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Homework Help: Linear algebra factorizations

  1. Jul 21, 2010 #1
    1. The problem statement, all variables and given/known data

    The problem is in the attached image, its easier to see this way

    2. Relevant equations



    3. The attempt at a solution

    I've never even heard of these factorizations so I have no idea what they mean. I'm completely lost on what to do for these. If someone could explain step by step that would really help, I just need to understand them.
     

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  3. Jul 21, 2010 #2

    hunt_mat

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    The first one is standard, it's the usual way of diagonalising a square matrix. It's all to do with computing eigenvalues and eigenvectors. Put the eigenvectors as columns in a matrix and call it P. Then you compute P^{T}AP, and this will be a diagonal matrix. You can also show that P^{T}P=PP^{T} and so P^{T}=P^{-1}. This is standard and you should look it up as it's quite important but the question doesn't require to know this, it is just matrix manipulation just like you've been doing before.

    As for the second question, R^{-1}=A^{-1}Q, it's just a matter of doing a little algebra.
     
  4. Jul 21, 2010 #3

    HallsofIvy

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    Because you are already given the diagonal matrix, you don't need to find the eigenvalues and eigenvectors that hunt_mat refers to yourself.

    Just use the fact that [itex](P^{-1}DP)^2= (P^{-1}DP)(P^{-1}
    DP)= (P^{-1}D)PP^{-1}(DP)[/itex][itex]= (P^{-1}D)(DP)= P^{-1}D^2P[/itex] and [itex](P^{-1}DP)^3[/itex][itex]= (P^{-1}DP)^2(P^{-1}DP)= (P^{-1}D^2P)(P^{-1}DP)[/itex], etc.. And, of course, it is easy to find powers of a diagonal matrix.
     
  5. Jul 21, 2010 #4
    I'm still not understanding these three. And yeah it shouldn't use eigenvalues and vectors because we haven't learned them yet. Could you explain them step by step? I have an exam in a few hours and I'm trying to understand these as much as possible before taking it.
     
  6. Jul 21, 2010 #5

    hunt_mat

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    The key thing is you don't need to understand them, you just need to be confidant with matrix algebra.
     
  7. Jul 21, 2010 #6
    I understand matrix algebra, I just have a hard time when its put into problems like these. Maybe I'm just not understanding exactly what its asking or something because i'm very lost with these.
     
  8. Jul 21, 2010 #7
    I think I can do 4.4 somewhat.
    Since (A[tex]^{T}[/tex])[tex]^{-1}[/tex] = A from the description
    (A[tex]^{T}[/tex])[tex]^{-1}[/tex] = U E V[tex]^{T}[/tex]
    and since (A[tex]^{T}[/tex])[tex]^{-1}[/tex] = (A[tex]^{-1}[/tex])[tex]^{T}[/tex]
    (A[tex]^{-1}[/tex])[tex]^{T}[/tex] = U E V[tex]^{T}[/tex]
    And I think I can take the transpose of both sides to get
    ((A[tex]^{-1}[/tex])[tex]^{T}[/tex])[tex]^{T}[/tex] = ( U E V[tex]^{T}[/tex])[tex]^{T}[/tex]
    and since (A[tex]^{T}[/tex])[tex]^{T}[/tex] = A
    A[tex]^{-1}[/tex] = (U E V[tex]^{T}[/tex])[tex]^{T}[/tex]

    I'm not sure if the above is correct or not so I would appreciate if someone checked it. I don't understand how to show that A in invertible though, and I still don't understand the other two problems. I only have a few hours left before my exam, if someone could please explain them to me I would feel a lot more confident taking an exam.
     
  9. Jul 21, 2010 #8
    actually, I think I read the question wrong so that doesn't work
     
  10. Jul 21, 2010 #9

    HallsofIvy

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    I pointed out above that if [itex]A= P^{-1}DP[/itex] then [itex]A^2= (P^{-1}DP)(P^{-1}AP)= P^{-1}D^2P[/itex] because the "[itex]P^{-1}P[/itex]" cancels. For higher powers, [itex]A^3= P^{-1}D^3P[/itex], and, in general, [itex]A^n= P^{-1}D^nP[/itex].

    Now, as an example, if
    [tex]D= \begin{bmatrix}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3\end{bmatrix}[/tex]
    what are [itex]D^2[/itex], [itex]D^3[/itex], [itex]D^4[/itex], or, in general, [itex]D^n[/itex]?
     
  11. Jul 21, 2010 #10
    Isn't the diagonal to the power of what D is
     
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