$A$ is normal matrix and has distinct eigenvalue, and $AB=0$. why $B$ is normal matrix? Let $A,B \in {M_n}$ . suppose $A$ is normal matrix and has distinct eigenvalue, and $AB=0$. why $B$ is normal matrix?
 A: If by a "distinct eigenvalue" you mean that all of them are distinct (plural), notice that this means that $A$ has at most one zero eigenvalue. Since $AB=0$, either $A$ is nonsingular, which implies that $B=0$ (which is normal), or $A$ has one zero eigenvalue.
Now, let $A = \operatorname{diag}(0, A')$, where $A'$ is a nonsingular normal matrix with distinct eigenvalues. It is easy to see that $A$ is also a normal matrix with distinct eigenvalues. Further, let $B = \begin{bmatrix} e & 0 & \cdots & 0\end{bmatrix}^T$, i.e., $B$ has all the elements of the first row equal to $1$, while the rest of them are $0$.
Notice that
$$AB = \begin{bmatrix} 0 \\[1ex] & \Huge A'\\ \mbox{} \end{bmatrix} \begin{bmatrix} 1 & 1 & \dots & 1 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix} = 0,$$
but
$$B^*B = \begin{bmatrix} 1 & 1 & \dots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{bmatrix} \color{red}{\ne} \begin{bmatrix} n & 0 & \dots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix} = BB^*,$$
so $B$ is not normal, which means that your statement is generally incorrect.
A: As Vedran noted, your assumptions do not imply that $B$ is normal.  However, if you assume $BA = AB = 0$, then it is true.  
Namely, suppose $A$ is singular.  Then as Vedran noted, $0$ is an eigenvalue of $A$ of multiplicity $1$.  Since $A$ is normal, $\ker(A) = \ker(A^*)$ (where $A^*$ is the conjugate transpose).  $B$ must map $\ker(A)$ into itself and 
$\ker(A)^\perp$ to $0$.  Thus $B = \lambda u u^*$ for some scalar $\lambda$ and $u \in \ker(A)$.  And then $B B^* = |\lambda|^2 (u^* u) u u^* = B^* B$,
so $B$ is normal.
