How do I show that $R_z(\theta)=e^{-iZ\theta/2}$?

Question

I know that an $R_z (\theta)$ gate is equivalent to the unitary transformation $e^{-iZ \theta/2}$ but I'm not sure how we get there.

I know that for every Hermitian matrix $H$ there is a corresponding Unitary matrix $ U = e^{iH} $ where the eigenvalues are exponentiated and the eigenstates remain the same. But I don't see how in this case, it leads to the matrix $$\begin{pmatrix}e^{-i \frac{\lambda}{2}} & 0 \\ 0 & e^{i \frac{\lambda}{2}}\end{pmatrix}$$ and not: $$\begin{pmatrix}e^{i} & 0 \\ 0 & e^{-i}\end{pmatrix}$$ I feel like I'm missing something really obvious

Answer 1

$$-iZ\frac{\theta}{2} = \begin{bmatrix}-i\frac{\theta}{2} &0\\0&i\frac{\theta}{2} \end{bmatrix}$$

This is alread diagonal, so now you just take the exponenital operation on each of the eigenvalues.

To get the matrix you thought you would get, you would need $$iZ=\begin{bmatrix}i&0\\0&-i\end{bmatrix}$$ which when exponeniated, would give you $$\begin{bmatrix}e^{i}&0\\0&e^{-i}\end{bmatrix}$$

I am assuming you meant to write $\theta$ and not $\lambda$ in your first matrix.