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I always understand the norm of operator as the supreme multiple of the operator 'stretching' vector. But I think the multiple of operator compressing vector is same important.

Why we don't define the norm of operator $A$ as $\inf\limits_{||u||=1}||Au||$?

What is I think : first , it will dissatisfy triangle inequality of norm .Then, the norm can't induce a metric. But I really can't realize the important of triangle inequality. Why we can't use a inverse triangle inequality to define metric ?

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    Also $\|A\|$ can be zero even if $A$ is not zero. – 2017-02-26
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    @Shalop Yes, I miss it .I think I should use $\inf\limits_{||Au||=1}\frac{1}{||u||}$. I don't know whether there will be another trouble ,if so . – 2017-02-26
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    If $\|{\cdot}\|$ dont have the porperties of a norm, as the triangle inequality, then it isnt a norm or a distance function. Take a look at [semimetrics](https://en.wikipedia.org/wiki/Metric_(mathematics)#Semimetrics). – 2017-02-26
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    The point is that any norm induces a topology on a given vector space. In order for this to be true, the norm needs to satisfy the triangle inequality. – 2017-02-26
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    @lanse7pty: This is bascially $1/\|A^{-1}\|^{-1}$, if the inverse $A^{-1}$ exists (and is bounded). – 2017-02-27

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