I'm having trouble with the following matrix:
$$\begin{bmatrix}2 & 1 & \frac{1}{\sqrt{2}}\\1 & 2 & -\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1\end{bmatrix}$$
I've already calculated the eigenvalues $(3, 1+i, 1-i)$ and the eigenvector for Eigenvalue=3, but I'm having real trouble getting the complex eigenvectors. I've tried an online calculator, but the method seems overly complex and I feel like I've I've missed a trick somewhere.
Can anyone help?
Cheers!
Andy
Edit: I've got as far as figuring out that $y_1 = -y_2$ for $(A-xI)y=0$; $x=(1+i) $ but i get no further
Edit 2: Not sure I'm making myself clear enough. I understand how I can solve for the complex Eigenvectors, but don't seem to be able to do it. I really feel like I missed something.
Edit 3: OK. Starting from $$\begin{bmatrix}2-(1+i) & 1 & \frac{1}{\sqrt{2}}\\1 & 2-(1+I) & -\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 1-(1+i)\end{bmatrix}$$
Then $$\begin{bmatrix}1-i & 1 & \frac{1}{\sqrt{2}}\\1 & 1-i & -\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -i\end{bmatrix}$$
I combined the first and second rows to eliminate $y_3$ to find that $y_1=-y_2$.
After that I tried substituting $y_1$ with$-y_2$ in every row but then can only solve for $y_3$
in terms of $y_2$ which I also don't know.
It has just occurred to me to try to add/subtract the substituted rows to try to eliminate $y_2$.
I'm on my phone so that took ages to type!
Best Answer
This problem is just a matter of obtaining the reduced row echelon form of the matrix. This is a standard technique for solving for the kernel of a matrix. Finding the kernel of $det(A - \lambda I)$ gives you the eigenvector associated with the eigenvalue $\lambda$. Since the eigenvector is complex, the other complex eigenvector is the complex conjugate of the first. Therefore, we only need to solve for one.
Using $\lambda = 1 + i$,
\begin{bmatrix} 1-i & 1 & \frac{1}{\sqrt{2}} \\ 1 & 1-i & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -i \\ \end{bmatrix}
Add the second row to the first, and multiply the bottom row by $-\sqrt{2}$:
\begin{bmatrix} 2-i & 2-i & 0 \\ 1 & 1-i & -\frac{1}{\sqrt{2}} \\ 1 & -1 & i\sqrt{2} \\ \end{bmatrix} multiply the top row by $\frac{1}{5}(2+i)$
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 1-i & -\frac{1}{\sqrt{2}} \\ 1 & -1 & i\sqrt{2} \\ \end{bmatrix}
subtract the first row from the other two rows:
\begin{bmatrix} 1 & 1 & 0 \\ 0 & -i & -\frac{1}{\sqrt{2}} \\ 0 & -2 & i\sqrt{2} \\ \end{bmatrix} multiply second row by $i$:
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & -\frac{i}{\sqrt{2}} \\ 0 & -2 & i\sqrt{2} \\ \end{bmatrix}
subtract the second row from the first, and add 2 times the second row to the third:
\begin{bmatrix} 1 & 0 & \frac{i}{\sqrt{2}} \\ 0 & 1 & -\frac{i}{\sqrt{2}} \\ 0 & 0 & 0 \\ \end{bmatrix} This is the reduced row echelon form. Therefore, the eigenvector associated with $\lambda = 1 + i$ is:
\begin{bmatrix} 1 \\ -1 \\ i \sqrt{2} \\ \end{bmatrix}
and so the eigenvector associated with $\lambda = 1-i$ is:
\begin{bmatrix} 1 \\ -1 \\ -i \sqrt{2} \\ \end{bmatrix}