August 30th, 2012, 06:56 AM  #1 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  Relation
 The relation for 2 \times 2 matrices such that A R B if and only if there exists a nonsingular matrix H such that AH=HB. Show that R is an equivalence relation. Given A is singular show that B is also singular. let R be a realtion on the set Z such that aRb iff for a,b is an element of integers. i) is it reflexive? ii) is it transitive? iii) is it symmetric? can anyone help me solve this? thank you 
August 30th, 2012, 09:53 AM  #2 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  Re: Relation For the matrix problem: To prove R is an equivalence relation, we must show that R is reflexive, symmetric, and transitive. Proof of reflexive: AI = IA for all matrices A, where I is the identity matrix, and I is nonsingular, so for all A, A R A. Proof of symmetric: If A R B, then there exists a nonsingular matrix H such that AH = HB. Now, since H is nonsingular, it has an inverse G. Multiplying the equation AH = HB on the left and on the right by G, we obtain: GAHG = GHBG, which simplifies to GA = BG, or BG = GA. Since G is nonsingular, this statement is equivalent to saying B R A. Proof of transitive: If A R B and B R C, then there exists a nonsingular matrix F such that (1) AF = FB and a nonsingular matrix G such that (2) BG = GC. Multiplying (1) on the right by G, we obtain AFG = FBG. Multiplying (2) on the left by F, we obtain FBG = FGC. Combining these two resultant equations yields AFG = FGC. Now, since both F and G are nonsingular, their product H is nonsingular, and we have AH = HC for some nonsingular matrix H; thus, A R C. Note that if A is singular and there exists a nonsingular matrix H such that AH = HB, B is singular. Suppose B is nonsingular. The product AH is a singular matrix, which means HB is a singular matrix, but the product of two nonsingular matrices is another nonsingular matrix, and a matrix cannot be both singular and nonsingular, so the assumption that B is nonsingular must be wrong. 
August 31st, 2012, 06:18 AM  #3  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Relation Quote:
Quote:
(i) (ii) is not true 0 times any integer is 0 so, for example 5R0 and 1R0 but 1R0 is NOT true. (iii) if then  
August 31st, 2012, 06:51 AM  #4 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Relation
I think what HallsofIvy means for (ii) is: 5R0 and 0R1 However, 5 is not related to 1 

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