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Math Help - Nullspace and basis

  1. #1
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    Nullspace and basis

    In my textbook there is a question where they ask for a basis for the 1x3 matrix:

    (10 20 10)

    Answer: an orthogonal basis is {(-1,0,1)^T,(1,-1,1)^T}.

    So, I know that the equation corresponding to the homogeneous system would be "10x+20y+10z=0", but I can't see how to solve this to get that basis. Could anyone please show me how we can get that answer?
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  2. #2
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    Could you please type up the book's statement of the problem, word-for-word? Thanks!
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  3. #3
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    Quote Originally Posted by demode View Post
    In my textbook there is a question where they ask for a basis for the 1x3 matrix:

    (10 20 10)

    Answer: an orthogonal basis is {(-1,0,1)^T,(1,-1,1)^T}.

    So, I know that the equation corresponding to the homogeneous system would be "10x+20y+10z=0", but I can't see how to solve this to get that basis. Could anyone please show me how we can get that answer?
    Do you mean "find a basis for the null space of the matrix?"

    Yes, 10x+ 20y+ 10z= 0 for any (x, y, z) in the null space. Dividing by 10, x+2y+ z= 0 which we can solve for any one of the variables in terms of the other two. For example, z= -x- 2y which tells us that (x, y, z)= (x, y, -x- 2y)= x(1, 0, -1)+ y(0, 1, -2). You should be able to see the basis for the null space from that. Of course, that's not an orthogonal basis.
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  4. #4
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    Do you mean "find a basis for the null space of the matrix?"

    Yes, 10x+ 20y+ 10z= 0 for any (x, y, z) in the null space. Dividing by 10, x+2y+ z= 0 which we can solve for any one of the variables in terms of the other two. For example, z= -x- 2y which tells us that (x, y, z)= (x, y, -x- 2y)= x(1, 0, -1)+ y(0, 1, -2). You should be able to see the basis for the null space from that. Of course, that's not an orthogonal basis.
    Yes, I meant a basis for its null space.

    Well, from that a basis would be {(1, 0, -1), (0, 1, -2)}. And I used the Gram-Shmidt process to orthogonalize them and I got {(1,0,-1), (√2, 1 -√2)}, which is wrong.

    So what do I need to do to get the right orthogonal basis? According to the answer they've given us it should be {(-1,0,1), (1,-1,1)}.

    Could you please type up the book's statement of the problem, word-for-word? Thanks!
    Well, this is part of the following optimization problem:

    Nullspace and basis-89933765.gif
    Nullspace and basis-57244772.gif
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  5. #5
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    Any subspace has an infinite number of bases and an infinite number of "orthogonal bases". The answer given is one of those. There is no way for you to know which is meant just from "find a basis".

    Of course, if (x, y, z)= a(-1,0,1)+ b(1,-1,1)= (b- a, -b, a+ b), then 10x+ 20y+ 10z= 10(b- a)+ 20(-b)+ 10(a+ b)= 10b- 10a- 20b+ 10a+ 10b= (10b- 20b+ 10b)+ (-10a+ 10a)= 0 so the given answer is a basis for the null space. And, since (-1, 0, 1)\cdot(1, -1, 1)= (-1)(1)+ 0(-1)+ (1)(1)= 0, it is an orthogonal basis.

    However, your orthogonalization is wrong. \{(1, 0, -1), (\sqrt{2}, 1, \sqrt{2})\} does NOT span this null space. In particular, (\sqrt{2}, 1,  \sqrt{2}) itself is not in the null space because 10x+ 20y+ 10z=  10\sqrt{2}+ 20+ 10\sqrt{2}\ne 0.

    The projection of (0, 1, -2) on (1, 0, -1) is \frac{(0, 1, -2)\cdot(1, 0, -1)}{|(1, 0, -1)^2}(1, 0, -1)= \frac{2}{2}(1, 0, -1)= (1, 0, -1) so the vector (0, 1, -2)- (1, 0, -1)= (-1, 1, -1) is itself orthogonal to (1, 0, -1) and in the span of (0, 1, -2) and (1, 0, -1).

    That gives {(1, 0, -1), (-1, 1, -1)} as a basis which is just the basis given except that each vector is multiplied by -1.
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