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Math Help - Matrix Questions!

  1. #1
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    Matrix Questions!

    Ok, I've got 3 questions.

    1) Given C and D are square matrices of order n, and inverse of C ( C^{-1}) exists, expand and simplify  (C^{-1} + CD)^2 as much as possible.

    My answer turns out to be

     (C^{-1})^2 + D + CDC^{-1} + C^{2}D^{2}

    Is this the most you can simplify it? And is the way I stated (C^{-1})^2 alright? Or should I do it like C^{-1}C^{-1}

    2) Let u be any unit vector in IR^n, and let A = I - uu ^T (where the T indicates that the matrix is transposed).

    Prove that

    a) A = A^T
    b) A^2 = A

    I can do a) but I cannot do b). Can anyone help me?

    3)
    a) For B a square matrix, expand the product (I + B)(I − B + B^2)
    b) Let B = \begin{array}{ccc}0&p&q\\0&0&r\\0&0&0\end{array}, and find B^2 and  B^3
    c) Using the matrix B and the result from part (b), simplify the formula in (a).
    d) Let A = \begin{array}{ccc}1&p&q\\0&1&r\\0&0&1\end{array} =  B + I for B in b).

    Use the results in (c) and (a) to find A^{-1} .
    Note: I represents the identity matrix. Also recall that if  AC = CA = I , if and only if ,  C = A^{-1}

    Ok, for this question I can do all the way up to c but then I am lost how to do d). I mean I can do it by row reducing A and then finding the inverse that way. But the question says to use my results from a and c so I am thinking there must be abother way to do it. Can anyone help me out here?

    My result for a) turns out to be I + B^3. Subsequently I find that B^3 is a 3x3 matrix with all zeros and thus my answer in c turns out to be just "I", the identity matrix.

    Thanks in advance for all your help!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Johnaloa View Post
    Ok, I've got 3 questions.

    1) Given C and D are square matrices of order n, and inverse of C ( C^{-1}) exists, expand and simplify  (C^{-1} + CD)^2 as much as possible.

    My answer turns out to be

     (C^{-1})^2 + D + CDC^{-1} + C^{2}D^{2}

    Check that last term. It's CDCD.

    Looks to me like that's all you can do.

    -Dan[/quote]
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Johnaloa View Post
    Ok, I've got 3 questions.

    1) Given C and D are square matrices of order n, and inverse of C ( C^{-1}) exists, expand and simplify  (C^{-1} + CD)^2 as much as possible.

    My answer turns out to be

     (C^{-1})^2 + D + CDC^{-1} + C^{2}D^{2}

    Is this the most you can simplify it? And is the way I stated (C^{-1})^2 alright? Or should I do it like C^{-1}C^{-1} As topsquark says, C^{2}D^{2} should be CDCD. Also, you can write C^{-2} for C^{-1})^2.

    2) Let u be any unit vector in IR^n, and let A = I - uu ^T (where the T indicates that the matrix is transposed).

    Prove that

    a) A = A^T
    b) A^2 = A

    I can do a) but I cannot do b). Can anyone help me? See below.

    3)
    a) For B a square matrix, expand the product (I + B)(I − B + B^2) I + B^3, correct.
    b) Let B = \begin{array}{ccc}0&p&q\\0&0&r\\0&0&0\end{array}, and find B^2 and  B^3 B^3 = 0, correct.
    c) Using the matrix B and the result from part (b), simplify the formula in (a). It becomes (I + B)(I − B + B^2) = I. You're still correct up to here.
    d) Let A = \begin{array}{ccc}1&p&q\\0&1&r\\0&0&1\end{array} =  B + I for B in b).

    Use the results in (c) and (a) to find A^{-1}. From c), I − B + B^2 is the inverse of I + B.
    Note: I represents the identity matrix. Also recall that if  AC = CA = I , if and only if ,  C = A^{-1}

    Ok, for this question I can do all the way up to c but then I am lost how to do d). I mean I can do it by row reducing A and then finding the inverse that way. But the question says to use my results from a and c so I am thinking there must be another way to do it. Can anyone help me out here?

    My result for a) turns out to be I + B^3. Subsequently I find that B^3 is a 3x3 matrix with all zeros and thus my answer in c turns out to be just "I", the identity matrix.
    For 2), let P = \mathbf{uu}^{\textsc{t}}. Then P^2 = \mathbf{uu}^{\textsc{t}}\mathbf{uu}^{\textsc{t}} = \mathbf{uu}^{\textsc{t}} = P (because u is a unit vector and so \mathbf{u}^{\textsc{t}}\mathbf{u}=1). Therefore A^2 = (I-P)^2 = \ldots (you do the rest).
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