How come dim row A = rank if dim Im A is also = rank?

Question

The following identities are true for a matrix $A$. \begin{align} \dim \mathrm{row}\, A &= \mathrm{rank}\,A \\ \dim \mathrm{Im}\, A &= \mathrm{rank}\, A \\ \dim \mathrm{row}\, A &= \dim \mathrm{Im}\, A^\mathrm{T} \end{align} Does this mean that $\dim \mathrm{Im}\, A^\mathrm{T} = \dim \mathrm{Im}\, A = \mathrm{rank}\,A$? How come?

Answer 1

$\text {Im}\ (A^{\mathrm T})$ is same as the row space of $A.$ Since row rank and column rank of a matrix are the same so we can conclude that $$\begin{align} \dim \mathrm{Im}\, (A^\mathrm{T}) & = \text {row rank}\ (A). \\ & = \text {column rank}\ (A). \\ & = \dim \mathrm{Im}\, (A). \\ & = \mathrm{rank}\,(A). \end{align}$$

Answer 2

If you want, you can define the rank of a matrix as the number of pivot entries, for the following reason:

The dimension of the column space is equal to the number of pivot columns, which is equal to the number of pivot rows (why?), which is equal to the dimension of the row space.