General Linear Group: A belongs to Rn×n, the real n×n matrices,...

Naruto

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General Linear Group: A belongs to Rn×n, the real n×n matrices,...

If [FONT=MathJax_Math]A belongs to [FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Math]n[/FONT], the real [FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Math]n[/FONT] matrices, and consider the sets

[/FONT]
[FONT=MathJax_Math]S[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]{[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Main]GL[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]}[/FONT][FONT=MathJax_Math]
S
[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]{[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Main]GL[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]}[/FONT][FONT=MathJax_Math]
S
[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]{[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Main]GL[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]}[/FONT][FONT=MathJax_Math]
S
[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]{[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Main]GL[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_AMS]Q[/FONT][FONT=MathJax_Main])
[/FONT][FONT=MathJax_Math]S[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]{[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]∣[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Main]GL[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]}[/FONT]
Which of these sets, if any, is actually a subspace of [FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Math]n[/FONT]?

I have tried assuming A & T to be a 2*2 Matrix, and verifiy that the matrices {TA. TAinv(T) etc} are invertible. I believe
None of these sets are actually subspaces of [FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Math]n[/FONT], unless [FONT=MathJax_Math]A[/FONT]is chosen to be the zero matrix. Notably, any subspace must contain the zero-element, which in this case is the matrix whose entries are all zero. It doesn't make sense to talk about the "subspaces of [FONT=MathJax_Math]G[/FONT][FONT=MathJax_Math]L[/FONT]", since [FONT=MathJax_Math]G[/FONT][FONT=MathJax_Math]L[/FONT] fails to be a vector space.
 
Last edited:
If A belongs to R dimension(n×n) Matrix and consider the sets

S1 = {TA | T belongs to Gln(R)}
S2 = {TAInv(T) | T belongs to Gln(R)}
S3 = {TA | T belongs to Gln(Q)}
S4 = {TAInv(T) | T belongs to Gln(Q)}
S5 = {AInv(T)| T belongs to Gln(R)}
Which of these sets, if any, is actually a subspace of Rn×n? Justify your answer.
Where Inv(T) is Inverse Of MATRIX T
What are your thoughts?

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Satisfies Matrix Multipication

Hi
So primarily, for all S1 to S6, assuming the A & T to be a 2*2 Matrix, I have tested the conditions for final matrix (TA or Tinv(A) etc) satisfies the General Linear Group Property for invertibility under usual Matrix Multipication.
 
What I believe

If [FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Math]M[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main])[/FONT], then [FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Math]M[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main])[/FONT] for [FONT=MathJax_Math]T[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_Math]G[/FONT][FONT=MathJax_Math]L[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]R[/FONT][FONT=MathJax_Main])[/FONT]. This seems obvious.
I have tried assuming A & T to be a 2*2 Matrix, and verifiy that the matrices {TA. TAinv(T) etc} are invertible.


 
My thoughts

None of these sets are actually subspaces of [FONT=MathJax_AMS]R[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]×[/FONT][FONT=MathJax_Math]n[/FONT], unless [FONT=MathJax_Math]A[/FONT]is chosen to be the zero matrix. Notably, any subspace must contain the zero-element, which in this case is the matrix whose entries are all zero.
It doesn't make sense to talk about the "subspaces of [FONT=MathJax_Math]G[/FONT][FONT=MathJax_Math]L[/FONT]", since [FONT=MathJax_Math]G[/FONT][FONT=MathJax_Math]L[/FONT] fails to be a vector space.
 
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