o1269652

given that AB^2 - A is invertible (A,B square matrices of the same size)
prove BA - A is invertible

been thinking about it for hours. could anybody please help?

johng40

Use the fact that a matrix X is invertible if and only if $\text{det}X\neq0$. Then show that $\text{det}(AB-A)\neq0$.

o1269652

Thanks
If somebody will ever want to solve the same problem, I found a cool solution.
If and only if XY is invertible, then both X and Y are.

AB^2 = A(B+I)(B-I)
so
(B-I)A=BA-A
is invertible.

SDK

While technically true, this triggers me on a fundamental level. I don't know of a single case where invoking the determinant in a proof is morally correct. Often it ends up being completely circular. In any case, it always obscures the beauty of linear algebra which makes it rub me the wrong way. In this case, I don't see how to compute $\det(AB - A)$ short of using the multiplication identity which then seems awfully circular.

The "right" way to approach this problem is to notice the following inclusions. If $A,B$ are any linear operators, then $\ker(A) \subset \ker(AB)$ and $\ker(B) \subset \ker(AB)$. Both of these are straightforward to prove, and give complete insight into why the main result is also true. Neither relies on matrices being square, or even requires matrix algebra at all.

Prokhartchin

OBS: $A(B+I)(B-I)=A(B^2 - I) = AB^2 - A \ne AB^2$ (your approach have mistake but its a good idea because all matrices are square by hypothesis, which assegure that we'll not have problem with determinants)

Follow your approach,

$\det(AB^2 - A) = \det(A (B^2 - I)) = \det(A)\det(B+I)\det(B-I) \ne0$ (by the determinant properties and hypothesis)

which implies

$\det(A),\det(B+I)$ and $\det(B-I)\ne 0$.

From this,

$\det(BA - A) = \det((B-I)A) = \det(B-I)\det(A)\ne0$ (by the properties deduced above).

And $\det(BA - A)\ne0$ is equivalent to $BA - A$ be invertible.

zylo

AB$\displaystyle ^2$-A=A(B-1)(B+1) is invertible, NS.
Therefore A, (B-1), and (B+!) are NS.
Therfore (BA-A)=(B-1)A is NS, invertible.

NS: Non-Singular