How Does Squaring a Matrix Affect Its Determinant?

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Discussion Overview

The discussion revolves around the relationship between squaring a matrix and its determinant, specifically addressing the condition where A² equals the identity matrix. Participants explore the implications of this condition for the determinant of matrix A, with a focus on whether it must equal ±1. The conversation includes attempts to clarify the mathematical reasoning and definitions involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks assistance in proving that if A² = I, then det A = ±1, expressing confusion about the problem's clarity.
  • Another participant suggests that matrix arithmetic may be simpler than scalar arithmetic for this problem.
  • A participant clarifies that the theorem applies to any n x n matrix, not just 2x2, questioning the assumption about the matrix size.
  • Taking the determinant of both sides of the equation A² = I is proposed as a potential approach.
  • Some participants discuss the algebraic manipulation of the equation a² = 1, noting that while it applies to scalars, it does not directly translate to matrices.
  • A counterexample is provided to challenge a previous argument, demonstrating that a specific matrix squared results in the identity matrix while having a determinant of 0.
  • There is a discussion about the meaning of "order" in relation to matrices, with clarification that it can refer to group order and implications for invertibility.

Areas of Agreement / Disagreement

Participants express differing views on the implications of A² = I for the determinant of A, with some supporting the idea that det A must equal ±1 and others questioning this conclusion based on counterexamples. The discussion remains unresolved with multiple competing perspectives on the definitions and implications involved.

Contextual Notes

Participants note the potential confusion arising from the term "order" as it applies to matrices, indicating that definitions may vary and affect the interpretation of the problem. There is also an acknowledgment of the limitations of applying scalar reasoning to matrix operations.

chapone
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(Please help!) If A² = I, prove det A = ±1

If A² = I, show that det A = ±1

This book is very unclear, but I am assuming by "I" they mean the identity matrix with a size of 2x2. I have tried putting in A for row 1 column 1 - B for 1,2 - C for 2,1 and D for 2,2 multiplying and setting the results equal to the values of the identity matrix. I thought I was close, but now am doubting that I am going about this the right way. Any help is MUCH appreciated! Thank you!
 
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Well, in this case, doing matrix arithmetic is going to be a lot easier than doing scalar arithmetic on the entries of the matrix.
 
Welcome to PF!

chapone said:
If A² = I, show that det A = ±1

This book is very unclear, but I am assuming by "I" they mean the identity matrix with a size of 2x2. I have tried putting in A for row 1 column 1 - B for 1,2 - C for 2,1 and D for 2,2 multiplying and setting the results equal to the values of the identity matrix. I thought I was close, but now am doubting that I am going about this the right way. Any help is MUCH appreciated! Thank you!

Hi chapone! Welcome to PF! :smile:

What makes you think they mean 2x2?

This theorem is true for any n x n matrix. :smile:

Do you know any formulas for determinants (for example, for det (AB))?
 
Take the determinant of both sides of A[tex]^{2}[/tex] = I.

(Any don't make any assumptions beyond what the book gives you.)
 


a² = 1
a² - 1 = 0
(a - 1) (a + 1) = 0
a=1 and a = -1
then a = ±1
 


john the gree said:
a² = 1
a² - 1 = 0
(a - 1) (a + 1) = 0
a=1 and a = -1
then a = ±1
This is all well and good for a real number a, but the OP is working with a matrix A, not a scalar. As such, A [itex]\neq[/itex] 1.
 


chapone got 3 good answers already, so I'll just add to what Mark44 said (also a good post, but not an answer for chapone) by providing a counterexample that shows that John's argument gets the wrong result for matrices:

[tex]\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}[/tex]
 


If A^2=1, then A is a matrix of order 2, which means that A is invertible.
 
Last edited by a moderator:


vigvig said:
If A^2=1, then A is a matrix of order 2
There are multiple meanings of the term "order", and your meaning here is not the meaning usually meant for matrices.
 
Last edited:
  • #10


D H said:
There are multiple meanings of the term "order", and your meaning here is not the meaning usually meant for matrices.

What do you mean? Order in the sense of "group order", meaning A generates a group of order 2. Implying A must be invertible
 

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