# How to find eigenvalue and eigenvectors of spin linear operator?

In Exercise 3.4 of Leonard Susskind's "Quantum Mechanics The Theoretical Minimum" it asks to compute the eigenvalues and eigenvectors of the spin linear operator given by: $$\sigma_n=\begin{pmatrix}\cos \theta & \sin \theta \ e^{-i\phi}\\\sin\theta\ e^{i\phi}&-\cos \theta\end{pmatrix}$$Which is the Hermitian operator describing spin of a system with spherical parameters $\theta$ and $\phi$,

It is quite simple to see that the eigenvalues must be $\lambda_1=1$ and $\lambda_2 = -1$, however finding the eigenvectors seems to be more of a challenge. I tried to resolve components of $|\lambda_1\rangle$ and $|\lambda_2\rangle$ using $\sigma_n |\lambda_k\rangle=\lambda_k|\lambda_k\rangle$ however I seem to get $$|\lambda_1\rangle = \begin{pmatrix}(1+e^{i2\phi}\tan^2\frac{\theta}2)^{-1/2}\\(1+e^{-i2\phi}\cot^2\frac{\theta}2)^{-1/2}\end{pmatrix}$$ and $$|\lambda_2\rangle = \begin{pmatrix}(1+e^{i2\phi}\cot^2\frac{\theta}2)^{-1/2}\\(1+e^{-i2\phi}\tan^2\frac{\theta}2)^{-1/2}\end{pmatrix}$$ However I have reason to believe that this is incorrect as it seems to be a much simpler answer.

Is there a good way to go about finding these eigenvectors? Any hints or observations are appreciated. I should also note that the eigenvectors are normalised such that $\langle \lambda_k | \lambda_k\rangle=1.$

This is the same as finding the nullspace of the matrix $$\sigma_n=\begin{pmatrix}\cos \theta -xI & \sin \theta \ e^{-i\phi}\\\sin\theta\ e^{i\phi}&-\cos \theta - xI\end{pmatrix}$$ when $x =1$ or $x=-1$. Finding a vector in the nullspace of a $2 \times 2$ matrix which you know to be non-invertible is (usually) easy: you just set the first entry equal to one, and then the second entry to whatever will make it dot to zero with the top row of the matrix. For instance, for $x=1$, $$\sigma_n=\begin{pmatrix}\cos \theta -1 & \sin \theta \ e^{-i\phi}\\\sin\theta\ e^{i\phi}&-\cos \theta -1 \end{pmatrix}$$ you just note that $\left [\begin{matrix} 1 \\ \frac{\cos \theta -1}{\sin \theta e^{-1\varphi}} \end{matrix}\right]$ is what it would have to be. (If you find you're dividing by zero, the fix should be easy.)

In the special case where the eigenspace has first entry equal to zero (and therefore no eigenvectors of the form $(1,r)$ exist), you can usually spot what the eigenvector should be anyway.

The eigenvector for $\lambda$ is in the kernel of $\sigma_n-\lambda$, so up to normalisation it can be read off the first row as

$$\pmatrix{ \sin\theta\,\mathrm e^{-\mathrm i\phi}\\ \lambda-\cos\theta }\;.$$

The squared magnitude of this vector is $\lambda^2-2\lambda\cos\theta+1=2-2\lambda\cos\theta$, so the normalised eigenvector is

$$\frac1{\sqrt{2-2\lambda\cos\theta}}\pmatrix{ \sin\theta\,\mathrm e^{-\mathrm i\phi}\\ \lambda-\cos\theta }\;.$$

For $\lambda=1$ this is

$$\pmatrix{ \cos\frac\theta2\,\mathrm e^{-\mathrm i\phi}\\ \sin\frac\theta2 }\;,$$

and for $\lambda=-1$ it's

$$\pmatrix{ \sin\frac\theta2\,\mathrm e^{-\mathrm i\phi}\\ -\cos\frac\theta2 }\;.$$