If I have a $n\times d$ matrix $A$, use the formula $(\operatorname{Im}A)^\perp$ = $\ker A^T$ to prove $\operatorname{rank}A=\operatorname{rank}A^T$
This is my attempt:
We know that $\operatorname{rank} A + \operatorname{null}A = d$, $\operatorname{rank} A^T + \operatorname{null} A^T = n$
Also, since $\operatorname{Im} A$ is a subspace, we have $\dim (\operatorname{Im} A) + \dim(\operatorname{Im}A)^\perp = d$.
Start with $\operatorname{rank}A^T = n - \operatorname{null}A^T$, that is $\operatorname{rank}A^T = n -\dim(\ker A^T)$.
By $(\operatorname{Im}A)^\perp$ = $\ker A^T$, we have $\operatorname{rank}A^T = n -\dim(\operatorname{Im}A)^\perp$
Substitute $\dim(\operatorname{Im} A)^\perp$ by $\dim(\operatorname{Im}A)^\perp = d - \dim(\operatorname{Im} A)$, we will get
$\dim(\operatorname{Im} A^T) = n -(d - \dim(\operatorname{Im} A))$ = $n - d + \operatorname{rank} A$
From here, I don't know how to go ahead, or is there any mistake I made in my previous steps?
Your mistake is this line $$\dim (\operatorname{Im} A) + \dim(\operatorname{Im}A)^\perp = d$$
It should be instead $$\dim (\operatorname{Im} A) + \dim(\operatorname{Im}A)^\perp = n$$ since $\operatorname{Im} A$ is a subspace of $\mathbb R^n$.
If you correct it, at the end you get $$\dim(\operatorname{Im} A^T) = n -n+ \operatorname{rank} A$$