0

My teacher told me that, in case of square matrices, binomial expansion holds true if and only if they commute. I am not able to figure out why they fail for other cases; any proof will be helpful.

J. W. Tanner
  • 60,406
imposter
  • 532

2 Answers2

1

Hint:

$(A+B)^2=(A+B)(A+B)=A(A+B)+B(A+B)=A^2+AB+BA+B^2$

J. W. Tanner
  • 60,406
1

Depends what you mean. $$(A+B)^3 = A^3 + 3 A^2 B + 3 A B^2 + B^3$$ also works for $$ A = \pmatrix{a_{11} & 0\cr a_{21} & a_{22}\cr},\ B = \pmatrix{b_{11} & 0\cr b_{21} & -a_{11}-2 a_{22}-2b_{11}\cr} $$ which in general do not commute.

Robert Israel
  • 448,999
  • So there is no specific condition? – imposter May 15 '20 at 08:04
  • 1
    If they commute, $(A+B)^n = \sum_{i=1}^n {n \choose i} A^i B^{n-i}$ is true for all $n$. For $n=2$, this is only true when they commute. For other $n$, there are likely to be some other matrices for which it turns out to be true. – Robert Israel May 15 '20 at 13:26