TITLE: Does $AB+BA=0$ imply $AB=BA=0$ when $A$ is real semidefinite matrix?
QUESTION [11 upvotes]: This is a general question that came to my mind while listening to a lecture(although its framing may make it look like a textbook question). 
Suppose that $A$ and $B$ be real matrices. $A$ is symmetric and positive semi-definite$(x^tAx\ge0\ \ \forall x\in \mathbb{R}^n)$.
Let $AB+BA=0$. Does this imply $AB=BA=0$?  If yes can you give me some example?

REPLY [2 votes]: Proposition. Let $A,B\in M_n(\mathbb{C})$; if $A$ is diagonalizable with non-negative eigenvalues and $AB+BA=0$, then 

$AB=BA=0$. 
If moreover $A$ is hermitian, then  $A,B$ are unitarily simultaneously similar to $A_1=diag(D_p,0_{n-p}),B_1=\begin{pmatrix}0_p&0\\0&T_{n-p}\end{pmatrix}$, where $D$ is diagonal $> 0$ and $T$ is upper triangular.

Proof. For 1. cf. the Hamed's answer.
For 2. Using the syeh_106's proof, $A,B$ are unitarily simultaneously similar to  $A_1=diag(D_p,0_{n-p})),B'=\begin{pmatrix}0_p&0\\0&U_{n-p})\end{pmatrix}$, where $D$ is diagonal $> 0$. It remains to triangularize $U$, using Schur's method.