Spectral Decomposition exponential with Pauli matrices Let $\sigma^0$ be the $2\times 2$ identity; $\sigma^1$, $\sigma^2$ and $\sigma^3$ the Pauli matrices:
$$ \sigma^1 = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix},\ \sigma^2 = \begin{pmatrix} 0 & -i \\ i & 0\end{pmatrix},\ \sigma^3 = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$$
From a vector $\overrightarrow{v}$ having real components, we define the matrix 
$$M = \overrightarrow{v}\cdot\overrightarrow{\sigma} = \sum\limits_{i=1}^3 v_i\sigma^i $$
that has the following eigenvalues
$$\mu_{\pm} = \pm|\overrightarrow{v}|=\pm\sqrt{v_1+v_2+v_3}$$
and the associated projectors
$$ P_{\pm} = \frac{1}{2}\left(\sigma^0\pm\frac{\overrightarrow{v}\cdot\overrightarrow{\sigma}}{|\overrightarrow{v}|}\right)$$
Which gives the following spectral decomposition:
$$ M = |\overrightarrow{v}|P_+ - |\overrightarrow{v}|P_-$$
If $\overrightarrow{w}$ is orthogonal to $\overrightarrow{v}$, i.e $\sum\limits_{i=1}^3v_iw_j=0$, I want to show the following relation:
$$ e^{\lambda\overrightarrow{v}\cdot\overrightarrow{\sigma}}\overrightarrow{w}\cdot\overrightarrow{\sigma} = \overrightarrow{w}\cdot\overrightarrow{\sigma} e^{-\lambda\overrightarrow{v}\cdot\overrightarrow{\sigma}}$$
I am tempted to use the spectral decomposition of $e^{\lambda\overrightarrow{v}\cdot\overrightarrow{\sigma}} = e^{\lambda |\overrightarrow{v}|}P_+ + e^{-\lambda |\overrightarrow{v}|}P_-$ but I'm stuck. Any idea how to prove the formula?
 A: This follows from the anticommutativity relation 
$$(v\cdot\sigma)(w\cdot\sigma)=-(w\cdot\sigma)(v\cdot\sigma)$$
valid when $v\cdot w=0$. You can apply this term by term after expanding $\exp(\lambda v\cdot\sigma)$.
The general formula for multiplying sums of Pauli matrices is
$$ (v\cdot\sigma)(w\cdot\sigma)=(v\cdot w)\sigma^0+i(v\times w)\cdot\sigma $$
where $\times$ is the 3D cross product. 
I only know this because of my familiarity with quaternions. In this setting, $\mathbb{H}$ is a 4D algebra whose elements are sums of scalars and 3D vectors. To multiply $(r+\mathbf{u})(s+\mathbf{v}$) one first uses the distributive property and then the relation $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$ which follows from inspecting the multiplication table for the basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$. If $S^3$ is the group of unit quaternions (which is a $3$-sphere, like $S^1$ in the complex numbers is a $1$-sphere i.e. a circle) then there is an isomorphism $S^3\to\mathrm{SU}(2)$ which extends to a (not onto) algebra homomorphism $\mathbb{H}\to M_2(\mathbb{C})$ given by $\mathbf{v}\mapsto v\cdot i\sigma$ (mixing up the notations a bit).
