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# 矩陣問題 Ax = 0

Let A denote an m x n matrix. Show that if Ax = 0 for any n-dimensional vector v, then A = [0]

令A為一 m x n 的矩陣. 證明如果 Ax = 0, 那對於所有n次向量 v, A必為 [0]

這就類似 假如兩實數 a, b 且 ab = 0, 如果 b 不為0, 則 a 一定為零

實數的證明方法就只需要兩邊各除以 b, 則 a就為 0

但如果是矩陣與任一向量相乘, 該如何證明呢?

請各位大大指教

謝謝

### 3 Answers

Rating

- myisland8132Lv 710 years agoFavorite Answer
Consider x_1 = [1, 0 , 0, ... 0]

From Ax = 0, we can conclude that a_11 = a_21 = ... = a_n1 = 0

Similarly, sub. x_i = [0, 0 ,... 1, 0, 0 ...0] (where 1 is on the i th position) we can conclude a_1i = a_2i = ... = a_ni = 0

So, A = [0]

- 小魚Lv 610 years ago
舉個例子:

A=

[1 -1]

[-1 1]

x=

[1]

[1]

則 Ax = 0 , 但 A不等零啊 !

所以要探討 A 的 rank !?

2011-05-29 06:32:33 補充：

補充一下:

Ax = 0 , 若 A 為滿秩(full rank) , 簡單講 A 可逆, 則 A 才會是 0. 若 A 不是滿秩,

則 A 不需要是零的 !

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