When is $rk(A+B)=rk(A)+rk(B)$? If $B^{*}A=0$ can we prove that $rk(A+B)=rk(A)+rk(B)$?
I tried to solve it I reached that $<Ax,Bx>=0$ and I didn't know how to continue.
I tried to prove that $rk(A)+rk(B) \le rk(A+B)$ since the other inequality is true, but I didnt know how to start 
 A: It's not just $\langle Ax,Bx\rangle = 0$; it's $\langle Ax,By\rangle = 0$ for any $x,y$. The full column spaces are orthogonal to each other.
But then, we still have trouble. Consider $A=\begin{bmatrix}0&0\\1&0\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$. Then $B^*=B$ and $B^*A = 0$. Each of $A$ and $B$ has rank $1$, but $A+B=\begin{bmatrix}1&0\\1&0\end{bmatrix}$ only has rank $1$.
We can't prove this proposition because it isn't true.
A: $\DeclareMathOperator\col{col}\DeclareMathOperator\rk{rk}$Denote by $\col(A)$ the column space of a matrix $A$ so that $\rk A = \dim(\col A)$. Note that we have
\begin{align*}
\rk(A+B) \le \dim(\col A + \col B) \le \dim(\col A)+\dim(\col B) = \rk A + \rk B.
\end{align*}
Hence in order to have equality, we need to have the following conditions on column spaces:


*

*$\col(A+B)=\col A + \col B$,

*$\col(A)\cap \col(B) = 0$.


Note that $B^* A = 0$ is only sufficient to guarantee the second condition (see jmerrys example), since in that case the subspaces $\col(A)$ and $\col(B)$ are orthogonal, in particular they intersect trivially.
For the first condition, note that we always have $\col(A+B)\subset \col A + \col B$ so that the condition is equivalent to $\col A \subset \col(A+B)$ and $\col B \subset \col(A+B)$. In other words, we have $\rk(A+B)=\rk A + \rk B$ if and only if


*

*The columns of $A$ and $B$ are linear combinations of the columns of $A+B$,

*The column spaces of $A$ and $B$ intersect trivially.

