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Math Help - Eigenvalues, eigenvectors, finding an orthogonal matrix

  1. #1
    Senior Member chella182's Avatar
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    Eigenvalues, eigenvectors, finding an orthogonal matrix

    Sorry if this is the wrong section - there seems to be a few matrices questions in here though. The question goes:

    Let A be the 2\times2 matrix given by
    A=\left( \begin{array}{cc} 0.8 & 0.2 \\ 0.2 & 0.8 \end{array} \right).
    i) Find the eigenvalues and eigenvectors of the matrix A.
    ii) Find an orthogonal matrix O, which diagonalizes A.
    iii) Compute A^{100}.

    I'm rubbish with stuff like this.
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  2. #2
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    Quote Originally Posted by chella182 View Post
    Sorry if this is the wrong section - there seems to be a few matrices questions in here though. The question goes:

    Let A be the 2\times2 matrix given by
    A=\left( \begin{array}{cc} 0.8 & 0.2 \\ 0.2 & 0.8 \end{array} \right).
    i) Find the eigenvalues and eigenvectors of the matrix A.
    ii) Find an orthogonal matrix O, which diagonalizes A.
    iii) Compute A^{100}.

    I'm rubbish with stuff like this.
    The characteristic equation is \left|\begin{array}{cc}.8- \lambda & 0.2 \\ 0.2 0.8- \lambda\end{array}\right|= (\lambda- .8)^2- .04= 0. Then [tex](\lambda-.8)^2= .04[tex] so \lambda- 0.8= \pm .2 and \lambda= .8+ .2= 1 or \lambda= .8- .2= .6.

    The find the eigenvectors corresponding to those, use the basic definition of "eigenvector" and "eigenvalue". Since 1 is an eigenvector there exist non-zero vectors \begin{bmatrix}x \\ y\end{bmatrix} such that \begin{bmatrix}.8 & .2 \\ .2 & .8\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix} which gives the two equations .8x+ .2y= x and .2x+ .8y= y. The first equation can be written .2y= .2x so y= x. The second equation can be written .2y= .2x also which is also satisfied by y= -x. We can write any eigenvector corresponding to \lambda= 1 as x<1, 1>. Since .6 is an eigenvector there exist non-zero vectors \begin{bmatrix}x \\ y\end{bmatrix} such that \begin{bmatrix}.8 & .2 \\ .2 & .8\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}.6x \\ .6y\end{bmatrix} which gives the two equations .8x+ .2y= .6x and .2x+ .8y= .6y. Both equations now reduce to .2y= -.2x or y= -x. Any eigenvector corresponding to eigenvalue x<1, -1>.

    Taking eigenvectors as columns gives a matrix P such that P^-1AP is the diagonal matrix with the eigenvalues 1 and .6. Since you want P and P^{-1} to be orthogonal, the columns must be "orthonormal". It is easy to see that <1, 1> and <1, -1> are orthogonal. To normalize them, find the length of each and divide each vector by its length.

    If P^{-1}AP= D, then A= PDP^{-1}. Now, A^2= (pDP^{-1})(PDP^{-1})= PD^2P^{-1} because the P^{-1}P in the middle cancel each other. Similarly, A^3= (PDP^{-1})A^2= PDP^{-1}(PD^2P^{-1})= PD^3P^{-1} and it is easy to see that continues: for any positive integer, n, A^n= PD^nP^{-1} so, in particular A^100= PD^{100}P^{-1} and D^{100} is easy: it is just the diagonal matrix with the 100th power of the eigenvalues on the diagonal.
    Last edited by Jameson; May 5th 2009 at 10:59 AM. Reason: hopefully fixing the errors... we'll see
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  3. #3
    Senior Member chella182's Avatar
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    Can you fix you LaTex errors? 'Cause this looks really helpful
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  4. #4
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    Quote Originally Posted by chella182 View Post
    Can you fix you LaTex errors? 'Cause this looks really helpful
    I think it's fixed now.
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  5. #5
    Senior Member chella182's Avatar
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    Cheers I'll have a look at this when I've tackled the other problems and see if I understand.
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