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KonradZuse

  • 3 years ago

Find a basis for the null space of A.

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  1. KonradZuse
    • 3 years ago
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    My book says that the null space is the solution space of Ax = b when b = 0. The example in the book shows this...

  2. KonradZuse
    • 3 years ago
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    @TuringTest @asnaseer @CliffSedge

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  3. KonradZuse
    • 3 years ago
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    Do I just solve it and that's all...?

  4. KonradZuse
    • 3 years ago
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    I'm also a bit confused how they got the column matrix's filled out... Maybe that's my issue...

  5. TuringTest
    • 3 years ago
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    pretty much, yeah just solve it and rewrite each variable as a new vector, i.e. one for r, one for s, and one for t

  6. KonradZuse
    • 3 years ago
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    hmm okay maybe this is easier than I'm thinking let me try it. My example is A = (Matrix(3, 4, {(1, 1) = 1, (1, 2) = 4, (1, 3) = 5, (1, 4) = 2, (2, 1) = 2, (2, 2) = 1, (2, 3) = 3, (2, 4) = 0, (3, 1) = -1, (3, 2) = 3, (3, 3) = 2, (3, 4) = 2})); print(`output redirected...`); # input placeholder

  7. KonradZuse
    • 3 years ago
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    woah... LOL

  8. KonradZuse
    • 3 years ago
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    1 4 5 2 2 1 3 0 -1 3 2 2

  9. KonradZuse
    • 3 years ago
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    now do I have to augment it with the 0 column-vector?

  10. TuringTest
    • 3 years ago
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    go ahead and row reduce it as much as you can brb, gotta go to get dinner

  11. KonradZuse
    • 3 years ago
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    the book does 2 eq's which is really annoying....

  12. KonradZuse
    • 3 years ago
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    how they would both be the same, so it seems like it's not an augmented matrix....

  13. TuringTest
    • 3 years ago
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    row reduce, what do you get?

  14. KonradZuse
    • 3 years ago
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    OOOOOOOOOOOOOOOo There's a Null Space/Nullity thing on Maple hehehe. I guess it isn't augmented then :D

  15. TuringTest
    • 3 years ago
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    no, it's not

  16. KonradZuse
    • 3 years ago
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    1 0 1 -2/7 0 1 1 4/7 0 0 0 0

  17. KonradZuse
    • 3 years ago
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    now I actually found something that says the num space is the subspace of vectors x satisfying A. x = 0

  18. KonradZuse
    • 3 years ago
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    it shows the null space as 2/7 -4/7 0 1 then another column matrix next to it. -1 -1 1 0..

  19. KonradZuse
    • 3 years ago
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    Is there a matrix thing in the equations..... This is stupid writing it out like this :P.

  20. KonradZuse
    • 3 years ago
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    oh I found it.... :)

  21. TuringTest
    • 3 years ago
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    setting up matrices with latex is a pain, you are doing it the easy way

  22. KonradZuse
    • 3 years ago
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    HJHAHAH

  23. KonradZuse
    • 3 years ago
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    \[\left(\begin{matrix}2/7 \\-4/7\\ 0\\ 1\end{matrix}\right), \left(\begin{matrix}-1 \\-1\\ 1\\ 0\end{matrix}\right)\]

  24. TuringTest
    • 3 years ago
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    yes, you get that from 1 0 1 -2/7 x1+r-2/7s=0->x1=-r+2/7s 0 1 1 4/7 x2+r+4/7s=0->x2=-r-4/7s collect the r's in one vector and the s's in another

  25. KonradZuse
    • 3 years ago
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    my eyes they burn.... :)

  26. KonradZuse
    • 3 years ago
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    you're so smart TT :Pe

  27. TuringTest
    • 3 years ago
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    first row of your matrix reduced: 1 0 1 -2/7 this is the same as the equation x1+r-2/7s=0 which leads to x1=-r+2/7s and thanks, but I need to get better at computer stuff now, so maybe I'll need your help sorry, back in 15

  28. KonradZuse
    • 3 years ago
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    yeah yeah yeah!

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