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I have had no problem while finding the eigen vectors for the x and y components of pauli matrix. However, while solving for the z- component, I got stuck. The eigen values are 1 and -1. While solving for the eigen vector corresponding to the eigen value 1 using $$(\sigma _z-\lambda I)X=0$$, I got $$\left( \begin{matrix} 0 & 0 \\ 0 & -2 \end{matrix} \right)\left( \begin{matrix} x \\ y\end{matrix} \right)=0$$ Now, how can I find the eigen vector for eigen value 1 with this relation since it will give me only $-2y=0$

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The math is telling you that $0 x = 0$, so $x$ can be any number, and $-2 y = 0$, so $y$ must be zero. For a unit length eigenvector, we usually choose $u=\{1,0\}^T$, but $v=\{e^{i\phi},0\}^T$, where $\phi$ is a real number, is also a unit length eigenvector with eigenvalue 1.

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