Vrftsjtryk

What does it mean for the column spaces of two matrices to span the same subspace?

It is equivalent to saying that the two matrices have the same image and range, but I don't really understand what the two matrices would have to have in common in order for their column space to span the same subspace.

Assume we are working with real vector spaces, so matrices have real numbers as their entries, and so vector spaces (of dim > 0) have infinitely many vectors.
And for every vector space there are infinitely many basis possible.

Now let us take na $mtimes n$ matrix $A$ whose column space $W$ has dimension $r$. Now consider any other basis of $W$. Take $mtimes r$ matrix $B$ with this new basis vectors as its columns. Now $B$ has the same column space as $A$.

As there are infinitely many possible bases for $W$ infinitely many choices of $B$ having same columns space as $A$.

Example: suppose the first two column vectors $u$ and $v$ of $A$ are linearly independent. Then the new matrix $B$ with same column space as $A$ can be obtained by replacing those two vectors by $u+ v$ and $u-v$, and retaining other columns. (Or you can take $u+av, u-av$ for any nonzero number $a$).

I think your terms are a little confused: the column space of a matrix is, by definition, the subspace spanned by the columns of that matrix.

For example, if you take the matrix of real numbers $A = begin{pmatrix} 1 & 1 & 5 \ 0 & 1 & 0 end{pmatrix}$, then the columns are
$begin{pmatrix} 1 \ 0 end{pmatrix}$,
$begin{pmatrix} 1 \ 1 end{pmatrix}$,
and $begin{pmatrix} 5 \ 0 end{pmatrix}$.
Thinking of these columns as vectors in the two-dimensional space $Bbb{R}^2$, the subspace they span is the set of vectors
$t_0 begin{pmatrix} 1 \ 0 end{pmatrix} +
t_1 begin{pmatrix} 1 \ 1 end{pmatrix} +
t_2 begin{pmatrix} 5 \ 0 end{pmatrix}$
for all real values $t_0$, $t_1$, $t_2$.
In fact (exercise) this subspace is the entire space $Bbb{R}^2$.

If we also take, say, $B = begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 0 end{pmatrix}$, we find that (exercise) the column space of $B$ is the same as the column space of $A$.

But if we take $C = begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 0 end{pmatrix}$, the column space of $C$ is different from the other two: it's the space of columns $begin{pmatrix} t \ 0 end{pmatrix}$ for all real values $t$.