The following is a statement which I believe to be correct but unable to prove (I have been trying to find a counterexample for a long time but never succeeded in doing so):
Let $X_1,X_2,X_3$ be $n\times n$ nonzero real matrices satisfying $X_{i}X_j=-X_jX_i\neq 0$ for $1\leq i<j\leq 3$ [as an example, $\{X_1,X_2,X_3\}=\{\sigma^x,i\sigma^y,\sigma^z\}$, the Pauli-matrices]. Prove that at least one of $X_1,X_2,X_3$ has a negative off-diagonal element.
The case $n=2$ is easy and straightforward: just expand $X_1,X_2,X_3$ in the basis of Pauli matrices and solve the equation $\{X_i,X_j\}=0$. But when $n$ becomes large, the coupled quadratic equations quickly becomes formidable, which I have no way to handle.
Notice also that the $\neq 0$ condition is important, otherwise we got a simple counter-example: $X_1=X_2=X_3=\sigma^+$, which mutually anti-commute but are all non-negative.
Best Answer
This isn't true, at least when the matrices are allowed to be singular. Let $$ D=\pmatrix{1&0\\ 0&-1}, \ S=\pmatrix{0&1\\ 1&0}. $$ Then $DS$ and $SD$ are nonzero but $DS+SD=0$. The following set of matrices now serves as a counterexample to your statement: $$ X_1=\pmatrix{D\\ &D\\ &&0}, \ X_2=\pmatrix{S\\ &0\\ &&D}, \ X_3=\pmatrix{0\\ &S\\ &&S}. $$