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 February 13th, 2016, 02:14 AM #1 Newbie   Joined: Feb 2016 From: Sweden Posts: 2 Thanks: 0 Coordinate systems, changing bases I can not get my head around this one. Note: e1, e2 does not seem to denote (1;0) and (0;1) respectively in this task, T is the transpose. _____________________________________________ Assume that we have two different bases in the plane e1=(-3;1) and e2=(1 1)T, and f1=(-3 0)T and f2=(-5 -2)T. (The coordinates of the vectors here are (of course) given in the standard basis.) (a) Determine the coordinates in the basis (e1,e2) of the vector with coordinates (3,-4)T (in the standard basis). (b) Determine the coordinates in the basis (e1,e2) of the vector with coordinates (3,-4)T in the basis (f1,f2). _____________________________________________ in a) I understand that we are looking for weights c1, c2 such that c1*(-3 1)T + c2*(1 1)T = (3,-4)T which is the same as forming a matrix A = [e1 e2 (3,-4)T] and row reducing... leading to weights c1 = -21/12 and c2 = -9/4. I dont understand the question in b) though. Can anyone tell me how to interpret the question? Last edited by wilfal; February 13th, 2016 at 02:17 AM. February 13th, 2016, 08:18 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Since you are told that "e1=(-3;1) and e2=(1 1)", the fact that you say "e1, e2 does not seem to denote (1;0) and (0;1)" indicates that you really do not understand the notation! You are told they are not! "Determine the coordinates in the basis (e1,e2) of the vector with coordinates (3,-4)T (in the standard basis)." Having coordinates (3, -4) "in the standard basis" means that the vector can be written 3(1, 0)- 4(0,1). Let "a" and "b" be the coefficients in the basis (-3,1), (1, 1). Then that vector is written a(3, -4)+ b(1, 1)= (3a, -4a)+ (b, b)= (3a+ b, -4a+ b). Since those represent the same vector, we have (3a+ b, -4a+ b)= (3, -4). And that means we must have 3a+ b= 3, -4a+ b= -4. Solve those two equations for a and b. Thanks from wilfal February 14th, 2016, 12:50 AM #3 Newbie   Joined: Feb 2016 From: Sweden Posts: 2 Thanks: 0 Thank you for your insights, your interpretation of the question was not correct either I am afraid. I simply feel this question is asked in a poorly manner. I just want to reply to the remark that I do not seem to understand the notation: I was so confused by this task that I started suspecting they had misprinted something, but I agree with you that of course e1 and e1 denotes what is given and nothing else. I finally found how to interpreted this correctly with help of Country Boy: In b) "the vector with coordinates (3,-4)T in the basis (f1,f2)" denotes the vector V which is the result of the following matrix multiplication (semicolon denotes new row): V = [f1; f2]*[3; -4] = [-3 -5; 0 -2]*[3; -4] = [11; 8] The question in b) is now the same as asking "Determine the coordinates in the basis (e1,e2) of the vector V". which is the same as solving a[e1] + b[e2] = V for a and b <=> a = -3/4, b = 35/4 Last edited by wilfal; February 14th, 2016 at 12:53 AM. February 17th, 2016, 07:37 AM   #4
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Quote:
 Originally Posted by wilfal Thank you for your insights, your interpretation of the question was not correct either I am afraid. I simply feel this question is asked in a poorly manner. I just want to reply to the remark that I do not seem to understand the notation: I was so confused by this task that I started suspecting they had misprinted something, but I agree with you that of course e1 and e1 denotes what is given and nothing else. I finally found how to interpreted this correctly with help of Country Boy: In b) "the vector with coordinates (3,-4)T in the basis (f1,f2)" denotes the vector V which is the result of the following matrix multiplication (semicolon denotes new row): V = [f1; f2]*[3; -4] = [-3 -5; 0 -2]*[3; -4] = [11; 8]
"The vector with coordinates (3, -4)T in the basis (f1, f2)" is given by 3f1- 4f2= 3(-3, 0)- 4(-5, -2)= (-9, 0)- (-20,-8 )= (11, 8 ) (in the "standard basis) just as you say.

Quote:
 The question in b) is now the same as asking "Determine the coordinates in the basis (e1,e2) of the vector V". which is the same as solving a[e1] + b[e2] = V for a and b <=> a = -3/4, b = 35/4
Yes, ae1+ be2= a(-3, 1)+ b(1, 1)= (-3a+ b, a+ b)= (11, 8 ) so -3a+ b= 11, a+ b= 8. Subtracting the second equation from the first, -4a= 3 so a= -3/4. Then b= 8- a= 8+ 3/4= 35/4. Well done! Tags bases, changing, coordinate, systems Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fazshah Linear Algebra 0 December 12th, 2014 07:17 AM Robertoo Calculus 1 June 19th, 2014 04:09 PM dtmowns Algebra 7 May 2nd, 2013 09:38 AM Ter Algebra 1 April 19th, 2012 11:09 AM jsdieorksw Linear Algebra 1 November 2nd, 2010 12:07 PM

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