We now prove one of the most surprising theorems about matrices. Notice the paucity of hypotheses compared to the precision of the conclusion.
Suppose $A$ is an $m\times n$ matrix. Then $\rank{A}=\rank{\transpose{A}}$.
This says that the row space and the column space of a matrix have the same dimension, which should be very surprising. It does not say that column space and the row space are identical. Indeed, if the matrix is not square, then the sizes (number of slots) of the vectors in each space are different, so the sets are not even comparable.
It is not hard to construct by yourself examples of matrices that illustrate Theorem RMRT, since it applies equally well to any matrix. Grab a matrix, row-reduce it, count the nonzero rows or the number of pivot columns. That is the rank. Transpose the matrix, row-reduce that, count the nonzero rows or the pivot columns. That is the rank of the transpose. The theorem says the two will be equal. Every time. Here is an example anyway.