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Canonical matrix question

  1. Jan 3, 2009 #1
    given A={(1,2,1),(2,4,2),(3,6,3)} findr and invertable matrices Q and P such that Q-1AP={(Ir,0),(0,0)} where each zero denotes a matrix of zeros not necessarily the same size

    paying special attension to the order of the vectors write down the bases of R3 with respect to which Q-1AP represents the mapping x->Ax

    i think i can do the first part getting row opps of r3-3r1 and r2-2r1 and then column opps of c2-2c1 and c3-c1 giving me {(1,0,0),(0,0,0),(0,0,0) and therefore r=1

    then i do I3 with the same row opps giving Q-1={(1,0,0),(-2,1,0),(-3,0,1)) giving Q=(Q-1)-1 = {(1,0,0),(1/2,1,0),(1/3,0,1)

    same with the column opps on P gives me {1.-2.-1),(0,1,0),(0,0,1)}=P

    i beleive this is right but have not done it in a while and may be messing up the method so a check wouldnt go a miss, also i dont know how to do the second part of the question, it looks slightly familia with the getting vectors in the right order but i cant remember where to start so help here would be aprichiated thanks. on a side note this is revision not homework so feal free to splurt it all out :P
     
    Last edited: Jan 3, 2009
  2. jcsd
  3. Jan 3, 2009 #2

    HallsofIvy

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    It's not difficult to check it yourself is it?

    If Q-1 and P are as you say then you want to calculate
    [tex]Q^{-1}AP= \begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & 1 \\ 2 & 4 & 2\\ 3 & 6 & 3\end{bmatrix}\begin{bmatrix}1 & -2 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}[/tex]

    Is that equal to
    [tex]\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex]?
     
    Last edited: Jan 3, 2009
  4. Jan 3, 2009 #3
    ah ye, silly me, what about the second part? no ideas where to start there
     
  5. Jan 4, 2009 #4
    how is this done or even started?
     
  6. Jan 4, 2009 #5
    what does this " with respect to which Q-1AP represents the mapping x-> Ax mean?
     
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