C*-algebras: Proving $(upu^{-1}-p)^+(uqu^{-1}-q)^-=0$, where $p\leq q$ are projections and $u$ is unitary. I'm studying group representations and got to the following question, just as in the title:

Question: Let $p,q$ be projections in a C*-algebra $A$ with $p\leq q$, and let $\theta\in\operatorname{Aut}(A)$. Is it true that
  $$(\theta(p)-p)^+(\theta(q)-q)^-=0?$$

As usual, $a^+$ and $a^-$ are the positive and negative parts of a self-adjoint element $a\in A$.
This is true in the commutative case: If $A=C(X)$, then $p=1_P$ and $q=1_Q$ for subsets $P\subseteq Q\subseteq X$ (where $1_Y$ denotes the characteristic function of $Y$), and $\theta$ is of the form $\theta(a)=a\circ f$ where $f:X\to X$ is a homeomorphism. Then
$$(\theta(p)-p)^+=1_{f^{-1}(P)\setminus P}\qquad\text{and}\qquad(\theta(q)-q)^-=1_{Q\setminus f^{-1}(Q)}$$
and since $P\subseteq Q$, the sets $f^{-1}(P)\setminus P$ and $Q\setminus f^{-1}(Q)$ are disjoint.
However in the non-commutative case, I can't handle positive and negative parts very well. Note that we can assume that $\theta$ is conjugation by some unitary: $\theta(p)=upu^{-1}$, by going to the crossed product $A\rtimes_\theta\mathbb{Z}$ if necessary.
An answer in the finite-dimensional case is sufficient for me.
 A: Let $A=M_3(\mathbb C)$, and 
$$
q=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{bmatrix},\ \ \
p=\begin{bmatrix}1/2&1/2&0\\1/2&1/2&0\\0&0&0\end{bmatrix},\ \ \ 
u=\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}.
$$
Then
$$
uqu^*=\begin{bmatrix}0&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix},\ \ \ 
upu^*=\begin{bmatrix}0&0&0\\ 0&1/2&1/2\\ 0&1/2&1/2\end{bmatrix}.
$$
It is clear that $$(uqu^*-q)^-=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}.$$
For $$upu^*-p=\begin{bmatrix} -1/2&-1/2&0\\-1/2&0&1/2\\0&1/2&1/2\end{bmatrix},$$
the positive eigenvalue is $\sqrt3/2$, with eigenvector $v_2=\begin{bmatrix}-2+\sqrt3 & -1+\sqrt3&1\end{bmatrix}^T$. So we have that the spectral decomposition is $$upu^*-p=-\frac{\sqrt3}2\,r_1+\frac{\sqrt3}2\,r_2,$$ where $r_1,r_2$ are respectively the orthogonal projections onto the one-dimensional subspaces spanned by $v_1$ and $v_2$. It follows that 
$$
(upu^*-p)^+=\frac{\sqrt3}2\,r_2=\frac{\sqrt3}{2\|v_2\|^2}\,v_2v_2^*
=\frac{\sqrt3}{2(12-6\sqrt3)}\begin{bmatrix} 7-4\sqrt3&5-3\sqrt3&-2+\sqrt3\\ 5-3\sqrt3&4-2\sqrt3&-1+\sqrt3\\ -2+\sqrt3&-1+\sqrt3&1\end{bmatrix}.
$$
It is now apparent that
$$(upu^*-p)^+(uqu^*-q)^-=\frac{\sqrt3}{2(12-6\sqrt3)} \begin{bmatrix} 7-4\sqrt3&0&0 \\ 5-3\sqrt3&0&0 \\ -2+\sqrt3&0&0\end{bmatrix}.$$
