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 December 2nd, 2018, 09:59 AM #1 Newbie   Joined: Dec 2018 From: a Posts: 2 Thanks: 0 Proving a matrix invertible 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? Last edited by o1269652; December 2nd, 2018 at 10:21 AM.
 December 2nd, 2018, 10:42 AM #2 Member   Joined: Jan 2016 From: Athens, OH Posts: 92 Thanks: 47 Use the fact that a matrix X is invertible if and only if $\text{det}X\neq0$. Then show that $\text{det}(AB-A)\neq0$.
 December 2nd, 2018, 01:09 PM #3 Newbie   Joined: Dec 2018 From: a Posts: 2 Thanks: 0 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.
December 2nd, 2018, 08:15 PM   #4
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Quote:
 Originally Posted by johng40 Use the fact that a matrix X is invertible if and only if $\text{det}X\neq0$. Then show that $\text{det}(AB-A)\neq0$.
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.

 December 19th, 2018, 05:26 AM #5 Member   Joined: May 2013 Posts: 31 Thanks: 3 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. Thanks from topsquark Last edited by Prokhartchin; December 19th, 2018 at 05:28 AM.
 January 15th, 2019, 12:11 PM #6 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 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

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