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Math Help - Find matrix B

  1. #1
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    Find matrix B

    B is a 2x2 matrix with the property that there is an invertible 2x2 matrix R such that

    R^-1BR = [(sqrt 3) 0 ] [ 1.2 -(sqrt 3)/2]
    [ 0 (sqrt 3) ] [(sqrt 3)/2 1/2 ]


    I know the first part is the scaling matrix and the second on in the rotation matrix.

    I'm suppose to find B^6 which is equal to
    B^6 = R^-1D^6R

    But I dont know how I would find the matrix for R or D.
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  2. #2
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    Are you saying that B is a matrix such that there exists invertible R such that

    R^{-1}BR=\begin{bmatrix}\sqrt{3}&0\\0&\sqrt{3}\end{bma  trix}<br />
\begin{bmatrix}1/2&-\sqrt{3}/2\\\sqrt{3}/2&1/2\end{bmatrix}?

    And you're asked to find B^{6}?
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  3. #3
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    Yea, that is what we were given and then asked to find B^6
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  4. #4
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    Hmm. Interesting problem. Well, the nice thing about diagonal matrices is that it's easy to find their nth power: just take the nth power of the elements on the main diagonal. Now the problem you've been given is not a typical diagonalization problem. Presumably, you've seen those already, so you know how you can always do this:

    If

    B=R^{-1}DR, then

    B^{n}=(R^{-1}DR)^{n}=\underbrace{R^{-1}DRR^{-1}DR\dots R^{-1}DR}_{n\times}=R^{-1}\underbrace{DD\dots D}_{n\times}R=R^{-1}D^{n}R.

    I played some "what if" games with your situation. Let's say you've got

    R^{-1}BR=\begin{bmatrix}\sqrt{3}&0\\0&\sqrt{3}\end{bma  trix}<br />
\begin{bmatrix}1/2&-\sqrt{3}/2\\\sqrt{3}/2&1/2\end{bmatrix}=C.

    So, C is known. Then

    B=RCR^{-1},

    and the R^{-1} is on the wrong side of C. We want to have

    B=R^{-1}DR. If I compare these two, it leads me to ask this question: what if I could find a diagonal D such that

    C=R^{-2}DR^{2}? Then

    B=RCR^{-1}=RR^{-2}DR^{2}R^{-1}=R^{-1}DR, as desired.

    So, can you diagonalize C=Q^{-1}DQ but then, when you find the invertible matrix Q, can you take its square root? There are ways to do that, I think, that aren't overly difficult, especially if you orthogonally diagonalize C.

    So, give this program a shot: orthogonally diagonalize C=Q^{-1}DQ. What do you get?
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  5. #5
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    Are you sure about that? I recommend you go back and reread the problem. I am inclined to suspect that it is R itself that is equal to \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2}  & \frac{1}{2}\end{bmatrix} and that then R^{-1}BR= \begin{bmatrix}\sqrt{3} & 0 \\ 0 & \sqrt{3}\end{bmatrix}
    which is then the same as saying that \begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}B= \bmatrix{\sqrt{3} & 0 \\ 0 & \sqrt{3}\end{bmatrix}\begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}.

    If that is the case, that you have R^{-1}BR= D, where I am using "D" to stand for that diagonal matrix, then you also have B= RDR^{-1}. Now notice that B^2= (RDR^{-1})(RDR^{-1})= (RD)(R^{-1}R)(DR^{-1})= (RD)(DR^{-1})= RD^2R^{-1}
    B^3= B(B^2)= (RDR^{=1})(RD^2R^{-1})= RD^3R^{-1}, etc.

    Further, since D is a diagonal matrix, it is easy to find powers of D:

    \begin{bmatrix}d & 0 \\ 0 & d\end{bmatrix}^n= \begin{bmatrix}d^n & 0 \\ 0 & d^n\end{bmatrix}.
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