Proof that Non-Singularity of I-AB Implies Non-Singularity of I-BA

[eyehategod]

Prove that if the matrix I-AB is non-singular, then so i I-BA.
This was one of my test questions and got 3 points off. Can anyone tell me what I did wrong.

my proof:
assume I-AB is nonsingular
then (I-AB)^{-1} exists

Let C=(I-AB)^{-1}
Consider (I+BCA)(I-BA)=I-BA+BCA-BCABA
=I-BA+BC(I-AB)A
=I-BA+B(I-AB)^{-1} (I-AB)A
=I-BA+BIA
=I-BA+BA
=I

 

[quasar987]

3 points over how many?

Maybe you lost some points because

i) You did not prove that if C is the left inverse of A, then it is also a right inverse. So you might have lost some points because you only showed that (I+BCA) is the left inverse of (I-BA).

ii) The definition of nonsingular in your book is that A is nonsingular if its determinant is nonzero. And so, you might have lost some points because you did not explain that since I-BA had an inverse, then its determinant was nonzero and hence I-BA is nonsingular.

But other than that, your proof looks very good! go see your instructor.