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anonymous
 4 years ago
Greetings, This question deals with linear algebra so it sucks here it is!
3a. Given the vector x and the basis B = [b1, b2, b3] of a subpace of R4 determine if x is in the span of B, and if so find the coordinates of x with respect to B and give the coordinate vector [x]B.
x:[1,1,1,1] b1: [1,0,2,0] b2:[0,1,3,0] b3:[0,0,4,1]
b. Given the basis C = [c1, c2, c3], determine if C is in the span of B. If so write the vector x using the C coordinates. and give the coordinate vector [x]C.
c1:[1,2,4,3] c2:[1,3,3,2] c3:[1,3,9,4]
anonymous
 4 years ago
Greetings, This question deals with linear algebra so it sucks here it is! 3a. Given the vector x and the basis B = [b1, b2, b3] of a subpace of R4 determine if x is in the span of B, and if so find the coordinates of x with respect to B and give the coordinate vector [x]B. x:[1,1,1,1] b1: [1,0,2,0] b2:[0,1,3,0] b3:[0,0,4,1] b. Given the basis C = [c1, c2, c3], determine if C is in the span of B. If so write the vector x using the C coordinates. and give the coordinate vector [x]C. c1:[1,2,4,3] c2:[1,3,3,2] c3:[1,3,9,4]

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0c. Find the change of coordinates matrix P from B to C. [x]C = P [x]B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There's something I don't quite understand. A basis for a 4 dimensional space should contain 4 elements.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 4th element i believe is x:[1,1,1,1] the instructor i have now is completely horrible and doesn't even teach any thing.. so i could be wrong lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The question doesn't make sense unless you have a basis for the vector space. A basis must have the same cardinality as the dimension of the vector space.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh and furthermore to ask of a vector is in the span of a basis is nonsensical. Of course any vector in the vector space can be expressed as a linear combination of a basis.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In short please make sure you copied the question correctly, as is it doesn't make sense.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is the question my professor stated word for word unfortunately and i have a hard enough time understanding him as it is lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In any case to check if a vector is in the span of that set, you can solve a system of equations.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You would use row reduction to verify if it is in the span.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Part a: Vector x is in the span of B if and only if it is a linear combination of the vectors in B. Then there are scalars \[x_{1} , x_{2} , x_{3}\] such that \[x_{1} b _{1} + x_{2} b_{2} + x_{3} b_{3} = x,\]or \[x_1 (1,0,2,0) + x_2 (0,1,3,0) + x_3 (0,0,4,1) = (1,1,1,1)\] a system of four simultaneous linear equations in unknowns \[x_1, x_2, x_3,\]which has the unique solution \[(x_1,x_2,x_3) = (1,1,1),\] so the vector x can be expressed as a linear combination if the vectors in B, and x is in the span of b. The coordinate vector [x]B is [1,1,1].

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0To setup this system of equations, start with the vectors in the set B as the columns of a matrix. You "augment" the matrix with the vector x.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just curious sorry if this question is lame but how did you acquire the results for x1, x2, x3 come from?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By solving the system of equations. You are essentially seeing if there exists a linear combination of the vectors in B such that the result of the linear combination is x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0To rephrase this you are seeing if x is in the image of A consisting of columns of B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do this using row reduction. I will demonstrate.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright, by the way abtrehearn thanks for the answer to part a it looks good

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Start with the augmented matrix: \[\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&1 \\ 2&3&4&1\\0&0&1&1\end {array} \right] \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now row reduce the matrix.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This would be the row reduced form? 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ sir alchemista please solve my liinear algebra questions

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, now do you see that the results of the row reduction yield the coefficients for the linear combination that yield x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh yah just pops right up there aha alrighty thanks to you both ill retype part be and c as a new question XD

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But its not magic, you should understand the theory behind it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yah just didnt realize it before until i was actually looking at the complete form, just something i over looked i guess aha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Right on, DevinBlade :^)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@sir abtrehearn please see my linear algebra questions i posted

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks very much abtrehearn, im reasking the question for part b so answer it on there i think this question is in deserves of multiple medals :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0last one is here http://openstudy.com/users/devinblade/updates/4e1d5ab10b8b4841c1a9b0d9* you should type the answer there so i can mark it answered but answering it here is fine too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0<<b. Given the basis C = [c1, c2, c3], determine if C is in the span of B. If so write the vector x using the C coordinates. and give the coordinate vector [x]C. c1:[1,2,4,3] c2:[1,3,3,2] c3:[1,3,9,4] >> C is in the span of B if and only if each vector in C can be expressed as a linear combination of the vectors in B. So we seek scalars \[x_1, x_2, x_3\]such that \[x_1 b_1 + x_2 b_2 + x_3 b_3 = c_1,\]scalars\[y_1, y_2, y_3\]such that\[y_1b_1 + y_2 b_2 + y_3 b_3 = c_2,\]and scalars \[z_1, z_2, z_3 \]such that\[z_1b_1+z_2b_2+z_3b_3=c_3.\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep, there is already a solution in the new thread.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/groups/mathematics/updates/4e1d5ab10b8b4841c1a9b0d9
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