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DevinBlade

  • 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]

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  1. DevinBlade
    • 4 years ago
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    c. Find the change of coordinates matrix P from B to C. [x]C = P [x]B

  2. Alchemista
    • 4 years ago
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    There's something I don't quite understand. A basis for a 4 dimensional space should contain 4 elements.

  3. DevinBlade
    • 4 years ago
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    the 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

  4. Alchemista
    • 4 years ago
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    The 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.

  5. Alchemista
    • 4 years ago
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    Oh 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.

  6. Alchemista
    • 4 years ago
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    In short please make sure you copied the question correctly, as is it doesn't make sense.

  7. DevinBlade
    • 4 years ago
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    this is the question my professor stated word for word unfortunately and i have a hard enough time understanding him as it is lol

  8. Alchemista
    • 4 years ago
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    In any case to check if a vector is in the span of that set, you can solve a system of equations.

  9. Alchemista
    • 4 years ago
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    You would use row reduction to verify if it is in the span.

  10. abtrehearn
    • 4 years ago
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    Part 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].

  11. Alchemista
    • 4 years ago
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    To 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.

  12. DevinBlade
    • 4 years ago
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    just curious sorry if this question is lame but how did you acquire the results for x1, x2, x3 come from?

  13. Alchemista
    • 4 years ago
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    By 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

  14. Alchemista
    • 4 years ago
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    To rephrase this you are seeing if x is in the image of A consisting of columns of B

  15. Alchemista
    • 4 years ago
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    Do this using row reduction. I will demonstrate.

  16. DevinBlade
    • 4 years ago
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    alright, by the way abtrehearn thanks for the answer to part a it looks good

  17. Alchemista
    • 4 years ago
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    Start with the augmented matrix: \[\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&1 \\ 2&3&4&1\\0&0&1&-1\end {array} \right] \]

  18. Alchemista
    • 4 years ago
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    Now row reduce the matrix.

  19. DevinBlade
    • 4 years ago
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    This would be the row reduced form? 1 0 0 1 0 1 0 1 0 0 1 -1 0 0 0 0

  20. luck
    • 4 years ago
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    @ sir alchemista please solve my liinear algebra questions

  21. Alchemista
    • 4 years ago
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    Yes, now do you see that the results of the row reduction yield the coefficients for the linear combination that yield x?

  22. DevinBlade
    • 4 years ago
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    ahh yah just pops right up there aha alrighty thanks to you both ill retype part be and c as a new question XD

  23. Alchemista
    • 4 years ago
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    But its not magic, you should understand the theory behind it.

  24. DevinBlade
    • 4 years ago
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    yah just didnt realize it before until i was actually looking at the complete form, just something i over looked i guess aha

  25. abtrehearn
    • 4 years ago
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    Right on, DevinBlade :^)

  26. luck
    • 4 years ago
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    @sir abtrehearn please see my linear algebra questions i posted

  27. abtrehearn
    • 4 years ago
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    Let me see...

  28. DevinBlade
    • 4 years ago
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    thanks 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

  29. DevinBlade
    • 4 years ago
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    last 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.

  30. abtrehearn
    • 4 years ago
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    <<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\[z_1b_1+z_2b_2+z_3b_3=c_3.\]

  31. Alchemista
    • 4 years ago
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    Yep, there is already a solution in the new thread.

  32. Alchemista
    • 4 years ago
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    http://openstudy.com/groups/mathematics/updates/4e1d5ab10b8b4841c1a9b0d9

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