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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!

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  • $\begingroup$ Just proceed like in the real case, you have to solve the linear system (A-x1)y=0 while setting x=3 $\endgroup$ Nov 21, 2016 at 7:58
  • $\begingroup$ My problem is in solving $(A-xI)y=0$ with $x=(1+i)$. I know i only need to solve for one, as the other is the complex conjugate, but i just can't seem to do it $\endgroup$
    – Andy Grey
    Nov 21, 2016 at 8:01
  • $\begingroup$ What did you try to solve $(A-(1+i)I)X=0$ for the vector $X$? $\endgroup$
    – Did
    Nov 21, 2016 at 8:02
  • $\begingroup$ Yup. See the comment above.. $\endgroup$
    – Andy Grey
    Nov 21, 2016 at 8:03
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    $\begingroup$ Since the eigenvector is not unique, we are free to choose the first non-zero component to be 1. So, we set $y_1=1$ and see if we get a solution for $y_2$ and $y_3$. If no, then $y_1$ is zero. If we require a normalized eigenvector, we choose the first non-zero component to be real. We can still multiply by all the components by an overall phase factor of $e^{i\theta}$ if we want without changing the normalization. If we wanted to choose some other non-zero component to be 1 instead of the first, that is okay, too. $\endgroup$
    – LouisB
    Nov 21, 2016 at 9:32

1 Answer 1

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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}

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