Re: [線代] det跟linearly dependent
※ 引述《mqazz1 (無法顯示)》之銘言:
: 已知 a(1)a(4) = a(2)a(3)
: [√[a(1)a(4)] 0 0 a(1) ]
: [ 0 √[a(1)a(4)] a(2) 0 ]
: [ 0 a(3) √[a(1)a(4)] 0 ]
: [ a(4) 0 0 √[a(1)a(4)]]
: 第1列和第4列為linearly dependent
: 第2列和第3列為linearly dependent
: 書上寫
: ([√[a(1)a(4)] a(2) ])
: det([ a(3) √[a(1)a(4)]]) = 0
: ([√[a(1)a(4)] a(1) ])
: det([ a(4) √[a(1)a(4)]]) = 0
: 請問為什麼可以這樣判斷?
: 謝謝
因為,如果第一列跟第四列線性相依(linearly depedent) 的話,
_____ _____
α[ √[a(1)a(4)] 0 0 a(1) ] = β[ a(4) 0 0 √[a(1)a(4)] ]
_____ _____
<=> α[ √[a(1)a(4)] a(1) ] = β[ a(4) √[a(1)a(4)] ]
_____
╭ √[a(1)a(4)] a(1) ╮
<=> det│ _____ │= 0 for some α,β 在體 |F.
╰ a(4) √[a(1)a(4)] ╯
同理, 由第二列和第三列線性相依可得
_____
╭ √[a(1)a(4)] a(2) ╮
det│ _____ │
╰ a(3) √[a(1)a(4)] ╯
再看不懂的話,用降階直接算(但也沒全部算就是了)
_____
╭ √[a(1)a(4)] 0 0 a(1) ╮
│ 0 √[a(1)a(4)] a(2) 0 │
det│ 0 a(3) √[a(1)a(4)] 0 │
╰ a(4) 0 0 √[a(1)a(4)] ╯
_____
_____ ╭ √[a(1)a(4)] a(2) 0 ╮
= √[a(1)a(4)] det│ a(3) √[a(1)a(4)] 0 │
╰ 0 0 √[a(1)a(4)] ╯
_____
╭ 0 √[a(1)a(4)] a(2) ╮
- a(1) det│ 0 a(3) √[a(1)a(4)] │
╰ a(4) 0 0 ╯
_____
_____ 2 ╭ √[a(1)a(4)] a(2) ╮
= (√[a(1)a(4)]) det│ _____ │
╰ a(3) √[a(1)a(4)] ╯
_____
╭ √[a(1)a(4)] a(2) ╮
- a(1)a(4) det│ _____ │.
╰ a(3) √[a(1)a(4)] ╯
_____
╭ √[a(1)a(4)] a(2) ╮ _____ 2
Since det│ _____ │= (√[a(1)a(4)]) - a(2)a(3)
╰ a(3) √[a(1)a(4)] ╯
= a(1)a(4) - a(2)a(3) = 0, the determinant is zero.
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