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 $.
The reason for this is because in quantum mechanics, states are always normalized. It is one of the postulates of quantum mechanics.
It is normalized by dividing with the modulus or magnitude which is sqrt of (eigenvalue1 * 2 + eigenvalue2 * 2) = sqrt(2)
