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Matrix Equation

  1. Aug 22, 2011 #1

    Hootenanny

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    I have a quick question regarding matrix equations. Usually, I would look this up but unfortunately I'm away from the office and library and it can't wait until I get back.

    Let [itex]A_1[/itex] and [itex]A_2[/itex] be [itex]n\times n[/itex] square matrices with real elements and let [itex]\boldsymbol{x}_1\;,\boldsymbol{x}_2\in\mathbb{R}^n[/itex]. Further, let [itex]A_1 \boldsymbol{x}_1 = \boldsymbol{0}[/itex]. What is the solvability condition for the following system?

    [tex]A_1\boldsymbol{x}_2 = A_2\boldsymbol{x}_1[/tex]

    The result would suggest [itex]\boldsymbol{x}_1^\text{T}A_2\boldsymbol{x}_1 = 0[/itex], but I'm clearly missing something. I fairly certain its something minor that I just can't see.

    Any help would be very much appreciated.
     
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  3. Aug 22, 2011 #2
    Can any of the x_{i} or A_{i} be inverted? (i.e., do you know anything about their determinants?)
     
  4. Aug 22, 2011 #3

    Hootenanny

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    [itex]\boldsymbol{x}_i[/itex] are vectors in [itex]\mathbb{R}^n[/itex], and [itex]A_i[/itex] are singular in general.
     
  5. Aug 22, 2011 #4
    Apologies, can now see the x_{i} are vectors. I'm at work and scanning articles when no-one is looking.

    I your reasoning has lead to to conclude that /boldsymbol{x_{1}^T}A_{1} = /boldsymbol{0} - How do you know this?
     
  6. Aug 22, 2011 #5

    Hootenanny

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    No problem :)

    I'm tracing back a result and I've found that the result would only hold if the above relation is true.

    I only asked because I assumed that solvability condition for an equation of the forum that I posted in my original post would be fairly well known, or at least established.
     
  7. Aug 22, 2011 #6

    micromass

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    Is your matrix symmetric by any chance?? In that case we have that

    The thing is that

    [tex]A_1x_2=A_2x_1[/tex]

    has a solution if and only if [itex]A_2x_1\in im(A_1)[/tex].
    But we know that [itex]im(A_1)=ker(A_1^T)^\bot[/itex].
    So the system has a solution if and only if [itex]A_2x_1\in ker(A_1^T)^\bot=ker(A_1)^\bot[/itex].

    So it must hold that [itex]x_1^TA_1x_1=0[/itex]. I fear that this is not a sufficient condition in general...
     
  8. Aug 22, 2011 #7

    Hootenanny

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    That was my first thought as well. Alas, there are no symmetry conditions on the matrices [itex]A_i[/itex].
     
  9. Aug 22, 2011 #8

    Hootenanny

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    Nevermind - I've figure it out :D
     
  10. Aug 22, 2011 #9

    micromass

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    Can you tell us the solution?? :smile:
     
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