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February 13th, 2016, 02:14 AM   #1
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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.
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February 13th, 2016, 08:18 AM   #2
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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.
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February 14th, 2016, 12:50 AM   #3
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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.
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February 17th, 2016, 07:37 AM   #4
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Quote:
Originally Posted by wilfal View Post
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!
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