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Math Help - Another question about large powers of matrices

  1. #1
    Senior Member chella182's Avatar
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    Another question about large powers of matrices

    So the matrix given is...

    A=\left(\begin{array}{cc}1-p & p \\ p & 1-p\end{array}\right)

    ...and I've established the eigenvectors are \left(\begin{array}{c}1 \\ 1\end{array}\right) and \left(\begin{array}{c}1 \\ -1\end{array}\right) with corresponding eigenvalues 1 and 1-2p. I'm then asked to show...

    A^n\left(\begin{array}{c}1 \\ 0\end{array}\right)=\frac{1}{2}\left(\begin{array}  {c}1 \\ 1\end{array}\right)+\frac{1}{2}(1-2p)^n\left(\begin{array}{c}1 \\ -1\end{array}\right)

    ...and I just can't seem to get it, the method I was taught seems to suggest I need to replace the \left(\begin{array}{c}1 \\ 0\end{array}\right) with the eignevalues and eigenvectors multiplied togeter and then added together, but I still can't quite seem to get that expression.
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  2. #2
    ynj
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    Quote Originally Posted by chella182 View Post
    So the matrix given is...

    A=\left(\begin{array}{cc}1-p & p \\ p & 1-p\end{array}\right)

    ...and I've established the eigenvectors are \left(\begin{array}{c}1 \\ 1\end{array}\right) and \left(\begin{array}{c}1 \\ -1\end{array}\right) with corresponding eigenvalues 1 and 1-2p. I'm then asked to show...

    A^n\left(\begin{array}{c}1 \\ 0\end{array}\right)=\frac{1}{2}\left(\begin{array}  {c}1 \\ 1\end{array}\right)+\frac{1}{2}(1-2p)^n\left(\begin{array}{c}1 \\ -1\end{array}\right)

    ...and I just can't seem to get it, the method I was taught seems to suggest I need to replace the \left(\begin{array}{c}1 \\ 0\end{array}\right) with the eignevalues and eigenvectors multiplied togeter and then added together, but I still can't quite seem to get that expression.
     A^nX=\lambda^n X,A^n\left(\begin{array}{c}1 \\ 0\end{array}\right)=\frac{1}{2}(A^n\left(\begin{ar  ray}{c}1 \\ 1\end{array}\right)+A^n\left(\begin{array}{c}1 \\ -1\end{array}\right))=\frac{1}{2}\lambda_1^n \left(\begin{array}{c}1 \\ 1\end{array}\right)+\frac{1}{2}\lambda_2^n\left(\b  egin{array}{c}1 \\ -1\end{array}\right)
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  3. #3
    Senior Member chella182's Avatar
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    Cheers, still not entirely sure why it's \frac{1}{2}, but I'm not sure that matters. Although I don't think the \frac{1}{2} was used in the other example I have... oh well.
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  4. #4
    ynj
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    Quote Originally Posted by chella182 View Post
    Cheers, still not entirely sure why it's \frac{1}{2}, but I'm not sure that matters. Although I don't think the \frac{1}{2} was used in the other example I have... oh well.
    (1,0)=0.5((1,1)+(1,-1))
    why not sure?
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  5. #5
    Senior Member chella182's Avatar
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    Because that's just repeating what was said and not explaining it, but s'ok, I figured it out in the exam anyway.
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  6. #6
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    This makes two threads in a row in which you have asked for help, got a response which you did not understand and then insulted a person who tried to help you. Do us all a favor and stop posting here.
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  7. #7
    ynj
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    Quote Originally Posted by chella182 View Post
    Because that's just repeating what was said and not explaining it, but s'ok, I figured it out in the exam anyway.
    well..may be it is not clear what i have said..sorry..
    A^nX=A^{n-1}AX=A^{n-1}\lambda X=\lambda A^{n-2}AX=\lambda^2 A^{n-2}X=...=\lambda^n X
    then you can substitude all the \lambda to the value given in the exercise.
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