Linear Alg.: Proving $\left(A^2\right)^{-1}\,=\,\left(A^{-1}\right)^2$

Click For Summary

Homework Help Overview

The discussion revolves around proving the relationship between the inverse of a squared matrix and the square of the inverse of that matrix, specifically addressing the equation \((A^2)^{-1} = (A^{-1})^2\). The context is linear algebra, focusing on matrix operations and properties of inverses.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the invertibility of \(A^2\) and question how to formally prove the stated relationship. Some suggest starting with the definition of the inverse and manipulating the expression \((A^2)^{-1} = (A A)^{-1}\). Others express uncertainty about the proof process and seek guidance on methods for proving mathematical statements.

Discussion Status

The discussion includes attempts to clarify the proof and explore different perspectives on the problem. Some participants provide insights into the nature of inverses and the simplicity of the proof, while others express confusion about the abstract nature of proofs and seek more structured guidance.

Contextual Notes

There are mentions of potential typographical errors and concerns about mixing variable notation, which may affect clarity. Additionally, some participants highlight the challenge of reasoning about abstract concepts in proofs.

VinnyCee
Messages
486
Reaction score
0
Given that [itex]A^2[/itex] is invertible, does it neccesarily mean that

[tex]\left(A^2\right)^{-1}\,=\,\left(A^{-1}\right)^2[/tex]?



I know that this is true, but I have no idea on where to even start a proof of this!

Maybe:

[tex]\left(A^2\right)^{-1}\,=\,\left(A\,A\right)^{-1}[/tex]

But how would I operate on infinite matrices (i.e. - [itex]a_{i\,j}[/itex])?
 
Last edited:
Physics news on Phys.org
[tex](A^2)^{-1}=(AA)^{-1}=A^{-1}A^{-1}=(A^{-1})^2[/tex]

This proves that the inverse of the square of the matrix is equal to the square of the inverse of the matrix.
I'm not sure if that's sufficient, because I'm not exactly sure what the little a is supposed to be.
 
Last edited:
X is the in inverse of Y if and only if XY=YX=1.
 
VinnyCee said:
Given that [itex]A^2[/itex] is invertible, does it neccesarily mean that

[tex]\left(A^2\right)^{-1}\,=\,\left(a^{-1}\right)^2[/tex]?



I know that this is true, but I have no idea on where to even start a proof of this!

Maybe:

[tex]\left(A^2\right)^{-1}\,=\,\left(A\,A\right)^{-1}[/tex]

But how would I operate on infinite matrices (i.e. - [itex]a_{i\,j}[/itex])?

Please do NOT mix "A" and "a"! matt grimes' suggestion is exactly what you need. Use the definition of inverse.
 
Sorry, a little typo I made!

It's fixed now.

How to prove something? Is there am algorithm for proofs? The concept is too abstract for me to reason about it or start a problem!

Can someone list the different methods for "proving" something? OIr maybe provide a good link to this type of information?
 
VinnyCee said:
How to prove something?

that would depend on the 'something'

[quote[Is there am algorithm for proofs?[/quote]

no, otherwise it would be a pointless subject.


The concept is too abstract for me to reason about it or start a problem!

Can someone list the different methods for "proving" something?

There is no such list. But in this case it is trivial. You only need to show something satisfies some rule, and that is trivially easy to do in this case.
 

Similar threads

Replies
5
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
12
Views
2K
Replies
1
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
Replies
0
Views
1K