Re: [線代] 矩陣 AB=I => BA = I 嗎

[motivic (Ian)]

※ 引述《kyoiku (生死間有大恐怖)》之銘言：
: 反方陣的定義
: AB = BA = I，則 B 稱為 A 的反矩陣
: 如果只有 AB = I 那是否必然 BA = I 呢
: 如何證明？

Assume AB=I.

To prove BA=I, it is suffice to show that B is a linear combination of
I,A,A^2,A^3,...A^k,...

Proof:

Consider the infinitely long sequence of matrices I,A,A^2,A^3,....
They are all in a vector space of dim n^2.
Hence they are linearly dependant:
so there are some coefficients c_0, c_1,c_2,...,c_s such that

c_0*I+c_1*A+c_2*A^2+..+c_s*A^s = 0.

By multiplying B from the left, we may assume c_0 is non-zero, says c_0=-1.

Thus, we have I=c_1*A+c_2*A^2+...+c_s*A^s.

Mutiply B from the left again, we get  B=c_1*I+c_2*A+....+c_s*A^{s-1}.

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→ WasabiSushi : Sorry,it seemed that you proved necessary
→ WasabiSushi : condition instead of sufficient one; Namely, you
→ WasabiSushi : assumed BA=I, and then showed B is a linear
→ WasabiSushi : combination of A^k for k non-negative integers.
→ motivic     : thx...change all left to right..than its fine
推 WasabiSushi : Yes,it is a good proof which aims to show A
→ WasabiSushi : commutes with B, so it is sufficient to show
→ WasabiSushi : that B is a polynomial of A.