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

Question

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?

KAJ226 · Answer

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.

Krishna Aditya · Answer

It is normalized by dividing with the modulus or magnitude which is sqrt of (eigenvalue1 * 2 + eigenvalue2 * 2) = sqrt(2)

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

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