Re: [線代] maxium norm

[armopen (八字-風水-姓名學)]

※ 引述《asdc20 (asdc20)》之銘言：
: Prove the following inequality used in verifying the maximum norm for
: R3.
: max{|x1|+|y1|,|x2|+|y2|,|x3|+|y3|} <= max{|x1|,|x2|,|x3|}+max{|y1|,|y2|,|y3|}
: 先謝謝各位的回覆......

Let |xj|+|yj| be the largest number of the left hand side.

Since |xj| <= max{|x1|,|x2|,|x3|} and |yj| <= max{|y1|,|y2|,|y3|},

we have |xj|+|yj| <= max{|x1|,|x2|,|x3|}+max{|y1|,|y2|,|y3|}.

This completes the proof.

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推 asdc20      :"="我可以理解，但為會啥小於? xj+yj已是最大數，
→ asdc20      :max{|x1|,|x2|,|x3|}和max{|y1|,|y2|,|y3|}
→ asdc20      :不就等於|xj|+|yj|?
→ suhorng     :why? xj+yj 是 {xk+yk} 裡的最大
→ suhorng     :不一定就 xj 是{xk}裡面最大且 yj 是 {yk} 裡最大
→ suhorng     :x1=0, x2=1, x3=1;     y1=100, y1=0, y3=0
→ asdc20      :謝謝您
→ armopen     :取 x1=0,x2=1,x3=2,y1=100,y2=99,y3=98等號不成立.