Exponential of the Pauli matrices My job is to prove:
$$\exp(i\theta \vec{v} \cdot \vec{ \sigma })=\cos(\theta)I+i\sin(\theta)\vec{v} \cdot \vec{ \sigma }$$
where $\theta \in \mathbb{R}$ and $\vec{v} \cdot \vec{ \sigma }=\Sigma^3_{i=1}v_i\sigma_i$ such that $\sigma_i$ are the Pauli matrices, and $\vec{v}$ is a three dimensional real vector.
My attempt:
$$\vec{v} \cdot \vec{ \sigma }=v_1
  \begin{bmatrix}
    0 & 1  \\
    1 & 0 
  \end{bmatrix}
+v_2  \begin{bmatrix}
    0 & -i  \\
    i & 0 
  \end{bmatrix}
+v_3  \begin{bmatrix}
    1 & 0  \\
    0 & -1 
  \end{bmatrix}
=  \begin{bmatrix}
    v_3 & v_1-iv_2  \\
    v_1+iv_2 & -v_3 
  \end{bmatrix} \tag{1}
$$
This is an Hermitian matrix as it is the sum of 3 Hermitian matrices because $v_1,v_2,v_3 \in \mathbb{R}$. Here's my plan: as $\vec{v} \cdot \vec{ \sigma }$ is Hermitian, then it is also diagonalizable, which by the spectral decomposition:
$$\vec{v} \cdot \vec{ \sigma }=\Sigma_i\lambda_i|i\rangle\langle i|\\
\exp(i\theta \vec{v} \cdot \vec{ \sigma })=\Sigma_i\exp(i\theta\lambda_i)|i\rangle\langle i|$$
where $\lambda_i$ are the eigenvalues of $\vec{v} \cdot \vec{ \sigma }$ and $|i\rangle\langle i|$ the outer product of its eigenvectors with themselves. Then work out from there. But from $(1)$ I get to:
$$
\exp(i\theta \vec{v} \cdot \vec{ \sigma })=e^{i\theta||\vec{v}||}
\begin{bmatrix}
    \frac{iv_2-v_1}{v_3-||\vec{v}||} \\
    1
  \end{bmatrix}
\begin{bmatrix}
    \frac{iv_2-v_1}{v_3-||\vec{v}||} & 1
  \end{bmatrix}
+e^{-i\theta||\vec{v}||}
\begin{bmatrix}
    \frac{iv_2-v_1}{v_3+||\vec{v}||} \\
    1
  \end{bmatrix}
\begin{bmatrix}
    \frac{iv_2-v_1}{v_3+||\vec{v}||} & 1
  \end{bmatrix}
$$
But this seems a bit excessive, and I don't know where I am mistaken, any help is appreciated.
Notes:


*

*This is the exercise 2.35 from Quantum Computation and Quantum Information by Nielsen and Chuang;

*I have seen this solution but I was unable to follow their reasoning. Also, could someone tell me which fundamentals I am missing in order to understand this solution?

 A: If I had read the exercise carefully I would have noticed that $||\vec{v}||=1$, thus resulting in the eigenvalues $\lambda_\pm=\pm1$. Now the problem reduces to:
$$\exp(i\theta\vec{v} \cdot\vec{\sigma})=e^{i\theta}|\lambda_+\rangle\langle\lambda_+|+e^{-i\theta}|\lambda_-\rangle\langle\lambda_-|\\
=\cos(\theta)(|\lambda_+\rangle\langle\lambda_+|+|\lambda_-\rangle\langle\lambda_-|)+i\sin(\theta)((|\lambda_+\rangle\langle\lambda_+|-|\lambda_-\rangle\langle\lambda_-|)$$
Now, as $I$ is an hermitian operator then it is also diagonalizable. $I$ has only $1$ eigenvalue($\lambda=1$), and $|\lambda_+\rangle,|\lambda_-\rangle$ are orthogonal, then:
$$I=|\lambda_+\rangle\langle\lambda_+|+|\lambda_-\rangle\langle\lambda_-|$$
As we have seen $\vec{v} \cdot\vec{\sigma}$ is Hermitian and has eigenvalues $\lambda_\pm=\pm1$, then:
$$\vec{v} \cdot\vec{\sigma}=|\lambda_+\rangle\langle\lambda_+|-|\lambda_-\rangle\langle\lambda_-|$$
Therefore:
$$\exp(i\theta \vec{v} \cdot \vec{ \sigma })=\cos(\theta)I+i\sin(\theta)\vec{v} \cdot \vec{ \sigma }$$
Credits to: goropikari
