Column and row space

Question

Let $M$ be a an $m\times n$ matrix whose rows are linearly independent. Suppose that the $k$ columns $c_1,c_2,\ldots,c_k$ of $M$ span the column space of $M$. Let $C$ be the matrix obtained from $M$ by deleting all columns except $c_1,c_2,\ldots,c_k$. Show that the rows of $C$ are also linearly independent. We know that $k<n$ and if $n$ rows are linearly independent, then $k-$rows are also linearly independently, basically. But how can we complete this proof formally?

Answer 1

Since the $m$ rows of matrix $M$ are linearly independent, then $\text{rank} (M) = m$. We know that

$\quad\quad\quad$ the dimension of the row space is equal to the dimension of the column space. $\quad (1)$

This means that there is a set of $m$ linearly independent column vectors that span the column space. Also, the set $\{c_1, \ldots, c_k\}$ spans the column space of $M$. Thus, we have that $m\le k$ and $$\dim \text{col } (A) = \dim \langle c_1,\ldots, c_k\rangle =m.$$

If we delete all the columns of $M$ except $c_1, \ldots, c_k$, then we have the $m\times k$ matrix $C$. But $\dim \langle c_1,\ldots, c_k \rangle = m =\dim \text{col } ( C )$. We take advantage again of the proposition $(1)$, thus the dimension of the row space of $C$ is equal to $m$. Thus, the $m$ rows of $C$ consist a basis of the row space of $C$, which implies that the $m$ rows of $C$ are linearly independent.