作者skyhigh8988 (Aesthetic)
看板Math
标题[线代] 这一题行列式值的题目
时间Mon Feb 14 17:39:05 2011
Let x and y be vectors in R^n n>1. Show that if A=xy^T then det(A)=0
解答是这样的:If A = xy^T then all of the rows of A are multiples of y^T .
It follows that if U is any row echelon form of A then U
can have at most one nonzero row. Since A is row equivalent to U and
det(U) = 0, it follows that det(A) = 0.
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我是知道xy^T是怎麽样的东西
但是我不知道为什麽会忽然的扯到了U 然後A和U列同价 所以行列式值一样
有请版上的大大帮忙解惑一下了
谢谢
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◆ From: 140.116.117.15
1F：→ sm008150204 :你把x=[x1 ... xn]^t y=[y1 ... yn]^t带进去就知道了 02/14 18:13
2F：→ sm008150204 :每一列都会是第一列的倍数 02/14 18:14
3F：→ skyhigh8988 :懂了懂了 谢谢您 02/14 20:06