# 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 $A^\mathrm{T}$? – Brian Mar 28 '19 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}