It is known that a and B are matrices of order n, and E-Ab is invertible. It is proved that e-Ba is invertible Counter proof: if e-Ba is irreversible, (e-Ba) x = 0, and the equation has non-zero solution, how to show that (E-Ab) x = 0 also has non-zero solution, and then the determinant of E-Ab is 0, which shows that E-Ab is irreversible and contradicts the known conditions, so the original proposition holds. How to show that (E-Ab) x = 0 also has non-zero solution?

It is known that a and B are matrices of order n, and E-Ab is invertible. It is proved that e-Ba is invertible Counter proof: if e-Ba is irreversible, (e-Ba) x = 0, and the equation has non-zero solution, how to show that (E-Ab) x = 0 also has non-zero solution, and then the determinant of E-Ab is 0, which shows that E-Ab is irreversible and contradicts the known conditions, so the original proposition holds. How to show that (E-Ab) x = 0 also has non-zero solution?

As long as we find a non-zero solution satisfying (E-Ab) y = 0, we can explain the contradiction with the problem,
Suppose e-Ba is irreversible, then (e-Ba) x = 0 has nonzero solution, then x = Bax can be obtained
And (E-Ab) AX = ax - abax = ax-ax = 0, that is, ax is a nonzero solution of (E-Ab) y = 0
Some people understand it this way
Because E-Ab is invertible, then there is invertible matrix C such that C (E-Ab) = e, then c-cab = E,
Multiply B on the left and multiply a on the right, with bca-bcaba = ba
There are BCA = (E + BCA) BA and (BCA + e) - E = (E + BCA) BA, and there are (BCA + e) (e-Ba) = E. according to the definition, e-Ba is reversible