# Why is the dimension of the column space of a matrix $A$ equal to the dimension of the row space of $A$?

The proof I have in the lecture notes is as follows:

\begin{align*} \dim(Im(A)) &= \#\text{ number of vectors in basis of } Im(A) \\ & =\#\text{ leading 1's in echelon matrix of }A \\ & =\#\text{ vectors in basis of }Im(A^T) \\ & =\dim(Im(A^T)) \end{align*}

Specifically, I have a problem following the second step, which says that the number of leading 1's in echelon matrix of $A$ is equal to the number of basis vectors for the row space $Im(A^T)$. I just cannot see where it comes from and all my attempts to prove or just understand why it is so failed.

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What is the transpose of the echelon matrix? Why must that be equivalent to the echelon matrix of the transpose? – Betty Mock Dec 10 '13 at 1:01
@BettyMock The transpose of the echelon matrix is not equivalent to the echelon form of the transpose. The transpose of the echelon matrix is not even in echelon form in most cases. – EuYu Dec 10 '13 at 18:15
Have I misunderstood what the echolon matrix is? If a matrix is in upper triangular form, the transpose will be in lower triangular form. The upper form is an arbitrary way to view things; we could as well have chosen lower triangular. – Betty Mock Dec 11 '13 at 19:19