# How come dim row A = rank if dim Im A is also = rank? [duplicate]

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?

• How do you define rank? Likewise, how you find the dimension of the row space and column space of a matrix? How is the row space of a matrix $A$ related to the column space of its transpose $A^\mathrm{T}$? Mar 28, 2019 at 2:26

$$\text {Im}\ (A^{\mathrm T})$$ 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}

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.