3
$\begingroup$

I need help with solving the problem about square matrices of equal size. I know that if $A + B = AB$, then $AB = BA$, but I can't prove this one. Please advise how to solve or think about this. Thanks in advance )

$\endgroup$
2
  • $\begingroup$ Multiply on the left and right by A. Why is that allowed? $\endgroup$ – Paul Feb 10 '20 at 10:47
  • $\begingroup$ Sorry for uncertainty, A and B both have size $n*$n $\endgroup$ – Tigran Petrosyan Feb 10 '20 at 10:56
7
$\begingroup$

Your equality is equivalent to $$(I-A^2)(I-B)=I$$ so $(I-B)= (I-A^2)^{-1}$. It is enough to check that $A$ commutes with $(I-B)$. But since $A$ commutes with $(I-A^2)$, it will also commute with its inverse.

$\endgroup$
2
5
$\begingroup$

Left-inverses are right-inverses. In your original problem, $$(I-A)(I-B) = I - A - B + AB = I$$ and so $(I-A)(I-B) = I = (I-B)(I-A)$ giving the result.

In your next example $$I=(I-A^2)(I-B)=(I+A)(I-A)(I-B) \\ \text{and} \\ I = (I-B)(I-A^2)= (I-B)(I-A)(I+A)$$ This gives two expressions for $(I+A)^{-1}$ and the result follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.