→ sm008150204 :你把x=[x1 ... xn]^t y=[y1 ... yn]^t帶進去就知道了 02/14 18:13
→ sm008150204 :每一列都會是第一列的倍數 02/14 18:14
→ skyhigh8988 :懂了懂了 謝謝您 02/14 20:06
Let x and y be vectors in R^n n>1. Show that if A=xy^T then det(A)=0
解答是這樣的: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是怎麼樣的東西
但是我不知道為什麼會忽然的扯到了U 然後A和U列同價 所以行列式值一樣
有請版上的大大幫忙解惑一下了
謝謝
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