The

*characteristic polynomial of A*is given by:

Remembering that for a 3 x 3 matrix

the determinant det|M| is the sum of three terms. The first is:

The other two are similar, except that second term, starting with a2 * ..., is multiplied by - 1. So, going back to our problem:

det|A = kI| is:

Setting this equal to zero, one solution is

We use the quadratic formula to obtain the

other two roots:

The solutions are: 4.414, 2, 1.586

To find the first eigenvector (for k = 3 + sqrt(2) ),

we have:

where v is a vector with values x,y,z:

For the first eigenvalue we have:

That is:

From the first equation:

If we choose

To normalize, we divide each by:

Check this is a unit vector:

In R:

## 1 comment:

Umm, this is tedious. How bout instead I put a u in front of another letter for the unit vector. I'm leaving out the square root and other mess.

uv=[1, 1+sqrt(2), (1+sqrt(2))^2]

vw= [(1/2-(1/2)*sqrt(5))^2, -(1/2-(1/2)*sqrt(5)), 1]

uv=vw

Basically it's [1, a, a^2] for a quadratic solution. You can get the other value with [a^2, -a, 1]

Notice I reversed the order and snuck in a negative sign? Btw, I found this by doing three consecutive Pell numbers. This works with Fibonacci number and any of the recursive number things.

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