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Linear Algebra - Find matrix A such that Ax is perpendicular to [1 2 3] for all x

  1. Feb 12, 2012 #1
    Find a nonzero 3x3 matrix A such that Ax is perpendicular to [1,2,3] for all x in R3

    A vector perpendicular to [1, 2, 3] could be, say, [-2, 1, 0]. But what would be a general approach to finding a solution that would satisfy all x? Thanks!
     
  2. jcsd
  3. Feb 12, 2012 #2
    what definition of perpendicular are you using?
     
  4. Feb 12, 2012 #3

    SammyS

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    [-2, 1, 0] is perpendicular to [1, 2, 3]. However there are many others which are perpendicular to [1, 2, 3] and are also linearly independent of [-2, 1, 0] .

    I suppose you need to take the scalar product of Ax and [-2, 1, 0] and set it to zero.
     
  5. Feb 13, 2012 #4

    Ray Vickson

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    If [itex] \vec{x} = (x_1, x_2, \ldots, x_n)^T [/itex] and A has columns [itex] \vec{v}_1, \ldots, \vec{v}_n, [/itex] then
    [tex] A \vec{x} = x_1 \vec{v}_1 + x_2 \vec{v}_2 + \cdots + x_n \vec{v}_n. [/tex] Since you want this to be perpendicular to [1,2,3] for all x, all the vectors [itex]\vec{v}_i[/itex] must be perpendicular to [1,2,3].

    PS: Does anyone know how to get the "\boldmath" command to work in the version of tex used here?

    RGV
     
  6. Feb 13, 2012 #5

    Deveno

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    i don't think such an A can be found whose determinant is non-zero.

    therefore, one may choose some of the columns of A to be 0 columns, which makes the problem rather simple.
     
  7. Feb 13, 2012 #6

    lanedance

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    [tex] \textbf{A}[/tex]
    does \textbf{} help?
     
  8. Feb 13, 2012 #7

    lanedance

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    the set of vectors perpindicular to (1,2,3)^T will form a plane through the origin, in fact they form a vector subspace

    If you could find an operator that maps any vector onto that plane, then you would have a suitable operator...
     
  9. Feb 13, 2012 #8

    Ray Vickson

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    Let's try it: [itex] \textbf{A}.[/itex] So, yes, it works. Thank you.

    RGV
     
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