My Math Forum Relation
 User Name Remember Me? Password

 Applied Math Applied Math Forum

 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 non-singular 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 $ab\ge 0$ 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:
 Originally Posted by gaussrelatz - The relation for 2 \times 2 matrices such that A R B if and only if there exists a non-singular matrix H such that AH=HB. Show that R is an equivalence relation.
icemanfan did this

Quote:
 Given A is singular show that B is also singular. If AH= HB then det(AH)= det(A)det(H)= det(HB)= det(H)det(B). Since H is non-singular its determinant is non-zero and so det(A)= det(B). If A is singular, its determinant is 0 so det(B)= 0 and B is singular. [quote:2ih0qu5k]-let R be a realtion on the set Z such that aRb iff $ab\ge 0$ 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
[/quote:2ih0qu5k]
(i) $aa= a^2 \ge 0$
(ii) is not true- 0 times any integer is 0 so, for example 5R0 and -1R0 but -1R0 is NOT true.
(iii) if $ab\ge 0$ then $ba\ge 0$

 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 0R-1 However, 5 is not related to -1

 Tags relation

### show that (ArB)'=A^B in matrix

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post RobRoc Algebra 1 November 2nd, 2011 06:46 AM jaredbeach Algebra 3 August 21st, 2011 12:18 PM tbone1209 Algebra 1 January 10th, 2010 03:47 PM jaredbeach Calculus 0 December 31st, 1969 04:00 PM gaussrelatz Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top