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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 $\text{det}(ABA)\neq0$.

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)(BI) so (BI)A=BAA is invertible. 
December 2nd, 2018, 08:15 PM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 562 Thanks: 325 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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: $A(B+I)(BI)=A(B^2  I) = AB^2  A \ne AB^2$ (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, $\det(AB^2  A) = \det(A (B^2  I)) = \det(A)\det(B+I)\det(BI) \ne0$ (by the determinant properties and hypothesis) which implies $\det(A),\det(B+I)$ and $\det(BI)\ne 0$. From this, $\det(BA  A) = \det((BI)A) = \det(BI)\det(A)\ne0$ (by the properties deduced above). And $\det(BA  A)\ne0$ is equivalent to $BA  A$ be invertible. 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(B1)(B+1) is invertible, NS. Therefore A, (B1), and (B+!) are NS. Therfore (BAA)=(B1)A is NS, invertible. NS: NonSingular 

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invertible, matrix, proving 
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