# Math Help - Problem #10 - Basic Linear Algebra

1. ## 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

2. ## 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