Is this a complete test to show that a matrix is orthogonal?

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Discussion Overview

The discussion revolves around the criteria for determining whether a matrix is orthogonal, specifically examining the relationships between the matrix and its transpose, as well as the implications of certain matrix products equating to the identity matrix. The scope includes theoretical aspects of linear algebra and definitions related to matrix properties.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose using the relationship \( M^T = M^{-1} \) as a definition of orthogonality, suggesting that proving \( MM^T = I \) could be an alternative method.
  • Others question whether it is possible for a matrix \( M \) to satisfy \( MM^T = I \) while not satisfying \( M^TM = I \), indicating a potential gap in understanding the implications of these conditions.
  • A later reply clarifies that \( MM^T = I \) does not necessarily imply \( M^TM = I \), introducing the concept of left-invertible and right-invertible matrices as relevant to the discussion.
  • One participant mentions that the result regarding left and right inverses is a significant concept in linear algebra, emphasizing its special nature for matrices.
  • Another participant expands on the implications of right inverses in algebraic structures, suggesting broader contexts where similar properties hold.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the relationships between matrix products and orthogonality, indicating that the discussion remains unresolved regarding the completeness of the proposed tests for orthogonality.

Contextual Notes

The discussion highlights the complexity of matrix properties and the conditions under which certain equalities hold, without reaching a consensus on the sufficiency of the proposed methods for proving orthogonality.

tamtam402
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I used to test orthogonality by using the definition MT = M-1, which means I always calculated the inverse of the matrices. However, isn't it true that if M is orthogonal, then MMT = I?

If we multiply both side by M-1, we get MT = M-1.

Can I use this to proof the orthogonality of a matrix M, instead of calculating it's (often tedious) inverse?
 
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tamtam402 said:
I used to test orthogonality by using the definition MT = M-1, which means I always calculated the inverse of the matrices. However, isn't it true that if M is orthogonal, then MMT = I?

If we multiply both side by M-1, we get MT = M-1.

Can I use this to proof the orthogonality of a matrix M, instead of calculating it's (often tedious) inverse?



Of course...it's exactly the same, right?!

DonAntonio
 
DonAntonio said:
Of course...it's exactly the same, right?!

DonAntonio

This might be a dumb question, but would it be possible to have a matrix M where MMT = I, yet MTM ≠ I? I'm pretty sure that by definition, proving either of these 2 equalities is enough to know that M-1 = MT, but I wanted to make sure.
 
tamtam402 said:
This might be a dumb question, but would it be possible to have a matrix M where MMT = I, yet MTM ≠ I? I'm pretty sure that by definition, proving either of these 2 equalities is enough to know that M-1 = MT, but I wanted to make sure.

That is actually a very good question. There is no reason why MM^T=I should imply M^TM=I.
But this is actually quite a deep result in linear algebra. The result is that every left-invertible matrix is invertible. And likewise with right-invertible.
So if AB=I, then it holds that BA=I. This is a special result that only holds for matrices.
 
micromass said:
That is actually a very good question. There is no reason why MM^T=I should imply M^TM=I.
But this is actually quite a deep result in linear algebra. The result is that every left-invertible matrix is invertible. And likewise with right-invertible.
So if AB=I, then it holds that BA=I. This is a special result that only holds for matrices.


Well, that holds in any group and in any monoid where every element has a right inverse, since then ( with \,a'=\, right inverse of \,a ):
a'a=a'a(a'a'')=a'(aa')a''=a'ea''=a'a''=e
so the right inverse is also the left one.

DonAntonio
 

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