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Math Help - Problem #10 - Basic Linear Algebra

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    Problem #10 - Basic Linear Algebra

    This one is going to take a lot of calculation. You might consider checking your work on something like Wolfram|Alpha. Only to check your work! You're on your honor. (Hey I managed it so you can too.)

    In any way you feel like, find the matrix, A, that generates the following eigenvectors and their associated eigenvalues.

    \lambda _1 = 6 \text{ and } | v_1 > = \left (  \begin{matrix} 2 \\ 0 \\ 1 \end{matrix} \right )

    \lambda _2 = 0 \text{ and } | v_2 > = \left ( \begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right )

    \lambda _3 = -6 \text{ and } | v_1 > = \left ( \begin{matrix} -1 \\ 1 \\ 2 \end{matrix} \right )

    Find a formula for A^n. and their eigenvectors and eigenvalues. (Hint: Consider even n and odd n. They have different formulas.)

    -Dan
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    Forum Admin topsquark's Avatar
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    Re: Problem #10 - Basic Linear Algebra

    Hmm...no action on this one. Here's the solution:

    First note that the three vectors I have given do span 3 space, even if they aren't orthogonal, so they do the trick for a basis. If you would like you can change the basis simply use
    e_1 = \frac{1}{3} |v_1 > + \frac{1}{6} |v_2 > - \frac{1}{6} |v_3 >

    e_2 = -\frac{1}{3} |v_1 > + \frac{5}{6} |v_2 > + \frac{1}{6} |v_3 >

    e_3 = \frac{1}{3} |v_1 > - \frac{1}{3} |v_2 > + \frac{1}{6} |v_3 >

    and use the usual orthonormal set of basis vectors, but where's the fun in that?

    Now, I don't know about you but messing around with non-orthogonal basis systems is a little out of my league. (I couldn't make it work anyway...I never was good with the crystal groups.) But we can always go back to the stand-by:
    A = \left ( \begin{matrix} a & b & c \\ d & e & f \\ g & h & j \end{matrix} \right )

    and we know that
    A | v_1 > = \lambda _1 |v_1 >
    etc.

    We get three sets of three equations and they're pretty easy to solve. I get
    A = \left ( \begin{matrix} 3 & -3 & 6 \\ 1 & -1 & -2 \\ 4 & -4 & -2 \end{matrix} \right )

    That's the east part. To find a pattern we need to do a bunch of calculations. If you do this on paper (c'mon! Feel the burn!) it takes a little while, but I get:
    A^3 = 36A, A^5 = 1296 A. This gives a pattern for the odds: A^n = 6^{n - 1} A.

    We will need the matrix A^2:
    A^2 = \left ( \begin{matrix} 30 & -30 & 12 \\ -6 & 6 & 12 \\ 0 & 0 & 6 \end{matrix} \right )

    We now find
    A^4 = 216 A^2, A^6 = 7776 A^2. This gives a pattern for the evens: A^n = 6^{n - 1} A^2

    -Dan
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