Intersection of Hyperplanes

Discussion in 'Physics & Math' started by oxymoron, Aug 30, 2003.

  1. oxymoron Registered Senior Member

    Messages:
    454
    If you have two hyperplanes in R^5 given as...
    H1 = (1, -2, -5, 0, 1)^T.x = 0
    H2 = (0, 1, 0, 1, -1)^T.x = 0
    (where ^T is transpose)
    Then S, which is a subspace of the 5th dimension, is simply the intersection of H1 and H2, ie. H1 n H2. I got this intuitively since the planes pass through the origin.

    1) How would you prove this?

    2) For any x the system holds as a solution to Ax = 0. But is it possible to row-reduce the system Ax = 0 to row-echelon form and use this to find a basis for S?

    Thankyou for any assistance or guidance on this matter.
     
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  3. James R Just this guy, you know? Staff Member

    Messages:
    31,341
    Putting the two hyperplanes into a matrix we have:

    (1 -2 -5 0 1)
    (0 1 0 1 -1)

    The solution space for this matrix is found by putting:

    x<sub>5</sub> = s
    x<sub>4</sub> = t
    x<sub>3</sub> = u

    Then

    x<sub>2</sub> = s - t
    x<sub>1</sub> = -s +5u +2x<sub>2</sub> = -s + 5u +2(s-t) = s - t + 5u

    So, we can write:

    <b>x<sup>T</sup></b> = s(-1,1,0,0,1) + t(-1,-1,0,1,0) + u(0,5,1,0,0)

    Thus, the intersection of the two hyperplanes is a three dimensional space spanned by the vectors (-1,1,0,0,1)<sup>T</sup>, (-1,-1,0,1,0)<sup>T</sup>, (0,5,1,0,0)<sup>T</sup>, which together form a basis for the solution space.
     
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  5. oxymoron Registered Senior Member

    Messages:
    454
    Thankyou James for your efforts. But don't go away just yet!

    Another thing, could you please explain how you came to this conclusion

    Remarks:
    You row-reduced the 2x5 matrix and obtained three vectors in R^5 namely (-1,1,0,0,1)^T, (-1,-1,0,1,0)^T and (0,5,1,0,0)^T. Since these three vectors are linearly independant (and are solutions to Ax = 0) you have proved that all you need to form a basis for the intersection of the given hyperplanes is these three vectors. Hence they form a basis for the intersection (ie. solution space). -->please tell me if this is a correct statement<--

    Also, does this tell me that the solution space has dimension 3?

    Further, how can you conclude that the solution space is a subspace of R^5. Is it that the bases are closed under vector addition and scalar multiplication, or is it more detailed than this?

    Finally, say you had the vector (5, 3, -2, 0, 1)^T how would you show that it is a solution?
     
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  7. oxymoron Registered Senior Member

    Messages:
    454
    Wait, I see.

    x1 = s - t + 5u
    x2 = s - t
    x3 = u
    x4 = t
    x5 = s

    = (x1, x2, x3, x4, x5)^T = (s-t+5u, s-t, u, t, s)^T
    = (s, s, 0, 0, s)^T + (-t, -t, 0, t, 0)^T + (5u, 0, u, 0, 0)^T

    Since x4 = t and x5 = s we can discard them since they can be constructed from x1 and x2.

    = s(1, 1, 0, 0, 1)^T + t(-1, -1, 0, 1, 0) + u(5, 0, 1, 0, 0)^T

    However slightly different to your answer.
    hmmm?
     
  8. James R Just this guy, you know? Staff Member

    Messages:
    31,341
    Oops. I made a couple of mistakes.

    The solution is, as you say:

    x<sub>1</sub> = s - t + 5u
    x<sub>2</sub> = s - t
    x<sub>3</sub> = u
    x<sub>4</sub> = t
    x<sub>5</sub> = s

    In vector form that's::

    <b>x<sup>T</sup></b> = s(1,1,0,0,1) + t(-1,-1,0,1,0) + u(5,0,1,0,0)

    which agrees with your answer.

    <i>You row-reduced the 2x5 matrix and obtained three vectors in R^5 ...Since these three vectors are linearly independant (and are solutions to Ax = 0) you have proved that all you need to form a basis for the intersection of the given hyperplanes is these three vectors. Hence they form a basis for the intersection (ie. solution space). -->please tell me if this is a correct statement<--</i>

    Yes, that's right.

    <i>Also, does this tell me that the solution space has dimension 3?</i>

    Yes. The number of basis vectors for a space is always equal to its dimension.

    <i>Further, how can you conclude that the solution space is a subspace of R^5. Is it that the bases are closed under vector addition and scalar multiplication, or is it more detailed than this?</i>

    It's no more complicated than that.

    <i>Finally, say you had the vector (5, 3, -2, 0, 1)^T how would you show that it is a solution?</i>

    Simple. Plug it into the original equations.

    (1 -2 -5 0 1).(5 3 -2 0 1) = 0 ????

    Doesn't work, so (5 3 -2 0 1) isn't a solution.
     

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