Prove that if $A$ ∈ R$^n$$^x$$^m$ then rank($A$) = rank($A^T$).
I'm trying to make a very mathematical proof. I've tried everything and I have no clue what to do. I have so far been able to use rank-nullity theorem to show that null($A$) ⊆ null($A^TA)$, but I don't know where to go from there.
On
Here's another strategy, although I like JCAA's explanation better.
Notice $[\text{Col}(A)]^{\perp}=N(A^T)$ and $\text{Col}(A) \oplus [\text{Col}(A)]^{\perp}=\mathbb{R}^n$. From this we immediately see $\text{nullity}(A^T)=n-\text{rank}(A)$ and so with rank nullity
$$n=\text{rank}(A^T)+\text{nullity}(A^T)=\text{rank}(A^T)+n-\text{rank}(A)$$ which boils down to $\text{rank}(A)=\text{rank}(A^T)$. Also, if you're looking to prove that $N(A^TA)\subseteq N(A)$ note $\vec{x}\in N(A^TA)$ implies $$0=\vec{x}^TA^TA\vec{x}=(A\vec{x})^T(A\vec{x})=||A\vec{x}||^2$$ and so $A\vec{x}=0$ i.e. $\vec{x}\in N(A)$.
Let $A$ be a matrix (not necessarily square). Reduce $A$ to its reduced row echelon form $A'$ using row transformations. Then the row rank of $A$ is the same as the row rank of $A'$, the same as the number of pivots is $A'$, and the same as the column rank of $A'$. But when we do a row transformation of a matrix, its column rank stays the same (easy to check because row operations preserve linear dependencies of columns). Hence the column rank of $A'$ is the same as the column rank of $A$ and the same as the row rank of $A$. But the row rank of $A$ is $\mathrm{rank}(A)$ while the column rank of $A$ is $\mathrm{rank}(A^T)$. QED