# Finding eigenprojections of an operator

I'm having trouble with the second part of this question.

Let $\vec v\in \mathbb{R}^3$ and define the operator $\vec v \cdot \vec \sigma := v_1 \sigma_1 + v_2 \sigma_2 + v_3 \sigma_3$ acting on $\mathbb{C}^2$, where the $\sigma_j$ are the three Pauli matrices.

Show the eigenvalues and eigenprojections of $\vec v \cdot \vec \sigma$ are $\pm 1$ and $P_{\pm} = (\textbf{1} \pm \vec v \cdot \vec \sigma)/2$.

It is important to know that $\vec v$ is a unit vector, although it is not stated. I have shown the eigenvalues are $\pm 1$, but I am stuck trying to find the eigenprojections. It is way too messy to just substitute the eigenvalues into the matrix and solve. Is there any easier way?

The matrix representing the operator is $V = \begin{bmatrix} v_3 & v_1 - iv_2 \\ v_1+iv_2 & -v_3 \end{bmatrix}$.

Thank you!

So you have $$(\vec v\cdot \vec\sigma)P_\pm = (\vec v\cdot \vec\sigma)(\textbf{1} \pm \vec v \cdot \vec \sigma)/2 = \left(\vec v\cdot\vec\sigma\pm(\vec v\cdot\vec\sigma)^2\right)/2$$ Now $$(\vec v\cdot \vec\sigma)^2 = \sum_{i=1}^3 \sum_{j=1}^3 v_i v_j \sigma_i \sigma_j$$ Using the equation $\sigma_i \sigma_j = \delta_{ij}\mathbf{1} + \mathrm i\sum_k\epsilon_{ijk} \sigma_k$ one notices that the sum above is symmetric when exchanging the two factors, but $\epsilon_{ijk}$ is antisymmetric, so that term cancels out, and thus $$(\vec v\cdot \vec\sigma)P_\pm = (\vec v\cdot \vec\sigma)(\textbf{1} \pm \vec v \cdot \vec \sigma)/2 = \frac12\left(\vec v\cdot \vec\sigma \pm \sum_{ij}v_iv_j\delta_{ij}\mathbf 1\right)$$ Now $\sum_{ij}v_iv_j\delta_{ij}=\vec v^2=1$ by assumption, so we finally arrive at $$(\vec v\cdot \vec\sigma)P_\pm = (\vec v\cdot\vec\sigma \pm \mathbf 1)/2 = \pm P_\pm \tag{*}$$
With an analogous calculation you can prove that $P_\pm^2=P_\pm$ which means that $P_pm$ is a projection. Alternatively observe that if $x_i$ are the eigenvalues of $X$, the eigenvalues of $(\mathbf 1+X)/1$ are $(1+x_i)/2$, and thus the eigenvalues of $P_\pm$ are $0$ and $1$, therefore $P_\pm$ is a projection.
Equation (*) together with the fact that $P_\pm$ are projections prove the claim.