# How to find the normalization factor of the eigenvectors of the $\sigma_x$ Pauli gate?

I'm trying to calcaute the eigenstates for the $$\sigma_x$$ gate, and I can follow the process up to finding eigenvalues $$\pm 1$$, but I don't understand where the $$\frac{1}{\sqrt{2}}$$ coefficient comes from for the answer:

$$\begin{bmatrix} -\lambda & 1\\ 1 & -\lambda \end{bmatrix}v = 0 \implies v = \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1 \end{bmatrix}$$

For the solution $$\lambda = 1$$, why does that $$\frac{1}{\sqrt{2}}$$ show up?

The $$\dfrac{1}{\sqrt{2}}$$ is the normalization constant to make sure the state/eigenvector is a unit vector.
Note that: if $$|\psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ then $$\bigg| \bigg| |\psi \rangle \bigg| \bigg| = |1/\sqrt{2}|^2 + |1/\sqrt{2}|^2 = 1$$.