My Math Forum  

Go Back   My Math Forum > College Math Forum > Linear Algebra

Linear Algebra Linear Algebra Math Forum


Thanks Tree2Thanks
  • 1 Post By SDK
  • 1 Post By Prokhartchin
Reply
 
LinkBack Thread Tools Display Modes
December 2nd, 2018, 08:59 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 09:21 AM.
o1269652 is offline  
 
December 2nd, 2018, 09:42 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 $\text{det}(AB-A)\neq0$.
johng40 is offline  
December 2nd, 2018, 12:09 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.
o1269652 is offline  
December 2nd, 2018, 07:15 PM   #4
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 598
Thanks: 366

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by johng40 View Post
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.
Thanks from topsquark
SDK is offline  
December 19th, 2018, 04:26 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 04:28 AM.
Prokhartchin is offline  
January 15th, 2019, 11:11 AM   #6
Banned Camp
 
Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 125

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
zylo is offline  
Reply

  My Math Forum > College Math Forum > Linear Algebra

Tags
invertible, matrix, proving



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
square of invertible matrix PrototypePHX Linear Algebra 6 January 25th, 2014 11:36 PM
Why Isn't This Matrix Invertible? Magnesium Linear Algebra 2 December 11th, 2013 02:09 AM
invertible matrix shine123 Linear Algebra 1 September 21st, 2012 08:47 AM
Invertible matrix problem Linear Algebra 3 August 31st, 2011 05:30 AM
Is this simple 4x4 matrix invertible? Thanks! Victorious Linear Algebra 1 January 5th, 2009 03:54 PM





Copyright © 2019 My Math Forum. All rights reserved.