sums of projections in C*-algebras

Question

Let $A$ be a $C$*-algebra with unit $e$. If $p$ and $q$ are projections such that $p+q+\lambda e$ is a projection for some $\lambda\in\mathbb{R}$, is it true that the only possible values for $\lambda$ can be $0$ or $-1$? Do $p$ and $q$ have to commute in this case?

Answer 1

It will depend on the algebra, but $\lambda\in(-1,0)$ is possible.

Let $A=M_2(\mathbb C)$, and put $$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ q=\frac14\,\begin{bmatrix}1&\sqrt3\\ \sqrt3&3\end{bmatrix},\ \ \lambda=-1/2. $$ Then $$ p+q-\frac12\,e=\frac14\,\begin{bmatrix}3&\sqrt 3\\ \sqrt 3&1\end{bmatrix}. $$

Note that in the example above $pq\ne qp$. When $\lambda=0$ or $\lambda=-1$, it is the case that $pq=qp$:

1) if $\lambda=0$, we have $r=p+q$; so $(p+q)^2=p+q$, which gives $pq=-qp$. Multiplying by $(e-p)$ on the right, $pq(e-p)=0$; so $pq=pqp$, and taking adjoints we get $pq=qp$. Combining with $pq=-qp$, we get $pq=0$, so $p$ and $q$ are mutually orthogonal.

2) If $\lambda=-1$, we have $r=p+q-e$. Then $p+q\geq e$, or $e-q\leq p$. This can be written as $p(e-q)=e-q$, so $e-q$ commutes with $p$, and then so does $q$.

In the case $\lambda=-1$, $p$ and $q$ need not be mutually orthogonal. For instance let $$ p=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix},\ \ q=\begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix}. $$ Then $$ p+q-e=\begin{bmatrix}0&0&0\\0&1&0\\0&0&0\end{bmatrix}=pq. $$