Connecting linear algebra concepts to groups

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Discussion Overview

The discussion revolves around the connections between linear algebra concepts and group theory, specifically focusing on properties related to determinants and group homomorphisms. Participants explore various mathematical statements and their implications in the context of invertible matrices and group structures.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests focusing on the determinant being non-zero due to invertibility, leading to a consideration of the statement "det(AB) = det(A)det(B)" as a potential answer.
  • Another participant expresses a strong inclination towards the determinant property but acknowledges the need for verification.
  • There is a clarification from a participant that they did not assert certainty about the determinant property but rather indicated a tendency to check it.
  • A later post introduces the definition of a group homomorphism as a critical aspect of the discussion, linking it to the determinant property.

Areas of Agreement / Disagreement

Participants show some agreement on the relevance of the determinant property, but there remains uncertainty about its correctness and the need for further verification. Disagreement exists regarding the confidence in selecting this option as the answer.

Contextual Notes

Participants express uncertainty about the correctness of the statements and the implications of group homomorphisms, highlighting the need for careful consideration of definitions and properties.

Who May Find This Useful

Readers interested in the interplay between linear algebra and group theory, particularly those exploring properties of determinants and group homomorphisms.

lemonthree
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The options are
$$rank(B)+null(B)=n$$
$$tr(ABA^{−1})=tr(B)$$
$$det(AB)=det(A)det(B)$$

I'm thinking that since it's invertible, I would focus on the determinant =/= 0. I believe the first option is out, because null (B) would be 0 which won't be helpful. The second option makes the point that $$AA^{−1}$$ is $$I$$, so it's suggesting invertibility. So I'm deciding between the second and last option. Does anyone have any tips?
 
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Since the problem specifically refers to "the map defined by taking determinants", I would almost automatically check "det(AB)= det(A)det(B)"! "Almost" because, of course, I would want to make sure it was correct.
 
Country Boy said:
Since the problem specifically refers to "the map defined by taking determinants", I would almost automatically check "det(AB)= det(A)det(B)"! "Almost" because, of course, I would want to make sure it was correct.

Agree, I was tempted to select det(AB)= det(A)det(B) but then again I can't be fully sure.

I guess since we both believe it to be correct, I guess I'm going for this answer then!
 
I didn't say I believe it to be correct! I said I would "almost" automatically check it. And I said "of course, I would want to make sure it was correct". Why can't you "be fully sure"?
 
Country Boy said:
I didn't say I believe it to be correct! I said I would "almost" automatically check it. And I said "of course, I would want to make sure it was correct". Why can't you "be fully sure"?

Good point! My gut feeling just told me that the determinant is the answer but my brain told me to be careful.
 
Let's take a closer look, and let's start with the crucial bit of information that we need: what is the definition of a group homomorphism?
 
Hi Klaas, indeed, it is det(AB) = det(A)det(B) because as per the definition of group homomorphism, φ (G) = H if φ (xy) = φ (x) φ (y)
 

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