# linear algebra: show that if \$ mathcal {N} (A) subseteq mathcal {N} (B) \$ then \$ mathcal {R} (B ^ intercal) subseteq mathcal {R} (A ^ intercal) \$

He was asked to try the following statement:

$$mathcal {N} (A) subseteq mathcal {N} (B) rightarrow mathcal {R} (B ^ intercal) subseteq mathcal {R} (A ^ intercal)$$

For me, it is intuitively true (partly because of the rank theorem), and I want to prove it by contradiction. So I'm assuming a $$z in mathcal {R} (B), not in mathcal {R} (A)$$, but I can not go beyond this. I am always rounding off the fact that $$z$$ It is a linear combination of the rows of $$B$$, that both the null space and the row space are based on the stepped form (although the first one is related to the reduced stepped form), but that's it. What additional properties should $$z$$ So that we can get a contradiction?