How Do I Verify Identity and Closure in a Matrix Subgroup?

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SUMMARY

The discussion focuses on verifying identity and closure within a specific matrix subgroup defined by matrices of the form [a, b; 0, c] where a and c are non-zero. Participants confirm that the identity matrix is [1, 0; 0, 1] and discuss the conditions under which closure is satisfied. Specifically, they emphasize that the determinant, given by ac, must not equal zero for inverses to exist. The conversation highlights the importance of ensuring that the lower left entry of the product matrix remains zero to maintain subgroup properties.

PREREQUISITES
  • Understanding of matrix multiplication and properties
  • Knowledge of identity matrices and their characteristics
  • Familiarity with determinants and their role in matrix inverses
  • Basic concepts of group theory as applied to matrix groups
NEXT STEPS
  • Study the properties of matrix groups, specifically the conditions for closure
  • Learn about determinants and their implications for matrix invertibility
  • Explore examples of identity matrices in various matrix forms
  • Investigate subgroup criteria in linear algebra and group theory
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Mathematicians, students of linear algebra, and anyone studying group theory in the context of matrices will benefit from this discussion.

elle
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Subgroup help please! Urgent!

Hi,
Can anyone help me with the following question? Many thanks! Again I'm working from a book so I'm must trying to apply theory to exercises -__-

http://i12.tinypic.com/3zsyuf8.jpg

How do I show that the identity lies in this? And also show that we have closure? Do I just choose any values of a, b and c and multiply two matrices? :confused:
 
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elle,

What conditions do you need on a, b, and c in order to make the identity? Are they within the constraints of this set? I think so. Just choose those values of a,b,c.

In order to show inverses exist for every element of the set, you know something regarding determinants that guarantees inverses.

Choose two exeplary elements from the matrix, ie a matrix with values a,b,c and another with d,e,f. Multiply them, and see if you get another matrix of the same form.
 
Okay, sorry I can't put this in math format but hopefully you understand what I mean :rolleyes:

In order for it to fulfill identity, I've put in ac not to be equal to 0 if a and c are not equal to 0..is that right?

And if say i take the given matrix in the question and another matrix which has elements d, e and f, lie in K, then when multiplied you get the products ad, ae+bf, cf in the new matrix? And ad, ae+bf and cf are not equal to 0 hence we have closure? :confused:
 
What is the identity matrix? Does it satisfy the rule that tells you which matrices belong to that set?
 
Right I can't type the identity matrix up in its correct format but its [ 1, 0, (first row) 0,1 (second row) ]. So it does satisfy the rule? :confused:

Is my evaluation on closure incorrect?
 
Why are you asking a question about the identity. What are the two rules it, the identity, must satisfy to be in the set. Are they both satisfied? Yes.

You composition one is not correct. You say that since ad, cf and ae+bf are not zero that it is in the group. What the upper right hand entry is is immaterial. All that matters is that the lower left is zero, and the the determinant (which is ac, for the matrix in post 1) is not zero.
 
The identity is
\left(\begin{array}{cc}1 && 0\\0 && 1\end{array}\right)
Your general matrix is
\left(\begin{array}{cc}a && b\\0 && c\end{array}\right)
with a and c not 0. What should a, b, c be to give the identity matrix?
 

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