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This is for an optimization problem? I haven't dealt with complex numbers(although I know the eigenvalues are going to give me a neutral circular node)

$\begin{bmatrix}-\sqrt{2}i & 2\\ -2 & -\sqrt{2}i\end{bmatrix}\begin{bmatrix}x_1 \\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

How to solve for $x_1$ & $x_2$?

Thanks

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  • $\begingroup$ @Amzoti I see, thank you very much. User Kaj in the chat also explained the linear algebra property that using $A-\lambda I$ will not give trivial solutions, so I realised I embarrassingly made a trivial error in my eigenvalues. $\endgroup$ – Question mark Nov 2 '14 at 2:54
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This matrix is invertible. Check the determinant: $(-\sqrt{2}i \cdot -\sqrt{2}i) - (-2 \cdot 2) = -2 + 4 = 2$. So the nullspace is trivial, and so $x_1 = x_2 = 0$.

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Since the system is homogeneous, the unique solution is the ordered pair $(x_1,x_2)=(0,0)$.

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Hint: If you do the matrix multiplication, you'll get a $2\times1$ matrix. The top and bottom entries must equal the top and bottom entries on the other side of the equals sign, so you get a system of equations you can solve.

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