证明由AB=E得,BA=E

前言
相信好多小伙伴在看到矩阵的逆的定义时都会有个小疑惑,为什么只需要证明 A B = E AB=E AB=E,则就可以说明B是A的逆,而无需再证 B A = E BA=E BA=E,这里给出个小证明。

由 A B = E AB=E AB=E得
∑ k = 1 n a i k b k j = { 1 , i = j 0 , i ≠ j \sum_{k=1}^{n}a_{ik}b_{kj}= \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases} k=1∑n​aik​bkj​={1,0,​i=ji​=j​

n为方阵的大小

记 A i j A_{ij} Aij​为矩阵A对应位置的代数余子式,结合上式(两边乘 A s i A_{si} Asi​)可构造以下等式
A s i ⋅ ∑ k = 1 n a s k b k j = { A j i , s = j 0 , s ≠ j A_{si} \cdot \sum_{k=1}^{n}a_{sk}b_{kj}= \begin{cases} A_{ji}, & s = j \\ 0, & s \neq j \end{cases} Asi​⋅k=1∑n​ask​bkj​={Aji​,0,​s=js​=j​
其中 s = 1 , 2 , . . . , n s=1,2,...,n s=1,2,...,n

对不同s求和
∑ s = 1 n A s i ⋅ ∑ k = 1 n a s k b k j = A j i \sum_{s=1}^{n}A_{si} \cdot \sum_{k=1}^{n}a_{sk}b_{kj}=A_{ji} s=1∑n​Asi​⋅k=1∑n​ask​bkj​=Aji​
交换求和顺序,并将 b k j b_{kj} bkj​提出得
∑ k = 1 n b k j ⋅ ∑ s = 1 n a s k A s i = A j i \sum_{k=1}^{n}b_{kj} \cdot \sum_{s=1}^{n}a_{sk}A_{si}=A_{ji} k=1∑n​bkj​⋅s=1∑n​ask​Asi​=Aji​
结合代数余子式的性质
∑ s = 1 n a s k A s i = { ∣ A ∣ , k = i 0 , k ≠ i \sum_{s=1}^{n}a_{sk}A_{si}= \begin{cases} \left | A\right | , & k = i \\ 0, & k \neq i \end{cases} s=1∑n​ask​Asi​={∣A∣,0,​k=ik​=i​可得
b i j ⋅ ∣ A ∣ = A j i b_{ij} \cdot \left | A\right | =A_{ji} bij​⋅∣A∣=Aji​
对 A B = E AB=E AB=E两边取行列式得: ∣ A ∣ ∣ B ∣ = 1 |A| |B|=1 ∣A∣∣B∣=1,即可推出: ∣ A ∣ ≠ 0 |A| \neq 0 ∣A∣​=0
故有以下重要等式
b i j = A j i ∣ A ∣ b_{ij} = \frac{A_{ji}}{\left | A\right | } bij​=∣A∣Aji​​
故有
∑ k = 1 n b i k a k j = ∑ k = 1 n A k i ∣ A ∣ ⋅ a k j = 1 ∣ A ∣ ⋅ ∑ k = 1 n A k i a k j = { 1 , i = j 0 , i ≠ j \sum_{k=1}^{n}b_{ik}a_{kj}=\sum_{k=1}^{n} \frac{A_{ki}}{\left | A\right | } \cdot a_{kj} = \frac{1}{\left | A\right | } \cdot \sum_{k=1}^{n}A_{ki}a_{kj}= \begin{cases} 1 , & i=j\\ 0, & i \neq j \end{cases} k=1∑n​bik​akj​=k=1∑n​∣A∣Aki​​⋅akj​=∣A∣1​⋅k=1∑n​Aki​akj​={1,0,​i=ji​=j​
既 B A = E BA=E BA=E,证明完毕。

参考

矩阵,如果AB=E,只用矩阵乘法的定义如何证明BA=E?

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