$\tau$ is a pure state and $\tau(a)=0$. Then $a=b+c$ for some $b,c$ with $\tau (b^*b)=\tau(cc^*)=0$. A pure state $\tau $ is a state such that every positive linear functional  $\rho\leq \tau$ satysfies $\rho=t\tau$ for some $t\in\mathbb R$.
$a$ is an element in a $C^*$-algebra and $\tau$ is a pure state. Prove that if $\tau(a)=0$ then there are $b,c$ such that $\tau(b^*b)=\tau(cc^*)=0$ and $a=b+c$.
I have no idea how I use pureness here. None of the propoties of pure state I know seem to help. Could someone please give me a hint?
 A: This is a very nice question. Believe it or not, the result follows from Kadison's transitivity theorem (I'm not sure if someone can avoid it).

Proposition: Let $A$ be a C$^*$-algebra and $\varphi\in S(A)$ a state. Set $N_\varphi:=\{a\in A:\varphi(a^*a)=0\}$ for the closed left ideal appearing in the GNS construction. Then $\varphi$ is a pure state if and only if $$\ker(\varphi)=N_\varphi+N_\varphi^*.$$

Proof:
The converse is easy and no big theorem is used, so I am leaving it as an exercise. Also, observe that the inclusion $N_\varphi+N_\varphi^*\subset\ker(\varphi)$ is true for any state $\varphi$. So the direct implication (which is what you are interested in) is about showing the other inclusion and it goes as follows.
Let $\varphi$ be a pure state and take the associated GNS representation $(H_\varphi,\pi_\varphi,\xi_\varphi)$, where $\xi_\varphi$ is the unique cyclic vector. Let $a\in\ker(\varphi)$. By the properties of the GNS construction we have that $$0=\varphi(a)=\langle\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle $$
so $\pi_\varphi(a)\xi_\varphi$ and $\xi_\varphi$ are two orthogonal vectors of $\mathcal{H}_\varphi$. Since $\varphi$ is pure, $\pi_\varphi(A)$ acts irreducibly on $\mathcal{H}_\varphi$, so we can apply Kadison's transitivity theorem: the vecotrs $\pi_\varphi(a)\xi_\varphi$ and $\xi_\varphi$ are linearly independent and the orthogonal projection $p\in\mathbb{B}(H_\varphi)$ onto the one-dimensional subspace $\text{span}\{\pi_\varphi(a)\xi_\varphi\}$ is a self-adjoint operator that maps $\pi_\varphi(a)\xi_\varphi$ to itself and it maps $\xi_\varphi$ to $0$. By Kadison's theorem, we can find a self-adjoint element $b\in A$ so that
$$\pi_\varphi(b)\big(\pi_\varphi(a)\xi_\varphi\big)=\pi_\varphi(a)\xi_\varphi\text{ and }\pi_\varphi(b)\big(\xi_\varphi\big)=0.$$
Now $b\in N_\varphi^*$: indeed, $\varphi(bb^*)=\varphi(b^2)=\langle\pi_\varphi(b^2)\xi_\varphi,\xi_\varphi\rangle=\|\pi_\varphi(b)\xi_\varphi\|^2=0$. Since $N_\varphi$ is a left ideal, $N_\varphi^*$ is a right ideal, thus $ba\in N_\varphi^*$. Now
$$a=(a-ba)+ba.$$
To conclude the proof, one observes that $a-ba\in N_\varphi$:
$$\varphi((a-ba)^*(a-ba))=\varphi((a^*-a^*b)(a-ba))=\varphi(a^*a-2a^*ba+a^*b^2a)= $$
$$=\langle\pi_\varphi(a^*a)\xi_\varphi-2\pi_\varphi(a^*)\pi_\varphi(b)\pi_\varphi(a)\xi_\varphi+\pi_\varphi(a^*)\pi_\varphi(b)^2\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle=$$
$$\langle\pi_\varphi(a)^*\pi_\varphi(a)\xi_\varphi-2\pi_\varphi(a^*)\pi_\varphi(a)\xi_\varphi+\pi_\varphi(a^*)\pi_\varphi(a)\xi_\varphi,\xi_\varphi\rangle=\langle0,\xi_\varphi\rangle=0.$$
References: I have seen this proposition in Ilijas Farah's book Combinatorial Set Theory of C$^*$-algebras. The idea of the proof is illustrated there.
