關於逆矩陣的,如何證明當AB=I時,BA=I?

AB=I
A-1 (AB) = A-1 (I)
A-1AB=A-1
(A-1A)B=A-1
IB=A-1
B=A-1
(B)A=A-1(A)
BA=A-1A
BA=I

2012-10-31 01:44:28 補充：
樓下：
當 AB=I，|A|, |B| 自然不會=0
|A| =/= 0 自然存在 A-1 使到 A-1 A = A A-1 =0
以上方法亦只用到這個概念，並未用到B=A-1 的概念。

2012-11-11 00:18:11 補充：
當 AB=I，|A|, |B| 自然不會=0
|A| =/= 0 自然存在 A-1 使到 A-1 A = A A-1 =I **
以上方法亦只用到這個概念，並未用到B=A-1 的概念。

先前給了很肉酸的意见，現在多寫一些東西當參考。
A=[1 0 0; 0 1 0] , B=[1 0; 0 1; 0 0]
AB=I , BA=/=I 且 det(BA)=0

AB=I
A-1 (AB) = A-1 (I)
A-1AB=A-1
(A-1A)B=A-1
IB=A-1
B=A-1
(B)A=A-1(A)
BA=A-1A
BA=I

|A|, |B|, |C| 均不為0是指引理 一 和 二

另外,可證明一下引理 一 嗎?

以上2位回答未夠嚴謹，因為我們在證明AB=I及BA=I之前 是不能寫A=B^-1以及利用逆矩陣的特性的。

引理 1：
如果AB = C 而 |C|不為0, 則|A| , |B| 均不為0

引理 2：
如果|A|, |B|不為0，存在C使得AC = B
(根據引理1, |C|也不為0)

設我們已知AB = I，由於| I | = 1 =/= 0, |A|, |B|不為0 (引理1)
這樣設|C| =/= 0 使得 BC = I (引理2)

BA
= BA (I)
= BA(BC)
= B(AB)C
= B(I)C
= BC = I

所以 AB = I 的話，BA也會等於I
這時候我們才可以寫A = B^-1
否則單以AB=I設A=B^-1去證明BA=I會出現循環謬誤。

以上證明是對於所有|A|, |B|=/=0的矩陣作出的證明(如果=0乘起來不會是I) 所以不會有AB=I但BA=/=I的情況.

2012-11-10 10:09:28 補充：
Sorry for my laziness to type English.

I used |M| =/= 0 instead of non-singular throughout the proof, this must be assumed which is used in the proof.

Lemma 1 is iff. This is the complete statement:

Suppose A, B is square matrix. AB is non-singular IFF BOTH A, B are non-signular.

2012-11-10 10:10:01 補充：
^ Therefore the example that 感冒唔駛食藥，飲盒仔茶就得啦。 is about SINGULAR matrix so they are not valid :>

Proof:
(=>)
Suppose B is singular. (A can be either singular or non-singular) Then there exist a vector z so that the system of linear equation Bz = 0 in which z is non-trivial.

2012-11-10 10:10:38 補充：
Consider the system of linear equation (AB)z = 0
(AB)z = A(Bz) = A0 = 0
However, z is a non-zero vector as indicated by Bz = 0, it implies that AB is non-singular

Suppose A is singular. (and B non-singular) There exist non-trivial y so that Ay = 0.

As B non-singular, Bw = y has an unique solution.

2012-11-10 10:10:53 補充：
If w = 0, y = Bw = 0, which is a contradiction.
If w =/= 0, (AB)w = A(Bw) = Ay = 0

Therefore AB is singular.

(<=)
Suppose both A and B is non-singular. Suppose x is a solution to (AB)x = 0.

2012-11-10 10:11:03 補充：
(AB)x = 0
A(Bx) = 0
Bx is a solution to Ay = 0.
By definition since A non-singular, y = 0.
Bx = 0
By definition since B non-singular, x = 0.
(AB)x = 0 implies x = 0 => AB non-singular.

2012-11-10 10:12:22 補充：
If A, B are singular, the proof concerning the system of linear equations are all false, hence wrong.

Lemma 2 is not coming from lemma 1, the idea involved using elemental row operations to transform A and B at the same time.

2012-11-10 10:12:52 補充：
I've posted the proof in the answering section.