Dimension of $\mathrm{ker}(\mathrm{A})$ and dimension of $mathrm{ker}(\mathrm{A}^T)$

Question

I read from a recent paper that for an $n\times n$ matrix $\mathrm{A}$, $\dim(\operatorname{ker(\mathrm{A})})=\dim(\operatorname{ker}(\mathrm{A}^T))$. Is this true? How did it happened? Thank you for those would give their 2 cents regarding the matter.

Answer 1

I think I get it now.

Note that for an $n\times n$ (square) matrix $A$, we have $\rho(A)=\rho(A^T)$ since the row rank of matrix $A$ is just the column rank of its transpose and we don't have any distinction between row rank and column rank as they both are equal to the rank of the matrix. By rank-nullity theorem, we have $\rho(A^T)+\eta(A^T)=n$ and thus we have

\begin{align*} \eta(A^T)&=n-\rho(A^T)\\ \eta(A^T)&=n-\rho(A) \end{align*}

Notice that the right hand side is just the $\eta(A)$ and thus we have $\eta(A^T)=\eta(A)$.