I am having some confusion that I hope people might be able to help me with. To be concrete I will use two (isomorphic) $\mathrm{C}^*$-algebras.
The first, $A_1$, is the $\mathrm{C}^*$-group algebra of the (amenable) infinite dihedral group $D_\infty$ with order two generators $u$ and $v$ (it is isomorphic to the free product of cyclic groups) $C_2\star C_2$. Consider it represented by the regular representation as $\pi_1(\mathrm{C}^*(D_\infty))\subset B(\ell^2(D_\infty))$. The particular element I am interested in is the sum of two projections:
$$a_1:=\frac12\delta^e+\frac12 \delta^u+\frac12\delta^ e+\frac12 \delta^v=\delta^e+\frac{1}{2}\delta^u+\frac{1}{2}\delta^v.$$
I understand (via the isomorphism with $A_2$ below that the spectrum $\sigma(a_1)=[0,2]$.
Question 1: is one an eigenvalue of $\pi_1(a_1)$? If yes what is an eigenvector?
The second $\mathrm{C}^*$-algebra $A_2$ is the universal $\mathrm{C}^*$-algebra generated by projections $p$ and $q$. It is isomorphic that it is isomorphic to $A_3$, where $A_3$ is the $\mathrm{C}^*$-algebra: $$A_3:=\left\{f\in C([0,1],M_2(\mathbb{C})):f(0) \text{ and }f(1)\text{ are diagonal}\right\}.$$
There is an isomorphism that carries $p$ and $q$, respectively, to $$p(x)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 0\end{array}\right)\quad \text{ and }\quad q(x)=\left(\begin{array}{cc}x & \sqrt{x(1-x)} \\ \sqrt{x(1-x)} & 1-x\end{array}\right).$$
This can be represented $\pi_2(A_3)\subset B(L^2([0,1])\oplus L^2([0,1]))$.
Consider $$a_2:=x\mapsto p(x)+q(x)=\left(\begin{array}{cc}1+x & \sqrt{x(1-x)} \\ \sqrt{x(1-x)} & 1-x\end{array}\right).$$
The determinant of $a_2(x)-\lambda I_{\pi(A_3)}$ is $\lambda^2-2\lambda-x+1$ which is an issue when $\lambda=1\pm\sqrt{x}$, and so the spectrum $\sigma(a_2)=[0,2]$. As far as I can see it $\pi_2(a_2)$ has no eigenvalues at all.
Question 2: Is this correct --- $\pi_2(a_2)$ has no eigenvalues?
This leads onto a third question (possibly giving an answer to this question).
Question 3: Unless I am mistaken, the spectrum of an element $a$ of a $\mathrm{C}^*$-algebra $A$ is the same in every faithful representation of $A$. Is it possible for a $\lambda\in \sigma(a)$ to be an eigenvalue in one representation $\pi(A)$ but not in another $\pi'(A)$?
Final question. Take a faithful representation of $A_1$, $A_2$ or $A_3$ (which I think are all isomorphic) and consider, as appropriate, the $a_1$ or $a_2$ given above. Let us call it by $a$. Is there any particular relation between an element of the spectrum $\lambda\in\sigma(a)$ being an eigenvalue or not, and the bounded operator given by the Borel functional calculus, $\mathbf{1}_{\{\lambda\}}(a)$? Is this $\mathbf{1}_{\{\lambda\}}$ a projection if $\lambda$ is not an eigenvalue?
In particular,
Question 4: For either $a_1$ or $a_2$ above, if it is true that $1\in\sigma(a)$ is not an eigenvalue, is $\mathbf{1}_{\{1\}}(a)$ equal to zero?
I fully expect there to be some errors/confusions above and would appreciate any help.
Here are a few answers. I'll be happy to elaborate more upon request.
I think you are correct. For almost all $x$, the matrix $a_2(x)-\lambda $ is invertible, so a potential eigenvector will have to vanish a.e.
Whenever $A$ is a C*-algebra, $a\in A$ is self adjoint, and $\lambda \in \sigma (A)$, there exista a faithful representation $\pi $ of $A$ such that $\lambda $ is an eigenvalue of $\pi (a)$. To prove it, start by considering a character on $C^*(1, a)$ sending $a$ to $\lambda $, extend this to a representation of $A$ (on a larger Hilbert space) and add on any faithful representation of $A$.
Whenever $T$ is a bounded self-adjoint operator on a Hilbert space, one has that $1_{\{\lambda \}}(T)$ is the projection onto the eigenspace $\text{Ker}(T−\lambda )$. So, yes, $1_{\{\lambda \}}(T)$ is nonzero iff $\lambda $ is an eigenvalue.