# Prove that R(A) = R(B) iff N$(B^T)$ = N$(A^T)$ where R(A) denotes the range of the matrix A.

This is a bonus question on my assignment and it bothered me quite a while.

I believe the range of a matrix is equivalent to the column space of the matric. That is to prove two matrices have the same row spaces iff their null spaces equal.

However, according to my understanding, the null space of a matrix is related to its column space(I doubted could it be the professor did some typographical error?). Ax means either the linear combination of the column vactors of A or the dot product of its row vectors. I tried to solve the question doing this to transfer between columns and rows(forgive my poor English), but it didn't work as I expected.

Then I run out of ideas

$$A^T$$ and $$B^T$$ are row equivalent, how is this indicate they have the same null space?

• Try proving the result $N(T^t)=[R(T)]^{\circ}$. (I'm denoting T for a linear operator , t for transpose and $\circ$ for the annihilator. math.stackexchange.com/questions/1532389/… solves your question along with this. Commented Mar 10, 2021 at 17:31
• @user600016 Thanks for your comment. It leads me to an insight into the solution. I believe that together with this link completes the answer. The main thought is to prove $N(A^T)=N(B^T) <=> R(A)^o=R(B)^o <=> R(A)=R(B)$ Commented Mar 12, 2021 at 1:39
• @user600016 I've made my own solution though another perspective in answers below Commented Mar 12, 2021 at 3:26

when $$N(A^T) \subseteq N(B^T)$$
let $$a_j$$ denotes the column vectors of A ; B = [$$b_1$$...$$b_n$$]
$$\iff$$ x$$\inN(B^T)$$ for x$$\inN(A^T)$$
$$\iff$$ $$(B^T,a_j^T)$$x = 0 and $$B^T$$x = 0
$$\iff$$ $$N(B^T) = N(B^T,a_j^T)$$
$$\iff$$ Rank($$B^T$$) = Rank$$(B^T,a_j^T)$$
$$\iff$$ Rank(B) = Rank$$(B,a_j)$$
$$\iff$$ $$a_j$$ = span($$b_1$$...$$b_n$$)
$$\iff$$ R(A) $$\subseteq$$ R(B)
$$\iff$$ $$N(A^T) = N(B^T)$$ gives R(A) = R(B)