Continuous function $f(\vartheta A)$ where $A$ is an Hermitian matrix Consider the matrices $A_{1}, A_{2}, A_{3}$ defined as:
$A_{1}=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right), A_{2}=\left(\begin{array}{cc}0 & -i \\ i & 0\end{array}\right), A_{3}=\left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right) .$
If a matrix $A$ is defined by
$A=n_{1} A_{1}+n_{2} A_{2}+n_{3} A_{3}$ where $n_{1}, n_{2}, n_{3}$ are real constants satisfying $n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1$

(a) Show that $A$ is an Hermitian matrix.


(b) Find the eigenvalues of the eigenvalue equation $A|\lambda\rangle=\lambda|\lambda\rangle .$ If we denote $|+\rangle$ and $|-\rangle$ be normalized eigenvectors, find $|+\rangle\langle+|$ and $|-\rangle\langle-|$ from the Spectral theorem.


(c) Show that


$f(\vartheta A)=\frac{f(\vartheta)+f(-\vartheta)}{2} I+\frac{f(\vartheta)-f(-\vartheta)}{2} A$


where $I=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$ and $f(x)$
is a continuous function.

 A: Note: the question makes use of the Dirac bra-ket notation. widely used (and very convenient) in quantum mechanics (but it is not strictly needed here and not much used by mathematicians, so I will try to avoid it).
a) The matrix $A$ is hermitian because it is a linear combination (with real coefficients) of hermitian matrices. 
b)  You did it, but let me state (for clarity) that this question uses the Dirac bra-ket notation: formally the normalized eigenvectors satisfy $ A |+\rangle\ =  |+\rangle\ $ and $ A |-\rangle\ = - |-\rangle\ $. According to this notation, the operators $P_-=|-\rangle\langle-|$ and $P_+=|+\rangle\langle+|$ are the projectors along the two eigenvectors. The identity can be expressed as $\mathbb{1} = P_+ + P_-$.
c)  Suppose you can expand $f$ as a power series (otherwise the question makes little sense),
$$
f(x) = \sum_m a_m x^m
$$
Therefore, using $A= P_+ - P_-$ and $P_-^m = P_-$ (and the same for $P_+$), you have
$$
f(t A) = \sum_m a_m t^m A^m = \sum_m a_m t^m [P_{+}^m -mP_{+}^{m-1} P_{-}^m+... + (-P_{-})^m]
$$
However, $P_- P_+ = 0 $ because they project on orthogonal subspaces, so that
$$
f(t A) = \sum_m a_m t^m A^m = \sum_m a_m t^m (P_{+}^m +(-P_{-})^m)
= \sum_m a_m t^m [ P_{+} + (-1)^m P_{-} ]
\\ =  \sum_m a_m t^m  P_{+}   + \sum_m a_m  (-t)^m P_{-} = f(t) P_+ + f(-t) P_- 
$$
Using $A= P_+ - P_-$ and $\mathbb{1} = P_+ + P_-$ you have the thesis.
