Let $H$ be a Hilbert space and let $a,b\in B(H)$ be such that
$${\rm Tr}(l(a))={\rm Tr}(r(a))<\infty , {\rm Tr}(l(b))={\rm Tr}(r(b))<\infty,$$
$${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha\_1 a +\alpha\_2 b ))$$
for any $0\ne \alpha\_1,\alpha\_2\in \mathbb{R}$, where ${\rm Tr}$ denotes the standard trace on $B(H)$ and $l(\cdot),r(\cdot)$ denote the left and the right-support respectively.
Can we say that $l(a)l(b)=r(a)r(b)=0$?
If we replace $B(H)$ with $\ell\_\infty$, then the above equality is clearly true. However, I am not sure about the case for $B(H)$.



Does ${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha_1 a +\alpha_2 b ))$ imply that $l(a)l(b)=r(a)r(b)=0$?
This already fails for $2 \times 2$ matrices. Take the rank one matrices $a = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $b = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$. For all $\alpha \neq 0$ and $\beta \neq 0$, the matrix $\alpha a + \beta b$ is invertible, so that $\text{Tr}(l(\alpha a + \beta b)) = 2$. Also $\text{Tr}(l(a) + l(b)) = 2$. But the ranges of $a$ and $b$ are not orthogonal.


