Is the row space of a matrix (order n by m, m < n) of full column rank equal to $\mathbb{R}^m$?

Question

Let $X$ be a matrix of order $n$ by $m$ where $m<n$ and let $C(X)$ denote the column space of $X$.

If rank($X$) = $m$ does this imply that $C(X')=\mathbb{R}^m$ (where $X'$ is $X$ transpose)

My gut tells me no, but all of the simple examples I've tried working out have failed to be counterexamples. Can anyone shed some light on this for me?

Answer 1

The rank of a matrix is equal to the dimension of the space generated by its columns and it is also equal to the dimension of the space generated by its rows.

Thus if your matrix has rows of length $m$ and rank $k$, then the space generated by the rows is a subspace of $\mathbb{R}^m$ of dimension $k$.

Now, in your case $k=m$. Thus, the space is a subspace of dimension $m$ of $\mathbb{R}^m$. And the only subspace of a finite dimensional vector-space with the same dimension is the space itself. Thus yes it is $\mathbb{R}^m$.