,|y|,|z|)}.$$

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  • $\begingroup$ why is the infinity norm $||v||_2$ and the 1-norm $||v||_{\infty}$? $\endgroup$
    – jerry2144
    Commented Oct 14, 2014 at 20:31
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    $\begingroup$ It comes from exponent in general formula $||v||_p=\sqrt[p]{x^p+y^p+z^p}$ (it's true in $1$ and $2$ case). When $p \to \infty$ then "$||v||_p \approx \max{|x|,|y|,|z|}$". $\endgroup$
    – agha
    Commented Oct 14, 2014 at 20:38
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It usually goes like this;
The 1-norm of a vector with components $x_n$ is $\sum_n |x_n|$
The 2-norm is the euclidean norm given by $\sqrt{\sum_n x_n^2}$
The p-norm is given by $\sqrt[p]{\sum |x_n|^p}$
The infinity norm is the limit as the powers of these things go to infinity which happens to have the nice form $max(|x_n|)$.
Here I'm using $|a|$ to mean the absolute value of a

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  • $\begingroup$ I've been reading about norms for a few months now and of course I find the most succinct and clear explanation on Stack Exchange. Finally understand them. Thank you! $\endgroup$
    – dijksterhuis
    Commented Nov 14, 2019 at 3:28

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