Rank of a Matrix that Lies in Left Null Space of Another Matrix

Question

Let $B \in R^{n \times m}$ with $n>m$, and let $D$ be the matrix that lies in the left null space of $B$. So, we have $D^TB=0$. Here is my question. If $\text{rank}(B)=m$, can we say that $\text{rank}(D)=n-m$ ?

Answer 1

I assume, that the columns of $D$ form a basis of the left null space of $B$. Note that the left null space of $B$ equals the null space of $B^T$. Hence, $$ rank(D) = \dim\ker B^T = n - rank(B^T) = n - rank(B) = n-m. $$ So, yes.