Projections on a Hilbert space Suppose $P$ and $Q$ are self-adjoint projections on a Hilbert space such that $P+Q+\lambda I$ is a self-adjoint projection for some $\lambda \in \mathbb{R}$. Does it follow that $P$ and $Q$ commute?
 A: The answer is no.
By Takesaki I, Theorem V.1.41, the von Neumann algebra generated by P and Q is the direct sum of an abelian part and a type $I_2$ part, in which the projections P and Q have the form
$$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} , \quad
Q = \begin{pmatrix} c^2 & cs \\ cs & s^2 \\ \end{pmatrix}, $$
where $c$ and $s$ satisfy $0 \leq c,s \leq 1$ and $c^2+s^2=1$ (if the Hilbert space is two-dimensional, $c$ and $s$ are the sine and cosine of the angle between the ranges of P and Q).
To answer the question it is clearly sufficient to answer the question in the type $I_2$ part. If $c$ and $s$ are real numbers, then the eigenvalues of $P+Q$ are easily seen to be $1+c$ and $1-c$, so if we take $c=1/2$ and $s = \sqrt{3}/2$, then $P+Q$ has eigenvalues $3/2$ and $1/2$ and so $P+Q-I/2$ is a projection.
So a counterexample with a two-dimensional Hilbert space is
$$ P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} , \quad
Q = \begin{pmatrix} 1/4 & \sqrt{3}/4 \\ \sqrt{3}/4 & 3/4 \\ \end{pmatrix}. $$
