Saladsamurai
- 3,009
- 7
X'=\left[\begin{array}{ccc}4 & 1 & 4\\ 1 & 7 & 1\\ 4 & 1 & 4\end{array}\right]X
After evaluating the determinant I get from the characteristic equation \lambda=0
\lambda=\frac{15\pm \sqrt{119}*i}{2}
Now here is where I am not sure where to go. I need to create an Eigenvector. I stared it off by plugging in for \lambda=0 giving the relationships:
4k_1+k_2+4k_3=0
k_1+7k_2+k_3=0
4k_1+k_2+4k_3=0
Now usually, in the case of a 2x2 matrix, I would solve for let's say k1 in terms of k2 and then just PICK some value for k2 and that would establish my Eigenvector.
I am a little confused as to how to do this with a 3x3. It would appear I have some sort of special case here since row 1 is the same as row 3, but I am unsure how to use that to my advantage?
Could someone just point me toward the next step?
Thanks!
After evaluating the determinant I get from the characteristic equation \lambda=0
\lambda=\frac{15\pm \sqrt{119}*i}{2}
Now here is where I am not sure where to go. I need to create an Eigenvector. I stared it off by plugging in for \lambda=0 giving the relationships:
4k_1+k_2+4k_3=0
k_1+7k_2+k_3=0
4k_1+k_2+4k_3=0
Now usually, in the case of a 2x2 matrix, I would solve for let's say k1 in terms of k2 and then just PICK some value for k2 and that would establish my Eigenvector.
I am a little confused as to how to do this with a 3x3. It would appear I have some sort of special case here since row 1 is the same as row 3, but I am unsure how to use that to my advantage?
Could someone just point me toward the next step?
Thanks!